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</div><h2>HL Paper 1</h2><div class="question">
<p class="p1">A point \(P\), relative to an origin \(O\), has position vector \(\overrightarrow {{\text{OP}}} = \left( {\begin{array}{*{20}{c}} {1 + s} \\ {3 + 2s} \\ {1 - s} \end{array}} \right),{\text{ }}s \in \mathbb{R}\).</p>
<p class="p1">Find the minimum length of \(\overrightarrow {{\text{OP}}} \).</p>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question">
<p class="p1"><strong>METHOD 1</strong></p>
<p class="p1">\(\left| {\overrightarrow {{\text{OP}}} } \right| = \sqrt {{{(1 + s)}^2} + {{(3 + 2s)}^2} + {{(1 - s)}^2}} \;\;\;\left( { = \sqrt {6{s^2} + 12s + 11} } \right)\) <strong><em>A1</em></strong></p>
<p class="p1"> </p>
<p class="p1"><strong>Note: </strong>Award <strong><em>A1 </em></strong>if the square of the distance is found.</p>
<p class="p1"> </p>
<p class="p1"><strong>EITHER</strong></p>
<p class="p1">attempt to differentiate: \(\frac{{\text{d}}}{{{\text{d}}s}}{\left| {\overrightarrow {{\text{OP}}} } \right|^2}\;\;\;( = 12s + 12)\) <strong><em>M1</em></strong></p>
<p class="p1">attempting to solve \(\frac{{\text{d}}}{{{\text{d}}s}}{\left| {\overrightarrow {{\text{OP}}} } \right|^2} = 0\;\;\;{\text{for }}s\) <strong><em>(M1)</em></strong></p>
<p class="p1">\(s = - 1\) (<strong><em>A1)</em></strong></p>
<p class="p1"><strong>OR</strong></p>
<p class="p1">attempt to differentiate: \(\frac{{\text{d}}}{{{\text{d}}s}}\left| {\overrightarrow {{\text{OP}}} } \right|\;\;\;\left( { = \frac{{6s + 6}}{{\sqrt {6{s^2} + 12s + 11} }}} \right)\) <strong><em>M1</em></strong></p>
<p class="p1">attempting to solve \(\frac{{\text{d}}}{{{\text{d}}s}}\left| {\overrightarrow {{\text{OP}}} } \right| = 0\;\;\;{\text{for }}s\) <strong><em>(M1)</em></strong></p>
<p class="p1">\(s = - 1\) (<strong><em>A1)</em></strong></p>
<p class="p1"><strong>OR</strong></p>
<p class="p1">attempt at completing the square: \(\left( {{{\left| {\overrightarrow {{\text{OP}}} } \right|}^2} = 6{{(s + 1)}^2} + 5} \right)\) <strong><em>M1</em></strong></p>
<p class="p1">minimum value <strong><em>(M1)</em></strong></p>
<p class="p1">occurs at \(s = - 1\) <strong>(<em>A1)</em></strong></p>
<p class="p1"><strong>THEN</strong></p>
<p class="p1">the minimum length of \(\overrightarrow {{\text{OP}}} \) is \(\sqrt 5 \) <strong><em>A1</em></strong></p>
<p class="p1"><strong>METHOD 2</strong></p>
<p class="p1">the length of \(\overrightarrow {{\text{OP}}} \) is a minimum when \(\overrightarrow {{\text{OP}}} \) is perpendicular to \(\left( {\begin{array}{*{20}{c}} 1 \\ 2 \\ { - 1} \end{array}} \right)\) <strong><em>(R1)</em></strong></p>
<p class="p1">\(\left( {\begin{array}{*{20}{c}} {1 + s} \\ {3 + 2s} \\ {1 - s} \end{array}} \right) \bullet \left( {\begin{array}{*{20}{c}} 1 \\ 2 \\ { - 1} \end{array}} \right) = 0\) <strong><em>A1</em></strong></p>
<p class="p1">attempting to solve \(1 + s + 6 + 4s - 1 + s = 0\;\;\;(6s + 6 = 0)\;\;\;{\text{for }}s\) <strong><em>(M1)</em></strong></p>
<p class="p1">\(s = - 1\) (<strong><em>A1)</em></strong></p>
<p class="p1">\(\left| {\overrightarrow {{\text{OP}}} } \right| = \sqrt 5 \) (<strong><em>A1)</em></strong></p>
<p class="p1"><strong><em>[5 marks]</em></strong></p>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question">
<p class="p1">Generally well done. But there was a significant minority who didn’t realise that they had to use calculus or completion of squares to minimise the length. Trying random values of <em>\(s\) </em>gained no marks. A number of candidates wasted time showing that their answer gave a minimum rather than a maximum value of the length.</p>
</div>
<br><hr><br><div class="question">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 19.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">Three distinct non-zero vectors are given by \(\overrightarrow {{\text{OA}}} \) = <strong><em>a</em></strong>, \(\overrightarrow {{\text{OB}}} \) = <strong><em>b</em></strong>, and \(\overrightarrow {{\text{OC}}} \) = <strong><em>c</em></strong> .</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 19.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">If \(\overrightarrow {{\text{OA}}} \) is perpendicular to \(\overrightarrow {{\text{BC}}} \) and \(\overrightarrow {{\text{OB}}} \) is perpendicular to \(\overrightarrow {{\text{CA}}} \) , show that \(\overrightarrow {{\text{OC}}} \) is perpendicular to \(\overrightarrow {{\text{AB}}} \).</span></p>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(\overrightarrow {{\text{BC}}} \) = <strong><em>c</em></strong> – <strong><em>b</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(\overrightarrow {{\text{CA}}} \) = <strong><em>a</em></strong> – <strong><em>c</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( \Rightarrow \) <strong><em>a</em></strong>\( \cdot \)(<strong><em>c</em></strong> – <strong><em>b</em></strong>) = 0 <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">and <strong><em>b</em></strong>\( \cdot \)(<strong><em>a</em></strong> – <strong><em>c</em></strong>) = 0 <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( \Rightarrow \) <strong><em>a</em></strong>\( \cdot \)<strong><em>c</em></strong> = <strong><em>a</em></strong>\( \cdot \)<strong><em>b</em></strong> <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">and <strong><em>a</em></strong>\( \cdot \)<strong><em>b</em></strong> = <strong><em>b</em></strong>\( \cdot \)<strong><em>c</em></strong> <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( \Rightarrow \) <strong><em>a</em></strong>\( \cdot \)<strong><em>c</em></strong> = <strong><em>b</em></strong>\( \cdot \)<strong><em>c</em></strong> <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( \Rightarrow \) <strong><em>b</em></strong>\( \cdot \)<strong><em>c</em></strong> – <strong><em>a</em></strong>\( \cdot \)<strong><em>c</em></strong> = 0</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>c</em></strong>\( \cdot \)(<strong><em>b</em></strong> – <strong><em>a</em></strong>) = 0 <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( \Rightarrow \) \(\overrightarrow {{\text{OC}}} \) is perpendicular to \(\overrightarrow {{\text{AB}}} \) , as <strong><em>b</em></strong> \( \ne \) <strong><em>a</em></strong> . <strong><em>AG</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[6 marks]</em></strong></span></p>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 17.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">Only the better candidates were able to make significant progress with this question. Many candidates understood how to begin the question, but did not see how to progress to the last stage. On the whole the candidates' use of notation in this question was poor.</span></p>
</div>
<br><hr><br><div class="question">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">Consider the vectors \(\overrightarrow {{\text{OA}}} \) = <strong><em>a</em></strong>, \(\overrightarrow {{\text{OB}}} \) = <strong><em>b</em></strong> and \(\overrightarrow {{\text{OC}}} \) = <strong><em>a</em></strong> + <strong><em>b</em></strong>. Show that if \(|\)<strong><em>a</em></strong>\(|\) = \(|\)<strong><em>b</em></strong>\(|\) then (<strong><em>a</em></strong> + <strong><em>b</em></strong>)\( \cdot \)(<strong><em>a</em></strong> − <strong><em>b</em></strong>) = 0. Comment on what this tells us about the parallelogram OACB.</span></p>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(<strong><em>a</em></strong> + <strong><em>b</em></strong>)\( \cdot \)(<strong><em>a</em></strong> – <strong><em>b</em></strong>) = <strong><em>a</em></strong>\( \cdot \)<strong><em>a</em></strong> + <strong><em>b</em></strong>\( \cdot \)<strong><em>a</em></strong> – <strong><em>a</em></strong>\( \cdot \)<strong><em>b</em></strong> – <strong><em>b</em></strong>\( \cdot \)<strong><em>b</em></strong> <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">= <strong><em>a</em></strong>\( \cdot \)<strong><em>a</em></strong> – <strong><em>b</em></strong>\( \cdot \)<strong><em>b</em></strong> <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">= \(|\)<strong><em>a</em></strong>\({|^2}\) – \(|\)<strong><em>b</em></strong>\({|^2}\) = 0 since \(|\)<strong><em>a</em></strong>\(|\) = \(|\)<strong><em>b</em></strong>\(|\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">the <strong>diagonals</strong> are perpendicular <strong><em>R1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>Note:</strong> Accept geometric proof, awarding <strong><em>M1</em></strong> for recognizing OACB is a rhombus, <strong><em>R1</em></strong> for a clear indication that (<strong><em>a</em></strong> + <strong><em>b</em></strong>) and (<strong><em>a</em></strong> – <strong><em>b</em></strong>) are the diagonals, <strong><em>A1</em></strong> for stating that diagonals cross at right angles and <strong><em>A1</em></strong> for “hence dot product is zero”.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">Accept solutions using components in 2 or 3 dimensions.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"> </span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[4 marks]</em></strong></span></p>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 23.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">Many candidates found this more abstract question difficult. While there were some correct statements, they could not “show” the result that was asked. Some treated the vectors as scalars and notation was poor, making it difficult to follow what they were trying to do. Very few candidates realized that <em>a</em> – <em>b</em> and <em>a</em> + <em>b</em> were the diagonals of the parallelogram which prevented them from identifying the significance of the result proved. A number of candidates were clearly not aware of the difference between scalars and vectors.</span></p>
</div>
<br><hr><br><div class="question">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">Consider the plane with equation \(4x - 2y - z = 1\) and the line given by the parametric equations</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"> \(x = 3 - 2\lambda \)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"> \(y = (2k - 1) + \lambda \)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"> \(z = - 1 + k\lambda .\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">Given that the line is perpendicular to the plane, find</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(a) the value of <em>k</em>;</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(b) the coordinates of the point of intersection of the line and the plane.</span></p>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 25.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(a) \(\boldsymbol{a} = \left( {\begin{array}{*{20}{c}}<br> 4 \\ <br> { - 2} \\ <br> { - 1} <br>\end{array}} \right) \bot \) to the plane \(\boldsymbol{e} = \left( {\begin{array}{*{20}{c}}<br> { - 2} \\ <br> 1 \\ <br> k <br>\end{array}} \right)\) is parallel to the line <strong><em>(A1)(A1)</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 25.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>Note:</strong> Award <strong><em>A1</em></strong> for each correct vector written down, even if not identified.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 25.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"> </span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 25.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">line \( \bot \) plane \( \Rightarrow \) \(\boldsymbol{e}\) parallel to \(\boldsymbol{a}\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 25.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">since \(\left( {\begin{array}{*{20}{c}}<br> 4 \\ <br> { - 2} \\ <br> { - 1} <br>\end{array}} \right) = t\left( {\begin{array}{*{20}{c}}<br> { - 2} \\ <br> 1 \\ <br> k <br>\end{array}} \right) \Rightarrow k = \frac{1}{2}\) <strong><em>(M1)A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 25.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em> </em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 25.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(b) \(4(3 - 2\lambda ) - 2\lambda - \left( { - 1 + \frac{1}{2}\lambda } \right) = 1\) <strong><em>(M1)(A1)</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 25.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>Note:</strong> <strong><em>FT</em></strong> their value of <em>k</em> as far as possible.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 25.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"> </span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 25.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(\lambda = \frac{8}{7}\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 25.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\({\text{P}}\left( {\frac{5}{7},\frac{8}{7}, - \frac{3}{7}} \right)\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 25.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[8 marks]</em></strong></span></p>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 25.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">Solutions to this question were often disappointing. In (a), some candidates found the value of <em>k</em>, incorrectly, by taking the scalar product of the normal vector to the plane and the direction of the line. Such candidates benefitted partially from follow through in (b) but not fully because their line turned out to be parallel to the plane and did not intersect it.</span></p>
</div>
<br><hr><br><div class="specification">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 12.0px Times;"><span style="font-family: 'times new roman', times; font-size: medium;">Two boats, <em>A </em>and <em>B </em>, move so that at time <em>t </em>hours, their position vectors, in kilometres, are <em><strong>r</strong></em>\(_A\) = (9<em>t</em>)<em><strong>i</strong></em> + (3 – 6<em>t</em>)<em><strong>j</strong></em> and <strong><em>r</em></strong>\(_B\) = (7 – 4<em>t</em>)<em><strong>i</strong></em> + (7<em>t</em> – 6)<em><strong>j</strong></em> .</span></p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 12.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">Find the coordinates of the common point of the paths of the two boats.</span></p>
<div class="marks">[4]</div>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 12.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">Show that the boats do not collide.</span></p>
<div class="marks">[2]</div>
<div class="question_part_label">b.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: 'times new roman', times; font-size: medium;"><strong>METHOD 1</strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(9{t_A} = 7 - 4{t_B}\) and</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(3 - 6{t_A} = - 6 + 7{t_B}\) <strong><em><strong>M1A1</strong></em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">solve simultaneously</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\({t_A} = \frac{1}{3},{\text{ }}{t_B} = 1\) <strong><em>A1</em></strong></span><strong style="font-family: 'times new roman', times; font-size: medium; line-height: 20px;"><em> </em></strong></p>
<p><span style="font-family: 'times new roman', times; font-size: medium;"><strong>Note: </strong>Only need to see one time for the <strong><em>A1</em></strong>.</span></p>
<p><span style="font-family: 'times new roman', times; font-size: medium;"> </span></p>
<p><span style="font-family: 'times new roman', times; font-size: medium;">therefore meet at (3, 1) </span><strong style="font-family: 'times new roman', times; font-size: medium;"><em>A1</em></strong></p>
<p><strong style="font-family: 'times new roman', times; font-size: medium;"><em>[4 marks]</em></strong></p>
<p><span style="font-family: 'times new roman', times; font-size: medium;"><strong>METHOD 2</strong></span></p>
<p><span style="font-family: 'times new roman', times; font-size: medium;">path of A is a straight line: \(y = - \frac{2}{3}x + 3\) <strong><em>M1A1</em></strong></span><strong style="font-family: 'times new roman', times; font-size: medium;"><em> </em></strong></p>
<p><span style="font-family: 'times new roman', times; font-size: medium;"><strong>Note:</strong> Award <strong><em>M1 </em></strong>for an attempt at simultaneous equations.</span></p>
<p><span style="font-family: 'times new roman', times; font-size: medium;"> </span></p>
<p><span style="font-family: 'times new roman', times; font-size: medium;">path of B is a straight line: \(y = - \frac{7}{4}x + \frac{{25}}{4}\) </span><strong style="font-family: 'times new roman', times; font-size: medium;"><em>A1</em></strong></p>
<p><span style="font-family: 'times new roman', times; font-size: medium;">\( - \frac{2}{3}x + 3 = - \frac{7}{4}x + \frac{{25}}{4}{\text{ }}( \Rightarrow x = 3)\)</span></p>
<p><span style="font-family: 'times new roman', times; font-size: medium;">so the common point is (3, 1) <strong><em>A1</em></strong></span></p>
<p><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[4 marks]</em></strong></span></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: 'times new roman', times; font-size: medium;"><strong>METHOD 1</strong></span></p>
<p><span style="font-family: 'times new roman', times; font-size: medium;">boats do not collide because the two times \(\left( {{t_A} = \frac{1}{3},{\text{ }}{t_B} = 1} \right)\) <strong><em>(A1)</em></strong></span></p>
<p><span style="font-family: 'times new roman', times; font-size: medium;">are different <strong><em>R1</em></strong></span></p>
<p><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[2 marks]</em></strong></span></p>
<p><span style="font-family: 'times new roman', times; font-size: medium;"><strong>METHOD 2</strong></span></p>
<p><span style="font-family: 'times new roman', times; font-size: medium;">for boat A, \(9t = 3 \Rightarrow t = \frac{1}{3}\) and for boat B, \(7 - 4t = 3 \Rightarrow t = 1\)</span></p>
<p><span style="font-family: 'times new roman', times; font-size: medium;">times are different so boats do not collide <strong><em>R1AG</em></strong></span></p>
<p><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[2 marks]</em></strong></span></p>
<p> </p>
<p> </p>
<div class="question_part_label">b.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 10.0px Arial;"><span style="font-family: 'times new roman', times; font-size: medium;">This was probably the least accessible question from section A. Most started by using the same value of <em>t </em>in attempting to find the common point, and so scored no marks. There were a number of very good candidates who set different parameters for <em>t</em> and correctly obtained (3,1) . There was slightly better understanding shown in part b), though some argued that the boats did not collide because their times were different, yet then provided incorrect times, or even no times at all.</span></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 10.0px Arial; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;">This was probably the least accessible question from section A. Most started by using the same value of <em>t </em>in attempting to find the common point, and so scored no marks. There were a number of very good candidates who set different parameters for <em>t</em> and correctly obtained (3,1) . There was slightly better understanding shown in part b), though some argued that the boats did not collide because their times were different, yet then provided incorrect times, or even no times at all.</span></p>
<div class="question_part_label">b.</div>
</div>
<br><hr><br><div class="question">
<p class="p1">O, A, B and C are distinct points such that \(\overrightarrow {{\text{OA}}} = \) <strong><em>a</em></strong>, \(\overrightarrow {{\text{OB}}} = \) <strong><em>b </em></strong>and \(\overrightarrow {{\text{OC}}} = \) <strong><em>c</em></strong><span class="s1">.</span></p>
<p class="p1"><span class="s1">It is given that </span><strong><em>c </em></strong>is perpendicular to \(\overrightarrow {{\text{AB}}} \) <span class="s1">and </span><strong><em>b </em></strong>is perpendicular to \(\overrightarrow {{\text{AC}}} \).</p>
<p class="p1"><span class="s1">Prove that </span><strong><em>a </em></strong>is perpendicular to \(\overrightarrow {{\text{BC}}} \).</p>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question">
<p class="p1"><span class="s1"><strong><em>c </em></strong>\( \bullet \)</span> (<span class="s1"><strong><em>b </em></strong></span>\( - \) <span class="s1"><strong><em>a</em></strong>) \( = 0\)</span> <strong><em>M1</em></strong></p>
<p class="p1"><strong>Note: <span class="Apple-converted-space"> </span></strong>Allow <span class="s1"><em>c</em></span> \( \bullet \) \(\overrightarrow {{\text{AB}}} = 0\) or similar for <strong><em>M1</em></strong>.</p>
<p class="p1"><span class="s2"><strong><em>c </em></strong></span>\( \bullet \) <span class="s2"><strong><em>b </em></strong></span>\( = \)<span class="s2"> <strong><em>c </em></strong></span>\( \bullet \)<span class="s2"> <strong><em>a <span class="Apple-converted-space"> </span></em></strong></span><strong><em>A1</em></strong></p>
<p class="p1"><span class="s2"><strong><em>b </em></strong></span>\( \bullet \)<span class="s2"> (<strong><em>c </em></strong></span>\( - \)<span class="s2"> <strong><em>a</em></strong>) </span>\( = 0\)</p>
<p class="p3"><strong><em>b </em></strong><span class="s3">\( \bullet \)</span> <strong><em>c </em></strong><span class="s3">\( = \)</span> <strong><em>b </em></strong><span class="s3">\( \bullet \)</span> <strong><em>a <span class="Apple-converted-space"> </span></em></strong><span class="s1"><strong><em>A1</em></strong></span></p>
<p class="p1"><span class="s1"><strong><em>c</em></strong></span> \( \bullet \) <span class="s1"><strong><em>a</em></strong></span> \( = \) <span class="s1"><strong><em>b</em></strong></span> \( \bullet \) <span class="s1"><strong><em>a <span class="Apple-converted-space"> </span></em></strong></span><strong><em>M1</em></strong></p>
<p class="p3">(<strong><em>c </em></strong><span class="s3">\( - \)</span> <strong><em>b</em></strong>) <span class="s3">\( \bullet \)</span> <strong><em>a </em></strong><span class="s3">\( = 0\)</span> <span class="Apple-converted-space"> </span><span class="s3"><strong><em>A1</em></strong></span></p>
<p class="p1">hence <span class="s1"><strong><em>a </em></strong></span>is perpendicular to \(\overrightarrow {{\text{BC}}} \) <span class="Apple-converted-space"> </span><strong><em>AG</em></strong></p>
<p class="p1"><strong>Note: <span class="Apple-converted-space"> </span></strong>Only award the final <strong><em>A1 </em></strong>if a dot is used throughout to indicate scalar product.</p>
<p class="p1">Condone any lack of specific indication that the letters represent vectors.</p>
<p class="p1"><strong><em>[5 marks]</em></strong></p>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question">
<p class="p1">This was generally poorly done. The recent syllabus change refers to ‘proof of geometrical properties using vectors’ and this is clearly a topic candidates are not entirely clear with at the moment. Despite the question clearly being written as a vector question some students tried to use a geometrical approach, assuming it was two-dimensional. Many did not seem to realise that vectors being perpendicular implies that their scalar product is zero.</p>
</div>
<br><hr><br><div class="specification">
<p>The points A, B, C and D have position vectors <em><strong>a</strong></em>, <em><strong>b</strong></em>, <em><strong>c</strong></em> and <em><strong>d</strong></em>, relative to the origin O.</p>
<p>It is given that \(\mathop {{\text{AB}}}\limits^ \to = \mathop {{\text{DC}}}\limits^ \to \).</p>
</div>
<div class="specification">
<p>The position vectors \(\mathop {{\text{OA}}}\limits^ \to \), \(\mathop {{\text{OB}}}\limits^ \to \), \(\mathop {{\text{OC}}}\limits^ \to \) and \(\mathop {{\text{OD}}}\limits^ \to \) are given by</p>
<p style="padding-left: 150px;"><em><strong>a</strong></em> = <em><strong>i</strong></em> + 2<em><strong>j</strong></em> − 3<em><strong>k</strong></em></p>
<p style="padding-left: 150px;"><em><strong>b</strong></em> = 3<em><strong>i</strong></em> − <em><strong>j</strong></em> + <em>p<strong>k</strong></em></p>
<p style="padding-left: 150px;"><em><strong>c</strong></em> = <em>q<strong>i</strong></em> + <em><strong>j</strong></em> + 2<em><strong>k</strong></em></p>
<p style="padding-left: 150px;"><em><strong>d</strong></em> = −<em><strong>i</strong></em> + <em>r<strong>j</strong></em> − 2<em><strong>k</strong></em></p>
<p>where <em>p</em> , <em>q</em> and <em>r</em> are constants.</p>
</div>
<div class="specification">
<p>The point where the diagonals of ABCD intersect is denoted by M.</p>
</div>
<div class="specification">
<p>The plane \(\Pi \) cuts the <em>x</em>, <em>y</em> and <em>z</em> axes at X , Y and Z respectively.</p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Explain why ABCD is a parallelogram.</p>
<div class="marks">[1]</div>
<div class="question_part_label">a.i.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Using vector algebra, show that \(\mathop {{\text{AD}}}\limits^ \to = \mathop {{\text{BC}}}\limits^ \to \).</p>
<div class="marks">[3]</div>
<div class="question_part_label">a.ii.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Show that <em>p</em> = 1, <em>q</em> = 1 and <em>r</em> = 4.</p>
<div class="marks">[5]</div>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Find the area of the parallelogram ABCD.</p>
<div class="marks">[4]</div>
<div class="question_part_label">c.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Find the vector equation of the straight line passing through M and normal to the plane \(\Pi \) containing ABCD.</p>
<div class="marks">[4]</div>
<div class="question_part_label">d.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Find the Cartesian equation of \(\Pi \).</p>
<div class="marks">[3]</div>
<div class="question_part_label">e.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Find the coordinates of X, Y and Z.</p>
<div class="marks">[2]</div>
<div class="question_part_label">f.i.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Find YZ.</p>
<div class="marks">[2]</div>
<div class="question_part_label">f.ii.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p>a pair of opposite sides have equal length and are parallel <em><strong>R1</strong></em></p>
<p>hence ABCD is a parallelogram <em><strong>AG</strong></em></p>
<p><em><strong>[1 mark]</strong></em></p>
<div class="question_part_label">a.i.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p>attempt to rewrite the given information in vector form <em><strong>M1</strong></em></p>
<p><em><strong>b</strong></em> − <em><strong>a</strong></em> = <em><strong>c</strong></em> − <em><strong>d</strong></em> <em><strong>A1</strong></em></p>
<p>rearranging <em><strong>d</strong></em> − <em><strong>a</strong></em> = <em><strong>c</strong></em> − <em><strong>b </strong> <strong>M1</strong></em></p>
<p>hence \(\mathop {{\text{AD}}}\limits^ \to = \mathop {{\text{BC}}}\limits^ \to \) <em><strong>AG</strong></em></p>
<p><strong>Note</strong>: Candidates may correctly answer part i) by answering part ii) correctly and then deducing there<br>are two pairs of parallel sides.</p>
<p><em><strong>[3 marks]</strong></em></p>
<div class="question_part_label">a.ii.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><strong>EITHER</strong></p>
<p>use of \(\mathop {{\text{AB}}}\limits^ \to = \mathop {{\text{DC}}}\limits^ \to \) <em><strong>(M1)</strong></em></p>
<p>\(\left( \begin{gathered}<br> 2 \hfill \\<br> - 3 \hfill \\<br> p + 3 \hfill \\ <br>\end{gathered} \right) = \left( \begin{gathered}<br> q + 1 \hfill \\<br> 1 - r \hfill \\<br> 4 \hfill \\ <br>\end{gathered} \right)\) <em><strong>A1A1</strong></em></p>
<p><strong>OR</strong></p>
<p>use of \(\mathop {{\text{AD}}}\limits^ \to = \mathop {{\text{BC}}}\limits^ \to \) <em><strong>(M1)</strong></em></p>
<p>\(\left( \begin{gathered}<br> - 2 \hfill \\<br> r - 2 \hfill \\<br> 1 \hfill \\ <br>\end{gathered} \right) = \left( \begin{gathered}<br> q - 3 \hfill \\<br> 2 \hfill \\<br> 2 - p \hfill \\ <br>\end{gathered} \right)\) <em><strong> A1A1</strong></em></p>
<p><strong>THEN</strong></p>
<p>attempt to compare coefficients of <em><strong>i</strong></em>, <em><strong>j</strong></em>, and <em><strong>k</strong></em> in their equation or statement to that effect <em><strong>M1</strong></em></p>
<p>clear demonstration that the given values satisfy their equation <em><strong>A1</strong></em><br><em>p</em> = 1, <em>q</em> = 1, <em>r</em> = 4 <em><strong>AG</strong></em></p>
<p><em><strong>[5 marks]</strong></em></p>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p>attempt at computing \(\mathop {{\text{AB}}}\limits^ \to \, \times \mathop {{\text{AD}}}\limits^ \to \) (or equivalent) <em><strong>M1</strong></em></p>
<p>\(\left( \begin{gathered}<br> - 11 \hfill \\<br> - 10 \hfill \\<br> - 2 \hfill \\ <br>\end{gathered} \right)\) <em><strong>A1</strong></em></p>
<p>area \( = \left| {\mathop {{\text{AB}}}\limits^ \to \, \times \mathop {{\text{AD}}}\limits^ \to } \right|\left( { = \sqrt {225} } \right)\) <em><strong>(M1)</strong></em></p>
<p>= 15 <em><strong>A1</strong></em></p>
<p><em><strong>[4 marks]</strong></em></p>
<div class="question_part_label">c.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p>valid attempt to find \(\mathop {{\text{OM}}}\limits^ \to = \left( {\frac{1}{2}\left( {a + c} \right)} \right)\) <em><strong>(M1)</strong></em></p>
<p>\(\left( \begin{gathered}<br> 1 \hfill \\<br> \frac{3}{2} \hfill \\<br> - \frac{1}{2} \hfill \\ <br>\end{gathered} \right)\) <em><strong>A1</strong></em></p>
<p>the equation is</p>
<p><em><strong>r</strong></em> = \(\left( \begin{gathered}<br> 1 \hfill \\<br> \frac{3}{2} \hfill \\<br> - \frac{1}{2} \hfill \\ <br>\end{gathered} \right) + t\left( \begin{gathered}<br> 11 \hfill \\<br> 10 \hfill \\<br> 2 \hfill \\ <br>\end{gathered} \right)\) or equivalent <em><strong>M1A1</strong></em></p>
<p><strong>Note</strong>: Award maximum <em><strong>M1A0</strong></em> if '<em><strong>r</strong></em> = …' (or equivalent) is not seen.</p>
<p><em><strong>[4 marks]</strong></em></p>
<div class="question_part_label">d.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p>attempt to obtain the equation of the plane in the form <em>ax</em> + <em>by</em> + <em>cz</em> = <em>d</em> <em><strong>M1</strong></em></p>
<p>11<em>x</em> + 10<em>y</em> + 2<em>z</em> = 25 <em><strong>A1A1</strong></em></p>
<p><strong>Note:</strong> <em><strong>A1</strong> </em>for right hand side, <em><strong>A1</strong></em> for left hand side.</p>
<p><em><strong>[3 marks]</strong></em></p>
<div class="question_part_label">e.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p>putting two coordinates equal to zero <em><strong>(M1)</strong></em></p>
<p>\({\text{X}}\left( {\frac{{25}}{{11}},\,0,\,0} \right),\,\,{\text{Y}}\left( {0,\,\frac{5}{2},\,0} \right),\,\,{\text{Z}}\left( {0,\,0,\,\frac{{25}}{2}} \right)\) <em><strong>A1</strong></em></p>
<p><em><strong>[2 marks]</strong></em></p>
<div class="question_part_label">f.i.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p>\({\text{YZ}} = \sqrt {{{\left( {\frac{5}{2}} \right)}^2} + {{\left( {\frac{{25}}{2}} \right)}^2}} \) <em><strong>M1</strong></em></p>
<p>\( = \sqrt {\frac{{325}}{2}} \left( { = \frac{{5\sqrt {104} }}{4} = \frac{{5\sqrt {26} }}{2}} \right)\) <em><strong> A1</strong></em></p>
<p><em><strong>[4 marks]</strong></em></p>
<div class="question_part_label">f.ii.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">a.i.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">a.ii.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">c.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">d.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">e.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">f.i.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">f.ii.</div>
</div>
<br><hr><br><div class="question" style="padding-left: 20px; padding-right: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">Show that the points \({\text{O}}(0,{\text{ }}0,{\text{ }}0)\), \({\text{ A}}(6,{\text{ }}0,{\text{ }}0)\), \({\text{B}}({6,{\text{ }}- \sqrt {24} ,{\text{ }}\sqrt {12} })\), \({\text{C}}({0,{\text{ }}- \sqrt {24} ,{\text{ }}\sqrt {12}})\) form a square.</span></p>
<div class="marks">[3]</div>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p><span style="font-family: 'times new roman', times; font-size: medium;">Find the coordinates of M, the mid-point of [OB].</span></p>
<div class="marks">[1]</div>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">Show that an equation of the plane </span><span style="font-family: 'times new roman', times; font-size: medium; background-color: #f7f7f7;"><span style="font-family: 'times new roman', times; font-size: medium;"><span style="line-height: normal;">\({\mathit{\Pi }}\), containing the square OABC, is \(y + \sqrt 2 z = 0\).</span></span></span></p>
<div class="marks">[3]</div>
<div class="question_part_label">c.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">Find a vector equation of the line \(L\), through M, perpendicular to the plane <span style="background-color: #f7f7f7;">\({\mathit{\Pi }}\)</span>.</span></p>
<div class="marks">[3]</div>
<div class="question_part_label">d.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">Find the coordinates of D, the point of intersection of the line \(L\) with the plane whose equation is \(y = 0\).</span></p>
<div class="marks">[3]</div>
<div class="question_part_label">e.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p><span style="font-family: 'times new roman', times; font-size: medium;">Find the coordinates of E, the reflection of the point D in the plane <span style="background-color: #f7f7f7;">\({\mathit{\Pi }}\)</span>.</span></p>
<div class="marks">[3]</div>
<div class="question_part_label">f.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(i) Find the angle \({\rm{O\hat DA}}\).</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(ii) State what this tells you about the solid OABCDE.</span></p>
<div class="marks">[6]</div>
<div class="question_part_label">g.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(\left| {\overrightarrow {{\text{OA}}} } \right| = \left| {\overrightarrow {{\text{CB}}} } \right| = \left| {\overrightarrow {{\text{OC}}} } \right| = \left| {\overrightarrow {{\text{AB}}} } \right| = 6\) (therefore a rhombus) <strong><em>A1A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman'; min-height: 25.0px;"> </p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>Note:</strong> Award <strong><em>A1 </em></strong>for two correct lengths, <strong><em>A2 </em></strong>for all four.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman'; min-height: 25.0px;"> </p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>Note:</strong> Award <strong><em>A1A0 </em></strong>for \(\overrightarrow {{\rm{OA}}} = \overrightarrow {{\rm{CB}}} = \left( \begin{array}{l}6\\0\\0\end{array} \right){\rm{ or \,\,} } \overrightarrow {{\rm{OC}}} = \overrightarrow {A{\rm{B}}} = \left( \begin{array}{c}0\\ - \sqrt {24} \\\sqrt {12} \end{array} \right)\) if no magnitudes are shown.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman'; min-height: 25.0px;"> </p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">\(\overrightarrow {{\rm{OA}}}\,\, {\rm{ g}}\overrightarrow {{\rm{OC}}} = \left( \begin{array}{l}6\\0\\0\end{array} \right){\rm{g}}\left( \begin{array}{c}0\\ - \sqrt {24} \\\sqrt {12} \end{array} \right) = 0 \) (therefore a square) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman'; min-height: 25.0px;"> </p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>Note:</strong> Other arguments are possible with a minimum of three conditions.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman'; min-height: 25.0px;"> </p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[3 marks]</em></strong></span></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: 'times new roman', times; font-size: medium;">\({\text{M}}\left( {3,{\text{ }} - \frac{{\sqrt {24} }}{2},{\text{ }}\frac{{\sqrt {12} }}{2}} \right)\left( { = \left( {3,{\text{ }} - \sqrt 6 ,{\text{ }}\sqrt 3 } \right)} \right)\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[1 mark]</em></strong></span></p>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>METHOD 1</strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(\overrightarrow {{\text{OA}}} \times \overrightarrow {{\text{OC}}} = \)\(\left( \begin{array}{l}6\\0\\0\end{array} \right) \times \left( \begin{array}{c}0\\ - \sqrt {24} \\\sqrt {12} \end{array} \right) = \left( \begin{array}{c}0\\ - 6\sqrt {12} \\ - 6\sqrt {24} \end{array} \right)\left( { = \left( \begin{array}{c}0\\ - 12\sqrt 3 \\ - 12\sqrt 6 \end{array} \right)} \right)\) <strong><em>M1A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman'; min-height: 25.0px;"> </p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>Note:</strong> Candidates may use other pairs of vectors.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman'; min-height: 25.0px;"> </p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">equation of plane is \( - 6\sqrt {12} y - 6\sqrt {24} z = d\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">any valid method showing that \(d = 0\) <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span><span style="background-color: #f7f7f7; line-height: normal;"><span style="font-family: 'times new roman', times; font-size: medium;">\(\mathit{\Pi} :y+\sqrt{2z}=0\)</span></span><span style="font-family: 'times new roman', times; font-size: medium;"><span style="line-height: normal;"> </span></span><em><strong><span style="font-family: 'times new roman', times; font-size: medium;">AG</span></strong></em></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em> </em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>METHOD 2</strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">equation of plane is \(ax + by + cz = d\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">substituting O to find \(d = 0\) <strong><em>(M1)</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">substituting two points (A, B, C or M) <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;"><em>eg</em></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">\(6a = 0,{\text{ }} - \sqrt {24} b + \sqrt {12} c = 0\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;"><span style="background-color: #f7f7f7;">\(\mathit{\Pi} :y+\sqrt{2z}=0\)</span> <strong><em>AG</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[3 marks]</em></strong></span></p>
<div class="question_part_label">c.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 22.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(\boldsymbol{r} = \left( \begin{array}{c}3\\ - \sqrt 6 \\\sqrt 3 \end{array} \right) + \lambda \left( \begin{array}{l}0\\1\\\sqrt 2 \end{array} \right)\) <strong><em>A1A1A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 22.0px Helvetica; min-height: 26.0px;"> </p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>Note:</strong> Award <strong><em>A1 </em></strong>for <strong><em>r </em></strong>= , <strong><em>A1A1 </em></strong>for two correct vectors.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman'; min-height: 25.0px;"> </p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[3 marks]</em></strong></span></p>
<div class="question_part_label">d.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">Using \(y = 0\) to find \(\lambda \) <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">Substitute their \(\lambda \) into their equation from part (d) <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">D has coordinates \(\left( {{\text{3, 0, 3}}\sqrt 3 } \right)\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[3 marks]</em></strong></span></p>
<div class="question_part_label">e.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: 'times new roman', times; font-size: medium;">\(\lambda \) for point E is the negative of the \(\lambda \) for point D <strong><em>(M1)</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman'; min-height: 25.0px;"> </p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>Note:</strong> Other possible methods may be seen.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman'; min-height: 25.0px;"> </p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">E has coordinates \(\left( {{\text{3, }} - 2\sqrt 6 ,{\text{ }} - \sqrt 3 } \right)\) <strong><em>A1A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman'; min-height: 25.0px;"> </p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>Note:</strong> Award <strong><em>A1 </em></strong>for each of the <em>y </em>and <em>z </em>coordinates.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman'; min-height: 25.0px;"> </p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[3 marks]</em></strong></span></p>
<div class="question_part_label">f.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(i) \(\overrightarrow {{\text{DA}}} {\text{ g}}\overrightarrow {{\text{DO}}} = \)\(\left( \begin{array}{c}3\\0\\ - 3\sqrt 3 \end{array} \right){\rm{g}}\left( \begin{array}{c} - 3\\0\\ - 3\sqrt 3 \end{array} \right) = 18\) <strong><em>M1A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(\cos {\rm{O\hat DA}} = \frac{{18}}{{\sqrt {36} \sqrt {36} }} = \frac{1}{2}\) <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">hence \({\rm{O\hat DA}} = 60^\circ \) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica; min-height: 25.0px;"> </p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>Note:</strong> Accept method showing OAD is equilateral.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman'; min-height: 25.0px;"> </p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(ii) OABCDE is a regular octahedron (accept equivalent description) <strong><em>A2</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica; min-height: 25.0px;"> </p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>Note:</strong> <strong><em>A2</em></strong> for saying it is made up of 8 equilateral triangles</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;"> Award <strong><em>A1 </em></strong>for two pyramids, <strong><em>A1 </em></strong>for equilateral triangles.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;"> (can be either stated or shown in a sketch – but there must be clear indication the triangles are equilateral)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman'; min-height: 25.0px;"> </p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[6 marks]</em></strong></span></p>
<div class="question_part_label">g.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">c.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">d.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">e.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">f.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">g.</div>
</div>
<br><hr><br><div class="question">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 30.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(a) Show that the two planes</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 30.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\[{\pi _1}:x + 2y - z = 1\]</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 30.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\[{\pi _2}:x + z = - 2\]</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 30.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">are perpendicular.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 30.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(b) Find the equation of the plane \({\pi _3}\) that passes through the origin and is perpendicular to both \({\pi _1}\) and \({\pi _2}\).</span></p>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 31.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(a) for using normal vectors <strong><em>(M1)</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 31.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(\left( {\begin{array}{*{20}{c}}<br> 1 \\ <br> 2 \\ <br> { - 1} <br>\end{array}} \right) \cdot \left( {\begin{array}{*{20}{c}}<br> 1 \\ <br> 0 \\ <br> 1 <br>\end{array}} \right) = 1 - 1 = 0\) <strong><em>M1A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 31.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">hence the two planes are perpendicular <strong><em>AG</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 31.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"> </span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 31.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(b) <strong>METHOD 1</strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 31.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>EITHER</strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 31.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(\left| {\begin{array}{*{20}{c}}<br> i&j&k \\ <br> 1&2&{ - 1} \\ <br> 1&0&1 <br>\end{array}} \right| = \) 2<strong>i</strong> – 2<em><strong>j</strong></em><em><strong>– </strong></em>2<em><strong>k</strong></em> <strong><em>M1A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 31.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>OR</strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 31.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">if \(\left( {\begin{array}{*{20}{c}}<br> a \\ <br> b \\ <br> c <br>\end{array}} \right)\) is normal to \({\pi _3}\), then</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 31.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(a + 2b - c = 0\) and \(a + c = 0\) <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 31.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">a solution is <em>a</em> = 1, <em>b</em> = –1, <em>c</em> = –1 <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 31.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>THEN</strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 31.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\({\pi _3}\) has equation \(x - y - z = d\) <strong><em>(M1)</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 31.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">as it goes through the origin, <em>d</em> = 0 so \({\pi _3}\) has equation \(x - y - z = 0\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 31.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>Note:</strong> The final <strong><em>(M1)A1</em></strong> are independent of previous working.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 31.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"> </span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 31.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>METHOD 2</strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 31.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(r = \left( {\begin{array}{*{20}{c}}<br> 0 \\ <br> 0 \\ <br> 0 <br>\end{array}} \right) + s\left( {\begin{array}{*{20}{c}}<br> 1 \\ <br> 2 \\ <br> { - 1} <br>\end{array}} \right) + t\left( {\begin{array}{*{20}{c}}<br> 1 \\ <br> 0 \\ <br> 1 <br>\end{array}} \right)\) <strong><em>A1(A1)A1A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 31.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[7 marks]</em></strong></span></p>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 24.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">Although many candidates were successful in answering this question, a surprising number showed difficulties in working with normal vectors. In part (b) there were several candidates who found the cross product of the vectors but were unable to use it to write the equation of the plane.</span></p>
</div>
<br><hr><br><div class="question">
<p class="p1">The following system of equations represents three planes in space.</p>
<p class="p1">\[x + 3y + z = - 1\]</p>
<p class="p1">\[x + 2y - 2z = 15\]</p>
<p class="p1">\[2x + y - z = 6\]</p>
<p class="p1">Find the coordinates of the point of intersection of the three planes.</p>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question">
<p class="p1"><strong>EITHER</strong></p>
<p class="p1">eliminating a variable, \(x\), for example to obtain \(y + 3z = - 16\)<span class="s1"> and \( - 5y - 3z = 8\) <span class="Apple-converted-space"> </span></span><strong><em>M1A1</em></strong></p>
<p class="p1">attempting to find the value of one variable <span class="Apple-converted-space"> </span><strong><em>M1</em></strong></p>
<p class="p2">point of intersection is \(( - 1,{\text{ }}2,{\text{ }} - 6)\) <span class="Apple-converted-space"> </span><span class="s2"><strong><em>A1A1A1</em></strong></span></p>
<p class="p1"><strong>OR</strong></p>
<p class="p1">attempting row reduction of relevant matrix, <em>eg</em><span class="s1">. <img src="images/Schermafbeelding_2017-01-27_om_09.53.16.png" alt="M16/5/MATHL/HP1/ENG/TZ2/01_01"> <span class="Apple-converted-space"> </span></span><strong><em>M1</em></strong></p>
<p class="p1">correct matrix with two zeroes in a column, <em>eg</em><span class="s1">. <img src="images/Schermafbeelding_2017-01-27_om_09.54.14.png" alt="M16/5/MATHL/HP1/ENG/TZ2/01_02"> <span class="Apple-converted-space"> </span></span><strong><em>A1</em></strong></p>
<p class="p1">further attempt at reduction <span class="Apple-converted-space"> </span><strong><em>M1</em></strong></p>
<p class="p2">point of intersection is \(( - 1,{\text{ }}2,{\text{ }} - 6)\) <span class="Apple-converted-space"> </span><span class="s2"><strong><em>A1A1A1</em></strong></span></p>
<p class="p3"> </p>
<p class="p1"><strong>Note: <span class="Apple-converted-space"> </span></strong>Allow solution expressed as \(x = - 1,{\text{ }}y = 2,{\text{ }}z = - 6\) for final <strong><em>A </em></strong>marks.</p>
<p class="p3"> </p>
<p class="p1"><strong><em>[6 marks]</em></strong></p>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question">
<p class="p1">This provided a generally easy start for many candidates. Most successful candidates obtained their answer through row reduction of a suitable matrix. Those choosing an alternative method often made slips in their algebra.</p>
</div>
<br><hr><br><div class="specification">
<p class="p1">The position vectors of the points \(A\), \(B\) and \(C\) are \(a\), \(b\) and \(c\) respectively, relative to an origin \(O\). The following diagram shows the triangle \(ABC\) and points \(M\), \(R\), \(S\) and \(T\).</p>
<p class="p1" style="text-align: center;"><img src="images/Schermafbeelding_2015-12-08_om_15.53.19.png" alt></p>
<p>\(M\) is the midpoint of [\(AC\)].</p>
<p>\(R\) is a point on [\(AB\)] such that \(\overrightarrow {{\text{AR}}} = \frac{1}{3}\overrightarrow {{\text{AB}}} \).</p>
<p>\(S\) is a point on [\(AC\)] such that \(\overrightarrow {{\text{AS}}} = \frac{2}{3}\overrightarrow {{\text{AC}}} \).</p>
<p>\(T\) is a point on [\(RS\)] such that \(\overrightarrow {{\text{RT}}} = \frac{2}{3}\overrightarrow {{\text{RS}}} \).</p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>(i) Express \(\overrightarrow {{\text{AM}}} \) in terms of <em>\(a\)</em> and \(c\).</p>
<p>(ii) Hence show that \(\overrightarrow {{\text{BM}}} = \frac{1}{2}\)\(a\) – \(b\)\( + \frac{1}{2}c\).</p>
<div class="marks">[4]</div>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>(i) Express \(\overrightarrow {{\text{RA}}} \) in terms of <em>\(a\) </em>and <em>\(b\)</em>.</p>
<p>(ii) Show that \(\overrightarrow {RT} = - \frac{2}{9}a - \frac{2}{9}b + \frac{4}{9}c\).</p>
<div class="marks">[5]</div>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p class="p1">Prove that \(T\) lies on [\(BM\)].</p>
<div class="marks">[5]</div>
<div class="question_part_label">c.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p>(i) \(\overrightarrow {{\text{AM}}} = \frac{1}{2}\overrightarrow {{\text{AC}}} \) <strong><em>(M1)</em></strong></p>
<p>\( = \frac{1}{2}\)(\(c\) – \(a\)) <strong><em>A1</em></strong></p>
<p>(ii) \(\overrightarrow {{\text{BM}}} = \overrightarrow {{\text{BA}}} + \overrightarrow {{\text{AM}}} \) <strong><em>M1</em></strong></p>
<p>\( = a - b + \frac{1}{2}\)\((c - a)\) <strong><em>A1</em></strong></p>
<p>\(\overrightarrow {{\text{BM}}} = \frac{1}{2}a - b + \frac{1}{2}c\) <strong><em>AG</em></strong></p>
<p><strong><em>[4 marks]</em></strong></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p>(i) \(\overrightarrow {{\text{RA}}} = \frac{1}{3}\overrightarrow {{\text{BA}}} \)</p>
<p>\( = \frac{1}{3}\)(<strong><em>a</em></strong> – <strong><em>b</em></strong>) <strong><em>A1</em></strong></p>
<p>(ii) \(\overrightarrow {{\text{RT}}} = \frac{2}{3}\overrightarrow {{\text{RS}}} \)</p>
<p>\( = \frac{2}{3}\left( {\overrightarrow {{\text{RA}}} + \overrightarrow {{\text{AS}}} } \right)\) <strong><em>(M1)</em></strong></p>
<p>\( = \frac{2}{3}\left( {\frac{1}{3}(a - b) + \frac{2}{3}(c - a)} \right)\;\;\;\)or equivalent. <strong><em>A1A1</em></strong></p>
<p>\( = \frac{2}{9}\)\(\left( {a - b} \right)\) \( + \frac{4}{9}\)\(\left( {c - a} \right)\) <strong><em>A1</em></strong></p>
<p>\(\overrightarrow {{\text{RT}}} = - \frac{2}{9}\)<em>\(a\)</em> – \( - \frac{2}{9}\)<em>\(b\)</em> \( + \frac{4}{9}\)<em>\(c\)</em> <strong><em>AG</em></strong></p>
<p><strong><em>[5 marks]</em></strong></p>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p>\(\overrightarrow {{\text{BT}}} = \overrightarrow {{\text{BR}}} + \overrightarrow {{\text{RT}}} \)</p>
<p>\( = \frac{2}{3}\overrightarrow {{\text{BA}}} + \overrightarrow {{\text{RT}}} \) <strong><em>(M1)</em></strong></p>
<p>\( = \frac{2}{3}a - \frac{2}{3}b - \frac{2}{9}a - \frac{2}{9}b + \frac{4}{9}c\) <strong><em>A1</em></strong></p>
<p>\(\overrightarrow {{\text{BT}}} = \frac{8}{9}\left( {\frac{1}{2}a - b + \frac{1}{2}c} \right)\) <strong><em>A1</em></strong></p>
<p>point \(B\) is common to \(\overrightarrow {{\text{BT}}} \) and \(\overrightarrow {{\text{BM}}} \) and \(\overrightarrow {{\text{BT}}} = \frac{8}{9}\overrightarrow {{\text{BM}}} \) <strong><em>R1R1</em></strong></p>
<p>so \(T\) lies on [\(BM\)] <strong><em>AG</em></strong></p>
<p><strong><em>[5 marks]</em></strong></p>
<p><strong><em>Total [14 marks]</em></strong></p>
<div class="question_part_label">c.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
<p class="p1">A fairly straightforward question for candidates confident in the use of and correct notation for relative position vectors. Sign errors were the most common, but the majority of candidates did not gain all the reasoning marks for part (c). In particular, it was necessary to observe that not only were two vectors parallel, but that they had a point in common.</p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">A fairly straightforward question for candidates confident in the use of and correct notation for relative position vectors. Sign errors were the most common, but the majority of candidates did not gain all the reasoning marks for part (c). In particular, it was necessary to observe that not only were two vectors parallel, but that they had a point in common.</p>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">A fairly straightforward question for candidates confident in the use of and correct notation for relative position vectors. Sign errors were the most common, but the majority of candidates did not gain all the reasoning marks for part (c). In particular, it was necessary to observe that not only were two vectors parallel, but that they had a point in common.</p>
<div class="question_part_label">c.</div>
</div>
<br><hr><br><div class="question">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 12.0px Times;"><span style="font-family: 'times new roman', times; font-size: medium;">Find the values of <em>x </em>for which the vectors \(\left( {\begin{array}{*{20}{c}}<br> 1 \\ <br> {2\cos x} \\ <br> 0 <br>\end{array}} \right)\) and \(\left( {\begin{array}{*{20}{c}}<br> { - 1} \\ <br> {2\sin x} \\ <br> 1 <br>\end{array}} \right)\) are perpendicular, \(0 \leqslant x \leqslant \frac{\pi }{2}\).</span></p>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Times;"><span style="font-family: 'times new roman', times; font-size: medium;">perpendicular when \(\left( {\begin{array}{*{20}{c}}<br> 1 \\ <br> {2\cos x} \\ <br> 0 <br>\end{array}} \right) \cdot \left( {\begin{array}{*{20}{c}}<br> { - 1} \\ <br> {2\sin x} \\ <br> 1 <br>\end{array}} \right) = 0\) <strong><em>(M1)</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Times;"><span style="font-family: 'times new roman', times; font-size: medium;">\( \Rightarrow -1 + 4\sin x\cos x = 0\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Times;"><span style="font-family: 'times new roman', times; font-size: medium;">\( \Rightarrow \sin 2x = \frac{1}{2}\) <strong> <em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Times;"><span style="font-family: 'times new roman', times; font-size: medium;">\( \Rightarrow 2x = \frac{\pi }{6},\frac{{5\pi }}{6}\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Times;"><span style="font-family: 'times new roman', times; font-size: medium;">\( \Rightarrow x = \frac{\pi }{{12}},\frac{{5\pi }}{{12}}\) <strong> <em>A1A1</em></strong></span><strong style="font-family: 'times new roman', times; font-size: medium;"><em> </em></strong></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Times;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>Note: </strong>Accept answers in degrees. </span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Times;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em> </em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Times;"><strong><em><span style="font-family: 'times new roman', times; font-size: medium;">[5 marks]</span><br></em></strong></p>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 10.0px Arial;"><span style="font-family: 'times new roman', times; font-size: medium;">Most candidates realised that the scalar product should be used to solve this problem and many obtained the equation \(4\sin x\cos x = 1\). Candidates who failed to see that this could be written as \(\sin 2x = 0.5\) usually made no further progress. The majority of those candidates who used this double angle formula carried on to obtain the solution \(\frac{\pi }{{12}}\) but few candidates realised that \(\frac{{5\pi }}{{12}}\) was also a solution.</span></p>
</div>
<br><hr><br><div class="specification">
<p class="p1"><span class="s1">Consider the vectors </span><strong><em>a</em></strong> \( = \) <strong><em>i</em></strong> \( - {\text{ }}3\)<strong><em>j</em></strong> \( - {\text{ }}2\)<strong><em>k</em></strong>, <strong><em>b</em></strong> \( = - {\text{ }}3\)<strong><em>j</em></strong> \( + {\text{ }}2\)<strong><em>k</em></strong><span class="s1">.</span></p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p class="p1"><span class="s1">Find </span><strong><em>a</em></strong> \( \times \) <strong><em>b</em></strong><span class="s2">.</span></p>
<div class="marks">[2]</div>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p class="p1">Hence find the Cartesian equation of the plane containing the vectors <span class="s1"><strong><em>a </em></strong></span>and <span class="s1"><strong><em>b</em></strong></span>, and passing through the point \((1,{\text{ }}0,{\text{ }} - 1)\).</p>
<div class="marks">[3]</div>
<div class="question_part_label">b.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p class="p1"><strong><em>a</em></strong> \( \times \) <strong><em>b</em></strong> \( = - 12\)<strong><em>i</em></strong> \( - {\text{ }}2\)<strong><em>j</em></strong> \( - {\text{ }}3\)<strong><em>k <span class="Apple-converted-space"> </span></em></strong><span class="s1"><strong><em>(M1)A1</em></strong></span></p>
<p class="p2"><strong><em>[2 marks]</em></strong></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1"><strong>METHOD 1</strong></p>
<p class="p2"><span class="Apple-converted-space">\( - 12x - 2y - 3z = d\) </span><span class="s1"><strong><em>M1</em></strong></span></p>
<p class="p3"><span class="Apple-converted-space">\( - 12 \times 1 - 2 \times 0 - 3( - 1) = d\) </span><span class="s1"><strong>(<em>M1)</em></strong></span></p>
<p class="p3"><span class="Apple-converted-space">\( \Rightarrow d = - 9\) </span><span class="s1"><strong><em>A1</em></strong></span></p>
<p class="p2">\( - 12x - 2y - 3z = - 9{\text{ }}({\text{or }}12x + 2y + 3z = 9)\)</p>
<p class="p1"><strong>METHOD 2</strong></p>
<p class="p2"><span class="Apple-converted-space">\(\left( {\begin{array}{*{20}{c}} x \\ y \\ z \end{array}} \right) \bullet \left( {\begin{array}{*{20}{c}} { - 12} \\ { - 2} \\ { - 3} \end{array}} \right) = \left( {\begin{array}{*{20}{c}} 1 \\ 0 \\ { - 1} \end{array}} \right) \bullet \left( {\begin{array}{*{20}{c}} { - 12} \\ { - 2} \\ { - 3} \end{array}} \right)\) </span><span class="s1"><strong><em>M1A1</em></strong></span></p>
<p class="p2"><span class="Apple-converted-space">\( - 12x - 2y - 3z = - 9{\text{ }}({\text{or }}12x + 2y + 3z = 9)\) </span><span class="s1"><strong><em>A1</em></strong></span></p>
<p class="p1"><strong><em>[3 marks]</em></strong></p>
<div class="question_part_label">b.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">b.</div>
</div>
<br><hr><br><div class="specification">
<p><span style="font-family: times new roman,times; font-size: medium;">The points A(1, 2, 1) , B(−3, 1, 4) , C(5, −1, 2) and D(5, 3, 7) are the vertices of a tetrahedron.</span></p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">Find the vectors \(\overrightarrow {{\text{AB}}} \)</span> <span style="font-family: times new roman,times; font-size: medium;">and \(\overrightarrow {{\text{AC}}} \).</span></p>
<div class="marks">[2]</div>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">Find the Cartesian equation of the plane \(\prod \) that contains the face ABC.</span></p>
<div class="marks">[4]</div>
<div class="question_part_label">b.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">\(\overrightarrow {{\text{AB}}} = \left( {\begin{array}{*{20}{c}}<br> { - 4} \\ <br> { - 1} \\ <br> 3 <br>\end{array}} \right)\), \(\overrightarrow {{\text{AC}}} = \left( {\begin{array}{*{20}{c}}<br> 4 \\ <br> { - 3} \\ <br> 1 <br>\end{array}} \right)\) <em><strong>A1A1</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"><strong>Note:</strong> Accept row vectors.</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"> </span></p>
<p><em><strong><span style="font-family: times new roman,times; font-size: medium;">[2 marks]</span></strong></em></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">\(\overrightarrow {{\text{AB}}} \times \overrightarrow {{\text{AC}}} = \left| {\begin{array}{*{20}{c}}<br> {\boldsymbol{i}}&{\boldsymbol{j}}&{\boldsymbol{k}} \\ <br> { - 4}&{ - 1}&3 \\ <br> 4&{ - 3}&1 <br>\end{array}} \right| = \left( {\begin{array}{*{20}{c}}<br> 8 \\ <br> {16} \\ <br> {16} <br>\end{array}} \right)\) </span><em><strong><span style="font-family: times new roman,times; font-size: medium;">M1A1</span></strong></em><br><span style="font-family: times new roman,times; font-size: medium;">normal \({\boldsymbol{n}} = \left( {\begin{array}{*{20}{c}}<br> 1 \\ <br> 2 \\ <br> 2 <br>\end{array}} \right)\) so \({\boldsymbol{r}} \cdot \left( {\begin{array}{*{20}{c}}<br> 1 \\ <br> 2 \\ <br> 2 <br>\end{array}} \right) = \left( {\begin{array}{*{20}{c}}<br> 1 \\ <br> 2 \\ <br> 1 <br>\end{array}} \right) \cdot \left( {\begin{array}{*{20}{c}}<br> 1 \\ <br> 2 \\ <br> 2 <br>\end{array}} \right)\)</span><span style="font-family: times new roman,times; font-size: medium;"> <em><strong>(M1)</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(x + 2y + 2z = 7\) <em><strong>A1</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"><strong>Note:</strong> If attempt to solve by a system of equations:</span><br><span style="font-family: times new roman,times; font-size: medium;">Award <em><strong>A1</strong></em> for 3 correct equations, <em><strong>A1</strong></em> for eliminating a variable and <em><strong>A2</strong></em> for </span><span style="font-family: times new roman,times; font-size: medium;">the correct answer.</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"> </span></p>
<p><em><strong><span style="font-family: times new roman,times; font-size: medium;">[4 marks]</span></strong></em></p>
<div class="question_part_label">b.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">Most candidates attempted this question and scored at least a few marks in (a) and (b). Part (c) was more challenging to many candidates who were unsure how to find the required distance. Part (d) was attempted by many candidates some of whom benefited from follow through marks due to errors in previous parts. However, many candidates failed to give the correct answer to this question due to the use of the simplified vector found in (b) showing little understanding of the role of the magnitude of this vector. Part (e) was poorly answered. Overall, this question was not answered to the expected level, showing that many candidates have difficulties with vectors and are unable to answer even standard questions on this topic.</span></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">Most candidates attempted this question and scored at least a few marks in (a) and (b). Part (c) was more challenging to many candidates who were unsure how to find the required distance. Part (d) was attempted by many candidates some of whom benefited from follow through marks due to errors in previous parts. However, many candidates failed to give the correct answer to this question due to the use of the simplified vector found in (b) showing little understanding of the role of the magnitude of this vector. Part (e) was poorly answered. Overall, this question was not answered to the expected level, showing that many candidates have difficulties with vectors and are unable to answer even standard questions on this topic.</span></p>
<div class="question_part_label">b.</div>
</div>
<br><hr><br><div class="specification">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">Consider the points \({\text{A(1, 0, 0)}}\), \({\text{B(2, 2, 2)}}\) and \({\text{C(0, 2, 1)}}\).</span></p>
</div>
<div class="specification">
<p><span style="font-family: 'times new roman', times; font-size: medium;">A third plane \({\Pi _3}\) is defined by the Cartesian equation \(16x + \alpha y - 3z = \beta \).</span></p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">Find the vector \(\overrightarrow {{\text{CA}}} \times \overrightarrow {{\text{CB}}} \).</span></p>
<div class="marks">[4]</div>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p><span style="font-family: 'times new roman', times; font-size: medium;">Find an exact value for the area of the triangle ABC.</span></p>
<div class="marks">[3]</div>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">Show that the Cartesian equation of the plane \({\Pi _1}\), containing the triangle ABC, is \(2x + 3y - 4z = 2\).</span></p>
<div class="marks">[3]</div>
<div class="question_part_label">c.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">A second plane \({\Pi _2}\) is defined by the Cartesian equation \({\Pi _2}:4x - y - z = 4\). \({L_1}\) is the line of intersection of the planes \({\Pi _1}\) and \({\Pi _2}\).</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">Find a vector equation for \({L_1}\).</span></p>
<div class="marks">[5]</div>
<div class="question_part_label">d.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p><span>Find the value of \(\alpha \) if all three planes contain \({L_1}\).</span></p>
<div class="marks">[3]</div>
<div class="question_part_label">e.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p><span style="font-family: 'times new roman', times; font-size: medium;">Find conditions on \(\alpha \) and \(\beta \) if the plane \({\Pi _3}\) does <strong>not </strong>intersect with \({L_1}\).</span></p>
<div class="marks">[2]</div>
<div class="question_part_label">f.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(\overrightarrow {{\rm{CA}}} = \left( \begin{array}{c}1\\ - 2\\ - 1\end{array} \right)\) <strong><em>(A1)</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(\overrightarrow {{\rm{CB}}} = \left( \begin{array}{c}2\\0\\1\end{array} \right)\) <strong><em>(A1)</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica; min-height: 25.0px;"> </p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>Note:</strong> If \(\overrightarrow {{\text{AC}}} \) and \(\overrightarrow {{\text{BC}}} \) found correctly award <strong><em>(A1) (A0)</em></strong>.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman'; min-height: 25.0px;"> </p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">\(\overrightarrow {{\rm{CA}}} \times \overrightarrow {{\rm{CB}}} = \left| {\begin{array}{*{20}{c}}i&j&k\\1&{ - 2}&{ - 1}\\2&0&1\end{array}} \right|\) <strong><em>(M1)</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman'; min-height: 25.0px;"> </p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">\(\left( \begin{array}{c} - 2\\ - 3\\4\end{array} \right)\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[4 marks]</em></strong></span></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: 'times new roman', times; font-size: medium;"><strong>METHOD 1</strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">\(\frac{1}{2}\left| {\overrightarrow {{\text{CA}}} \times \overrightarrow {{\text{CB}}} } \right|\) <strong><em>(M1)</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">\( = \frac{1}{2}\sqrt {{{( - 2)}^2} + {{( - 3)}^2} + {4^2}} \) <strong><em>(A1)</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">\( = \frac{{\sqrt {29} }}{2}\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>METHOD 2</strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">attempt to apply \(\frac{1}{2}\left| {{\text{CA}}} \right|\left| {{\text{CB}}} \right|\sin C\) <strong><em>(M1)</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">\({\text{CA.CB}} = \sqrt 5 .\sqrt 6 \cos C \Rightarrow \cos C = \frac{1}{{\sqrt {30} }} \Rightarrow \sin C = \frac{{\sqrt {29} }}{{\sqrt {30} }}\) <strong><em>(A1)</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">\({\text{area}} = \frac{{\sqrt {29} }}{2}\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[3 marks]</em></strong></span></p>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>METHOD 1</strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>r</strong>.\(\left( \begin{array}{c} - 2\\ - 3\\4\end{array} \right) = \left( \begin{array}{l}1\\0\\0\end{array} \right) \bullet \left( \begin{array}{c} - 2\\ - 3\\4\end{array} \right)\) <strong><em>M1A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( \Rightarrow - 2x - 3y + 4z = - 2\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( \Rightarrow 2x + 3y - 4z = 2\) <strong><em>AG</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>METHOD 2</strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( - 2x - 3y + 4z = d\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">substituting a point in the plane <strong><em>M1A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\({\text{d}} = - 2\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( \Rightarrow - 2x - 3y + 4z = - 2\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( \Rightarrow 2x + 3y - 4z = 2\) <strong><em>AG</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman'; min-height: 25.0px;"> </p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>Note:</strong> Accept verification that all 3 vertices of the triangle lie on the </span><span style="font-family: 'times new roman', times; font-size: medium;">given plane.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman'; min-height: 25.0px;"> </p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[3 marks]</em></strong></span></p>
<div class="question_part_label">c.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>METHOD 1</strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span><span style="background-color: #f7f7f7; line-height: normal;"><span style="font-family: 'times new roman', times; font-size: medium;">\(\left| {\begin{array}{*{20}{c}}{\mathbf{i}}&{\mathbf{j}}&{\mathbf{k}}\\2&3&{ - 4}\\4&{ - 1}&{ - 1}\end{array}} \right| = \left( \begin{array}{c} - 7\\ - 14\\ - 14\end{array} \right)\)</span></span><span style="font-family: 'times new roman', times; font-size: medium;"><span style="line-height: normal;"> </span></span><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>M1A1</em></strong></span></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\({\mathbf{n}} = \left( \begin{array}{l}1\\2\\2\end{array} \right)\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(z = 0 \Rightarrow y = 0,{\text{ }}x = 1\) <strong><em>(M1)(A1)</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\({L_1}:{\mathbf{r}} = \left( \begin{array}{l}1\\0\\0\end{array} \right) + \lambda \left( \begin{array}{l}1\\2\\2\end{array} \right)\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica; min-height: 25.0px;"> </p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>Note:</strong> Do not award the final <strong><em>A1 </em></strong>if \(\mathbf{r} =\) is not seen.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman'; min-height: 25.0px;"> </p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>METHOD 2</strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">eliminate 1 of the variables, <em>eg x</em> <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">\( - 7y + 7z = 0\) <strong><em>(A1)</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">introduce a parameter <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">\( \Rightarrow z = \lambda \),</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">\(y = \lambda {\text{, }}x = 1 + \frac{\lambda }{2}\) <strong><em>(A1)</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">\({\mathbf{r}} = \left( \begin{array}{l}1\\0\\0\end{array} \right) + \lambda \left( \begin{array}{l}1\\2\\2\end{array} \right)\) or equivalent <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman'; min-height: 25.0px;"> </p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>Note:</strong> Do not award the final <strong><em>A1 </em></strong>if \(\mathbf{r} =\) is not seen.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman'; min-height: 25.0px;"> </p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>METHOD 3</strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">\(z = t\) <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">write <em>x </em>and <em>y </em>in terms of \(t \Rightarrow 4x - y = 4 + t,{\text{ }}2x + 3y = 2 + 4t\) or equivalent <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">attempt to eliminate <em>x </em>or <em>y</em> <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">\(x,{\text{ }}y,{\text{ }}z\) expressed in parameters</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">\( \Rightarrow z = t\),</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">\(y = t,{\text{ }}x = 1 + \frac{t}{2}\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">\({\mathbf{r}} = \left( \begin{array}{l}1\\0\\0\end{array} \right) + t \left( \begin{array}{l}1\\2\\2\end{array} \right)\) or equivalent <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman'; min-height: 25.0px;"> </p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>Note:</strong> Do not award the final <strong><em>A1 </em></strong>if \(\mathbf{r} =\) is not seen.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman'; min-height: 25.0px;"> </p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[5 marks]</em></strong></span></p>
<div class="question_part_label">d.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>METHOD 1</strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">direction of the line is perpendicular to the normal of the plane</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(\left( \begin{array}{c}16\\\alpha \\ - 3\end{array} \right) \bullet \left( \begin{array}{l}1\\2\\2\end{array} \right) = 0\) <strong><em>M1A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(16 + 2\alpha - 6 = 0 \Rightarrow \alpha = - 5\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>METHOD 2</strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">solving line/plane simultaneously</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(16(1 + \lambda ) + 2\alpha \lambda - 6\lambda = \beta \) <strong><em>M1A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(16 + (10 + 2\alpha )\lambda = \beta \)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( \Rightarrow \alpha = - 5\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>METHOD 3</strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(\left| {\begin{array}{*{20}{c}}2&3&{ - 4}\\4&{ - 1}&{ - 1}\\{16}&\alpha &{ - 3}\end{array}} \right| = 0\) <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(2(3 + \alpha ) - 3( - 12 + 16) - 4(4\alpha + 16) = 0\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( \Rightarrow \alpha = - 5\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>METHOD 4</strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">attempt to use row reduction on augmented matrix <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">to obtain \(\left( {\begin{array}{*{20}{c}}2&3&{ - 4}\\0&{ - 1}&1\\0&0&{\alpha + 5}\end{array}\left| \begin{array}{c}2\\0\\\beta - 16\end{array} \right.} \right)\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( \Rightarrow \alpha = - 5\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[3 marks]</em></strong></span></p>
<div class="question_part_label">e.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: 'times new roman', times; font-size: medium;">\(\alpha = - 5\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(\beta \ne 16\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[2 marks]</em></strong></span></p>
<div class="question_part_label">f.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 19.0px Arial;"><span style="font-family: 'times new roman', times; font-size: medium;">Part a) proved an easy start, though a few (weaker) candidates still believe \(\overrightarrow {{\text{CA}}} \) to be \(\overrightarrow {{\text{OC}}} - \overrightarrow {{\text{OA}}} \).</span></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 19.0px Arial;"><span style="font-family: 'times new roman', times; font-size: medium;">Part b) was an easy 3 marks and incorrect answers were rare.</span></p>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 19.0px Arial;"><span style="font-family: 'times new roman', times; font-size: medium;">Part c) was answered well, though reasoning sometimes seemed sparse, especially given that this was a ‘show that’ question.</span></p>
<div class="question_part_label">c.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 19.0px Arial;"><span style="font-family: 'times new roman', times; font-size: medium;">Part d) proved more challenging, despite being a very standard question. Many candidates gained only 2 marks, either through correctly calculating the direction vector, or by successfully eliminating one of the variables. A number of clear fully correct solutions were seen, though the absence of ‘<strong><em>r</em></strong> =’ is still prevalent, and candidates might be reminded of the correct form for the vector equation of a line.</span></p>
<div class="question_part_label">d.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 19.0px Arial;"><span style="font-family: 'times new roman', times; font-size: medium;">Part e) proved a puzzle for most, though an attempt to use row reduction on an augmented matrix seemed to be the choice way for most successful candidates.</span></p>
<div class="question_part_label">e.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 19.0px Arial;"><span style="font-family: 'times new roman', times; font-size: medium;">Only the very best were able to demonstrate a complete understanding of intersecting planes and thus answer part f) correctly.</span></p>
<div class="question_part_label">f.</div>
</div>
<br><hr><br><div class="specification">
<p class="p1" style="text-align: center;"><img src="images/Schermafbeelding_2016-01-29_om_08.02.43.png" alt></p>
<p>Consider the triangle \(ABC\). The points \(P\), \(Q\) and \(R\) are the midpoints of the line segments [\(AB\)], [\(BC\)] and [\(AC\)] respectively.</p>
<p>Let \(\overrightarrow {{\text{OA}}} = {{a}}\), \(\overrightarrow {{\text{OB}}} = {{b}}\) and \(\overrightarrow {{\text{OC}}} = {{c}}\).</p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p class="p1">Find \(\overrightarrow {{\text{BR}}} \) <span class="s1">in terms of </span>\({{a}}\), \({{b}}\) <span class="s1">and </span>\({{c}}\).</p>
<div class="marks">[2]</div>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>(i) Find a vector equation of the line that passes through \(B\) and \(R\) in terms of \({{a}}\), \({{b}}\) and \({{c}}\) and a parameter \(\lambda \).</p>
<p>(ii) Find a vector equation of the line that passes through \(A\) and \(Q\) in terms of \({{a}}\), \({{b}}\) and \({{c}}\) and a parameter \(\mu \).</p>
<p>(iii) Hence show that \(\overrightarrow {{\text{OG}}} = \frac{1}{3}({{a}} + {{b}} + {{c}})\) given that \(G\) is the point where [\(BR\)] and [\(AQ\)] intersect.</p>
<div class="marks">[9]</div>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p class="p1">Show that the line segment <span class="s1">[\(CP\)] </span>also includes the point <span class="s1">\(G\)</span>.</p>
<div class="marks">[3]</div>
<div class="question_part_label">c.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p class="p1">The coordinates of the points <span class="s1">\(A\)</span>, <span class="s1">\(B\)</span> and <span class="s1">\(C\)</span> are \((1,{\text{ }}3,{\text{ }}1)\), \((3,{\text{ }}7,{\text{ }} - 5)\) and \((2,{\text{ }}2,{\text{ }}1)\) respectively.</p>
<p class="p1">A point <span class="s1">\(X\)</span> is such that [<span class="s1">\(GX\)</span>] is perpendicular to the plane <span class="s1">\(ABC\)</span>.</p>
<p class="p1">Given that the tetrahedron <span class="s1">\(ABCX\)</span> has volume \({\text{12 unit}}{{\text{s}}^{\text{3}}}\), find possible coordinates</p>
<p class="p1">of <span class="s1">\(X\).</span></p>
<div class="marks">[9]</div>
<div class="question_part_label">d.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p>\(\overrightarrow {{\text{BR}}} = \overrightarrow {{\text{BA}}} + \overrightarrow {{\text{AR}}} \;\;\;\left( { = \overrightarrow {{\text{BA}}} + \frac{1}{2}\overrightarrow {{\text{AC}}} } \right)\) <strong><em>(M1)</em></strong></p>
<p>\( = ({{a}} - {{b}}) + \frac{1}{2}({{c}} - {{a}})\)</p>
<p>\( = \frac{1}{2}{{a}} - {{b}} + \frac{1}{2}{{c}}\) <strong><em>A1</em></strong></p>
<p><strong><em>[2 marks]</em></strong></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p>(i) \({{\text{r}}_{{\text{BR}}}} = {{b}} + \lambda \left( {\frac{1}{2}{{a}} - {{b}} + \frac{1}{2}{{c}}} \right)\;\;\;\left( { = \frac{\lambda }{2}{{a}} + (1 - \lambda ){{b}} + \frac{\lambda }{2}{{c}}} \right)\) <strong><em>A1A1</em></strong></p>
<p> </p>
<p><strong>Note: </strong>Award <strong><em>A1A0 </em></strong>if the \({\text{r}} = \) is omitted in an otherwise correct expression/equation.</p>
<p>Do not penalise such an omission more than once.</p>
<p> </p>
<p>(ii) \(\overrightarrow {{\text{AQ}}} = - {{a}} + \frac{1}{2}{{b}} + \frac{1}{2}{{c}}\) <strong><em>(A1)</em></strong></p>
<p>\({{\text{r}}_{{\text{AQ}}}} = {{a}} + \mu \left( { - {{a}} + \frac{1}{2}{{b}} + \frac{1}{2}{{c}}} \right)\;\;\;\left( { = (1 - \mu ){{a}} + \frac{\mu }{2}{{b}} + \frac{\mu }{2}{{c}}} \right)\) <strong><em>A1</em></strong></p>
<p> </p>
<p><strong>Note: </strong>Accept the use of the same parameter in (i) and (ii).</p>
<p> </p>
<p>(iii) when \(\overrightarrow {{\text{AQ}}} \) and \(\overrightarrow {{\text{BP}}} \) intersect we will have \({{\text{r}}_{{\text{BR}}}} = {{\text{r}}_{{\text{AQ}}}}\) <strong><em>(M1)</em></strong></p>
<p> </p>
<p><strong>Note: </strong>If the same parameters are used for both equations, award at most <strong><em>M1M1A0A0M1</em></strong>.</p>
<p> </p>
<p>\(\frac{\lambda }{2}{{a}} + (1 - \lambda ){{b}} + \frac{\lambda }{2}{{c}} = (1 - \mu ){{a}} + \frac{\mu }{2}{{b}} + \frac{\mu }{2}{{c}}\)</p>
<p>attempt to equate the coefficients of the vectors \({{a}}\), \({{b}}\) and \({{c}}\) <strong><em>M1</em></strong></p>
<p>\(\left. {\begin{array}{*{20}{c}} {\frac{\lambda }{2} = 1 - \mu } \\ {1 - \lambda = \frac{\mu }{2}} \\ {\frac{\lambda }{2} = \frac{\mu }{2}} \end{array}} \right\}\) <strong><em>(A1)</em></strong></p>
<p>\(\lambda = \frac{2}{3}\) or \(\mu = \frac{2}{3}\) <strong><em>A1</em></strong></p>
<p>substituting parameters back into one of the equations <strong><em>M1</em></strong></p>
<p>\(\overrightarrow {{\text{OG}}} = \frac{1}{2} \bullet \frac{2}{3}{{a}} + \left( {1 - \frac{2}{3}} \right){{b}} + \frac{1}{2} \bullet \frac{2}{3}{{c}} = \frac{1}{3}({{a}} + {{b}} + {{c}})\) <strong><em>AG</em></strong></p>
<p> </p>
<p><strong>Note</strong>: Accept solution by verification.</p>
<p><strong><em>[9 marks]</em></strong></p>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p>\(\overrightarrow {{\text{CP}}} = \frac{1}{2}{{a}} + \frac{1}{2}{{b}} - {{c}}\) <strong><em>(M1)A1</em></strong></p>
<p>so we have that \({{\text{r}}_{{\text{CP}}}} = {{c}} + \beta \left( {\frac{1}{2}{{a}} + \frac{1}{2}{{b}} - {{c}}} \right)\) and when \(\beta = \frac{2}{3}\) the line passes through</p>
<p>the point <span class="s1">\(G\)</span> (<em>ie</em>, with position vector \(\frac{1}{3}({{a}} + {{b}} + {{c}})\)) <strong><em>R1</em></strong></p>
<p>hence [<span class="s1">\(AQ\)</span>], [<span class="s1">\(BR\)</span>] and [<span class="s1">\(CP\)</span>] all intersect in <span class="s1">\(G\)</span> <strong><em>AG</em></strong></p>
<p><strong><em>[3 marks]</em></strong></p>
<div class="question_part_label">c.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p>\(\overrightarrow {{\text{OG}}} = \frac{1}{3}\left( {\left( {\begin{array}{*{20}{c}} 1 \\ 3 \\ 1 \end{array}} \right) + \left( {\begin{array}{*{20}{c}} 3 \\ 7 \\ { - 5} \end{array}} \right) + \left( {\begin{array}{*{20}{c}} 2 \\ 2 \\ 1 \end{array}} \right)} \right) = \left( {\begin{array}{*{20}{c}} 2 \\ 4 \\ { - 1} \end{array}} \right)\) <strong><em>A1</em></strong></p>
<p> </p>
<p><strong>Note: </strong>This independent mark for the vector may be awarded wherever the vector is calculated.</p>
<p>\(\overrightarrow {{\text{AB}}} \times \overrightarrow {{\text{AC}}} = \left( {\begin{array}{*{20}{c}} 2 \\ 4 \\ { - 6} \end{array}} \right) \times \left( {\begin{array}{*{20}{c}} 1 \\ { - 1} \\ 0 \end{array}} \right) = \left( {\begin{array}{*{20}{c}} { - 6} \\ { - 6} \\ { - 6} \end{array}} \right)\) <strong><em>M1A1</em></strong></p>
<p>\(\overrightarrow {{\text{GX}}} = \alpha \left( {\begin{array}{*{20}{c}} 1 \\ 1 \\ 1 \end{array}} \right)\) <strong><em>(M1)</em></strong></p>
<p>volume of Tetrahedron given by \(\frac{1}{3} \times {\text{Area ABC}} \times {\text{GX}}\)</p>
<p>\( = \frac{1}{3}\left( {\frac{1}{2}\left| {\overrightarrow {{\text{AB}}} \times \overrightarrow {{\text{AC}}} } \right|} \right) \times {\text{GX}} = 12\) <strong><em>(M1)(A1)</em></strong></p>
<p> </p>
<p><strong>Note: </strong>Accept alternative methods, for example the use of a scalar triple product.</p>
<p> </p>
<p>\( = \frac{1}{6}\sqrt {{{( - 6)}^2} + {{( - 6)}^2} + {{( - 6)}^2}} \times \sqrt {{\alpha ^2} + {\alpha ^2} + {\alpha ^2}} = 12\) <strong><em>(A1)</em></strong></p>
<p>\( = \frac{1}{6}6\sqrt 3 |\alpha |\sqrt 3 = 12\)</p>
<p>\( \Rightarrow |\alpha | = 4\) <strong><em>A1</em></strong></p>
<p> </p>
<p><strong>Note: </strong>Condone absence of absolute value.</p>
<p> </p>
<p>this gives us the position of <span class="s1">\(X\)</span> as \(\left( {\begin{array}{*{20}{c}} 2 \\ 4 \\ { - 1} \end{array}} \right) \pm \left( {\begin{array}{*{20}{c}} 4 \\ 4 \\ 4 \end{array}} \right)\)</p>
<p>\({\text{X}}(6,{\text{ }}8,{\text{ }}3)\) or \(( - 2,{\text{ }}0,{\text{ }} - 5)\) <strong><em>A1</em></strong></p>
<p> </p>
<p><strong>Note: </strong>Award <strong><em>A1 </em></strong>for either result.</p>
<p><em><strong>[9 marks]</strong></em></p>
<p><em><strong>Total [23 marks]</strong></em></p>
<div class="question_part_label">d.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">c.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">d.</div>
</div>
<br><hr><br><div class="specification">
<p class="p1">Consider the lines \({l_1}\) and \({l_2}\) defined by</p>
<p class="p1">\({l_1}:\) <em><strong>r</strong></em> \( = \left( {\begin{array}{*{20}{c}} { - 3} \\ { - 2} \\ a \end{array}} \right) + \beta \left( {\begin{array}{*{20}{c}} 1 \\ 4 \\ 2 \end{array}} \right)\) and \({l_2}:\frac{{6 - x}}{3} = \frac{{y - 2}}{4} = 1 - z\) where \(a\) is a constant.</p>
<p class="p1">Given that the lines \({l_1}\) and \({l_2}\) intersect at a point P,</p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p class="p1">find the value of \(a\);</p>
<div class="marks">[4]</div>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p class="p1">determine the coordinates of the point of intersection <span class="s1">P</span><span class="s2">.</span></p>
<div class="marks">[2]</div>
<div class="question_part_label">b.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p class="p1"><strong>METHOD 1</strong></p>
<p class="p1">\({l_1}:\)<strong><em>r</em></strong> \( = \left( {\begin{array}{*{20}{c}} { - 3} \\ { - 2} \\ a \end{array}} \right) = \beta \left( {\begin{array}{*{20}{c}} 1 \\ 4 \\ 2 \end{array}} \right) \Rightarrow \left\{ {\begin{array}{*{20}{l}} {x = - 3 + \beta } \\ {y = - 2 + 4\beta } \\ {z = a + 2\beta } \end{array}} \right.\) <span class="Apple-converted-space"> </span><strong><em>M1</em></strong></p>
<p class="p1"><span class="Apple-converted-space">\(\frac{{6 - ( - 3 + \beta )}}{3} = \frac{{( - 2 + 4\beta ) - 2}}{4} \Rightarrow 4 = \frac{{4\beta }}{3} \Rightarrow \beta = 3\) </span><strong><em>M1A1</em></strong></p>
<p class="p1"><span class="Apple-converted-space">\(\frac{{6 - ( - 3 + \beta )}}{3} = 1 - (a + 2\beta ) \Rightarrow 2 = - 5 - a \Rightarrow a = - 7\) </span><strong><em>A1</em></strong></p>
<p class="p1"><strong>METHOD 2</strong></p>
<p class="p1"><span class="Apple-converted-space">\(\left\{ {\begin{array}{*{20}{l}} { - 3 + \beta = 6 - 3\lambda } \\ { - 2 + 4\beta = 4\lambda + 2} \\ {a + 2\beta = 1 - \lambda } \end{array}} \right.\) </span><strong><em>M1</em></strong></p>
<p class="p1">attempt to solve <span class="Apple-converted-space"> </span><strong><em>M1</em></strong></p>
<p class="p1"><span class="Apple-converted-space">\(\lambda = 2,{\text{ }}\beta = 3\) </span><strong><em>A1</em></strong></p>
<p class="p1"><span class="Apple-converted-space">\(a = 1 - \lambda - 2\beta = - 7\) </span><strong><em>A1</em></strong></p>
<p class="p1"><strong><em>[4 marks]</em></strong></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1"><span class="Apple-converted-space">\(\overrightarrow {{\text{OP}}} = \left( {\begin{array}{*{20}{c}} { - 3} \\ { - 2} \\ { - 7} \end{array}} \right) + 3 \bullet \left( {\begin{array}{*{20}{c}} 1 \\ 4 \\ 2 \end{array}} \right)\) </span><span class="s1"><strong><em>(M1)</em></strong></span></p>
<p class="p1"><span class="Apple-converted-space">\( = \left( {\begin{array}{*{20}{c}} 0 \\ {10} \\ { - 1} \end{array}} \right)\) </span><span class="s1"><strong><em>A1</em></strong></span></p>
<p class="p1">\(\therefore {\text{P}}(0,{\text{ 10, }} - 1)\)</p>
<p class="p2"><strong><em>[2 marks]</em></strong></p>
<div class="question_part_label">b.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">b.</div>
</div>
<br><hr><br><div class="specification">
<p><span style="font-family: 'times new roman', times; font-size: medium;">Consider the plane \({\mathit{\Pi} _1}\), parallel to both lines \({L_1}\) and \({L_2}\). Point C lies in the plane \({\mathit{\Pi} _1}\).</span></p>
</div>
<div class="specification">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">The line \({L_3}\) has vector equation </span><span style="font-family: 'times new roman', times; font-size: medium; background-color: #f7f7f7;"><span style="font-family: 'times new roman', times; font-size: medium;"><span style="line-height: normal;">\(\boldsymbol{r} = \left( \begin{array}{l}3\\0\\1\end{array} \right) + \lambda \left( \begin{array}{c}k\\1\\ - 1\end{array} \right)\).</span></span></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">The plane \({\mathit{\Pi} _2}\) has Cartesian equation \(x + y = 12\).</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">The angle between the line \({L_3}\) and the plane \({\mathit{\Pi} _2}\) is 60°.</span></p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">Given the points A(1, 0, 4), B(2, 3, −1) and C(0, 1, − 2) , </span><span style="font-family: 'times new roman', times; font-size: medium;">find the vector equation of the line \({L_1}\) passing through the points A and B.</span></p>
<div class="marks">[2]</div>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">The line \({L_2}\) has Cartesian equation \(\frac{{x - 1}}{3} = \frac{{y + 2}}{1} = \frac{{z - 1}}{{ - 2}}\).</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">Show that \({L_1}\) and \({L_2}\) are skew lines.</span></p>
<div class="marks">[5]</div>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">Find the Cartesian equation of the plane \({\Pi _1}\).</span></p>
<div class="marks">[4]</div>
<div class="question_part_label">c.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">(i) Find the value of \(k\).</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">(ii) Find the point of intersection P of the line \({L_3}\) and the plane \({\mathit{\Pi} _2}\).</span></p>
<div class="marks">[7]</div>
<div class="question_part_label">d.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">direction vector \(\overrightarrow {{\rm{AB}}} = \left( \begin{array}{c}1\\3\\ - 5\end{array} \right)\) or \(\overrightarrow {{\rm{BA}}} = \left( \begin{array}{c} - 1\\ - 3\\5\end{array} \right)\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(\boldsymbol{r} = \left( \begin{array}{l}1\\0\\4\end{array} \right) + t\left( \begin{array}{c}1\\3\\ - 5\end{array} \right)\) or \(\boldsymbol{r} = \left( \begin{array}{c}2\\3\\ - 1\end{array} \right) + t\left( \begin{array}{c}1\\3\\ - 5\end{array} \right)\) or equivalent <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman'; min-height: 25.0px;"> </p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>Note:</strong> Do not award final <strong><em>A1 </em></strong>unless ‘\(\boldsymbol{r} = {\text{K}}\)’ (or equivalent) seen.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;"> Allow FT on direction vector for final <strong><em>A1</em></strong>.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman'; min-height: 25.0px;"> </p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[2 marks]</em></strong></span></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">both lines expressed in parametric form:</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">\({L_1}\):</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">\(x = 1 + t\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">\(y = 3t\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">\(z = 4 - 5t\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">\({L_2}\):</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">\(x = 1 + 3s\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">\(y = - 2 + s\) <strong><em>M1A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">\(z = - 2s + 1\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman'; min-height: 25.0px;"> </p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>Notes:</strong> Award <strong><em>M1 </em></strong>for an attempt to convert \({L_2}\) from Cartesian to parametric form.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;"> Award <strong><em>A1 </em></strong>for correct parametric equations for \({L_1}\) and \({L_2}\).</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;"> Allow <strong><em>M1A1 </em></strong>at this stage if same parameter is used in both lines.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman'; min-height: 25.0px;"> </p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">attempt to solve simultaneously for <em>x </em>and <em>y</em>: <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">\(1 + t = 1 + 3s\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">\(3t = - 2 + s\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">\(t = - \frac{3}{4},{\text{ }}s = - \frac{1}{4}\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">substituting both values back into <em>z </em>values respectively gives \(z = \frac{{31}}{4}\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">and \(z = \frac{3}{2}\) so a contradiction <strong><em>R1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">therefore \({L_1}\) and \({L_1}\) are skew lines <strong><em>AG</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[5 marks]</em></strong></span></p>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">finding the cross product:</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(\left( \begin{array}{c}1\\3\\ - 5\end{array} \right) \times \left( \begin{array}{c}3\\1\\ - 2\end{array} \right)\) <strong><em>(M1)</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">= – <strong><em>i</em></strong> – 13<strong><em>j</em></strong> – 8<strong><em>k A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica; min-height: 25.0px;"> </p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>Note:</strong> Accept <em><strong>i</strong></em> + 13<em><strong>j</strong></em> + 8<em><strong>k</strong></em><br></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.5px 'Times New Roman'; min-height: 25.0px;"> </p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.5px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">\( - 1(0) - 13(1) - 8( - 2) = 3\) <strong><em>(M1)</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.5px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">\( \Rightarrow - x - 13y - 8z = 3\) or equivalent <em><strong>A1</strong></em></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.5px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[4 marks]</em></strong></span></p>
<div class="question_part_label">c.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: 'times new roman', times; font-size: medium;">(i) \((\cos \theta = )\frac{{\left( \begin{array}{c}k\\1\\ - 1\end{array} \right) \bullet \left( \begin{array}{l}1\\1\\0\end{array} \right)}}{{\sqrt {{k^2} + 1 + 1} \times \sqrt {1 + 1} }}\) <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman'; min-height: 25.0px;"> </p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>Note:</strong> Award <strong><em>M1 </em></strong>for an attempt to use angle between two vectors formula.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman'; min-height: 25.0px;"> </p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">\(\frac{{\sqrt 3 }}{2} = \frac{{k + 1}}{{\sqrt {2({k^2} + 2)} }}\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">obtaining the quadratic equation</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">\(4{(k + 1)^2} = 6({k^2} + 2)\) <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.5px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">\({k^2} - 4k + 4 = 0\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.5px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">\({(k - 2)^2} = 0\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.5px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">\(k = 2\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.5px 'Times New Roman'; min-height: 25.0px;"> </p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>Note:</strong> Award <strong><em>M1A0M1A0 </em></strong>if \(\cos 60^\circ \) is used \((k = 0{\text{ or }}k = - 4)\).</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman'; min-height: 25.0px;"> </p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 22.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(ii) \(r = \left( \begin{array}{l}3\\0\\1\end{array} \right) + \lambda \left( \begin{array}{c}2\\1\\ - 1\end{array} \right)\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">substituting into the equation of the plane \({\Pi _2}\):</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(3 + 2\lambda + \lambda = 12\) <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(\lambda = 3\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">point P has the coordinates:</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(9, 3, –2) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica; min-height: 25.0px;"> </p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>Notes:</strong> Accept 9<em><strong>i</strong></em> + 3<em><strong>j</strong></em> – 2<em><strong>k</strong></em> and \(\left( \begin{array}{l}9\\3\\- 2\end{array} \right)\).</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;"> Do not allow FT if two values found for <em>k</em>.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman'; min-height: 25.0px;"> </p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[7 marks]</em></strong></span></p>
<div class="question_part_label">d.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">c.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">d.</div>
</div>
<br><hr><br><div class="question">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">Given any two non-zero vectors <strong><em>a</em></strong> and <strong><em>b</em></strong> , show that \(|\)<strong><em>a</em></strong> \( \times \) <strong><em>b</em></strong>\({|^2}\) = \(|\)<strong><em>a</em></strong>\({|^2}\)\(|\)<strong><em>b</em></strong>\({|^2}\) – (<strong><em>a</em></strong> \( \cdot \) <strong><em>b</em></strong>)\(^2\).</span></p>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 31.0px Times; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>METHOD 1</strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 31.0px Times; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;">Use of \(|\)<strong><em>a</em></strong> \( \times \) <strong><em>b</em></strong>\(|\) = \(|\)<strong><em>a</em></strong>\(|\)\(|\)<strong><em>b</em></strong>\(|\)\(\sin \theta \) <strong><em>(M1)</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 31.0px Times; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;">\(|\)<strong><em>a</em></strong> \( \times \) <strong><em>b</em></strong>\({|^2}\) = \(|\)<strong><em>a</em></strong>\({|^2}\)\(|\)<strong><em>b</em></strong>\({|^2}\)\({\sin ^2}\theta \) <strong><em>(A1)</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 31.0px Times; color: #3f3f3f;"><strong style="font-family: 'times new roman', times; font-size: medium;">Note:</strong><span style="font-family: 'times new roman', times; font-size: medium;"> Only one of the first two marks can be implied.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 31.0px Times; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;"> </span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 31.0px Times; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;">= \(|\)</span><strong style="font-family: 'times new roman', times; font-size: medium;"><em>a</em></strong><span style="font-family: 'times new roman', times; font-size: medium;">\({|^2}\)\(|\)</span><strong style="font-family: 'times new roman', times; font-size: medium;"><em>b</em></strong><span style="font-family: 'times new roman', times; font-size: medium;">\({|^2}\)\((1 - {\cos ^2}\theta )\) </span><strong style="font-family: 'times new roman', times; font-size: medium;"><em>A1</em></strong></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 31.0px Times; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;">= \(|\)<strong><em>a</em></strong>\({|^2}\)\(|\)<strong><em>b</em></strong>\({|^2}\) – \(|\)<strong><em>a</em></strong>\({|^2}\)\(|\)<strong><em>b</em></strong>\({|^2}\)\({\cos ^2}\theta \) <strong><em>(A1)</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 31.0px Times; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;">= \(|\)<strong><em>a</em></strong>\({|^2}\)\(|\)<strong><em>b</em></strong>\({|^2}\) – \(\left( | \right.\)<strong><em>a</em></strong>\(|\)\(|\)<strong><em>b</em></strong>\(|\)\({\left. {\cos \theta } \right)^2}\) <strong><em>(A1)</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 31.0px Times; color: #3f3f3f;"><strong style="font-family: 'times new roman', times; font-size: medium;">Note:</strong><span style="font-family: 'times new roman', times; font-size: medium;"> Only one of the above two </span><strong style="font-family: 'times new roman', times; font-size: medium;"><em>A1</em></strong><span style="font-family: 'times new roman', times; font-size: medium;"> marks can be implied.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 31.0px Times; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;"> </span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 31.0px Times; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;">= \(|\)</span><strong style="font-family: 'times new roman', times; font-size: medium;"><em>a</em></strong><span style="font-family: 'times new roman', times; font-size: medium;">\({|^2}\)\(|\)</span><strong style="font-family: 'times new roman', times; font-size: medium;"><em>b</em></strong><span style="font-family: 'times new roman', times; font-size: medium;">\({|^2}\) – (</span><strong style="font-family: 'times new roman', times; font-size: medium;"><em>a</em></strong><span style="font-family: 'times new roman', times; font-size: medium;"> \( \cdot \) </span><strong style="font-family: 'times new roman', times; font-size: medium;"><em>b</em></strong><span style="font-family: 'times new roman', times; font-size: medium;">)\(^2\) </span><strong style="font-family: 'times new roman', times; font-size: medium;"><em>A1</em></strong></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 31.0px Times; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;">Hence LHS = RHS <strong><em>AG</em></strong> <strong><em>N0</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 31.0px Times; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[6 marks]</em></strong></span><strong style="font-family: 'times new roman', times; font-size: medium;"><em> </em></strong></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 31.0px Times; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>METHOD 2</strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 31.0px Times; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;">Use of <strong><em>a</em></strong> \( \cdot \) <strong><em>b </em></strong>= \(|\)<strong><em>a</em></strong>\(|\)\(|\)<strong><em>b</em></strong>\(|\)\(\cos \theta \) <strong><em>(M1)</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 31.0px Times; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;">\(|\)<strong><em>a</em></strong>\({|^2}\)\(|\)<strong><em>b</em></strong>\({|^2}\) – (<strong><em>a</em></strong> \( \cdot \) <strong><em>b</em></strong>)\(^2\) = \(|\)<strong><em>a</em></strong>\({|^2}\)\(|\)<strong><em>b</em></strong>\({|^2}\) – \(\left( | \right.\)<strong><em>a</em></strong>\(|\)\(|\)<strong><em>b</em></strong>\(|\)\({\left. {\cos \theta } \right)^2}\) <strong><em>(A1)</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 31.0px Times; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;">= \(|\)<strong><em>a</em></strong>\({|^2}\)\(|\)<strong><em>b</em></strong>\({|^2}\) – \(|\)<strong><em>a</em></strong>\({|^2}\)\(|\)<strong><em>b</em></strong>\({|^2}\) \({\cos ^2}\theta \) <strong><em>(A1)</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 31.0px Times; color: #3f3f3f;"><strong style="font-family: 'times new roman', times; font-size: medium;">Note:</strong><span style="font-family: 'times new roman', times; font-size: medium;"> Only one of the above two </span><strong style="font-family: 'times new roman', times; font-size: medium;"><em>A1</em></strong><span style="font-family: 'times new roman', times; font-size: medium;"> marks can be implied.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 31.0px Times; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;"> </span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 31.0px Times; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;">= \(|\)</span><strong style="font-family: 'times new roman', times; font-size: medium;"><em>a</em></strong><span style="font-family: 'times new roman', times; font-size: medium;">\({|^2}\)\(|\)</span><strong style="font-family: 'times new roman', times; font-size: medium;"><em>b</em></strong><span style="font-family: 'times new roman', times; font-size: medium;">\({|^2}\)\((1 - {\cos ^2}\theta )\) </span><strong style="font-family: 'times new roman', times; font-size: medium;"><em>A1</em></strong></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 31.0px Times; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;">= \(|\)<strong><em>a</em></strong>\({|^2}\)\(|\)<strong><em>b</em></strong>\({|^2}\)\({\sin ^2}\theta \) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 31.0px Times; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;">= \(|\)<strong><em>a</em></strong> \( \times \) <strong><em>b</em></strong>\({|^2}\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 31.0px Times; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;">Hence LHS = RHS <strong><em>AG</em></strong> <strong><em>N0</em></strong></span><span style="font-family: 'Helvetica Neue', Arial, 'Lucida Grande', 'Lucida Sans Unicode', sans-serif; font-size: 31px;"> </span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 31.0px Times; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>Notes:</strong> Candidates who independently correctly simplify both sides and show that LHS = RHS should be awarded full marks.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 31.0px Times; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;">If the candidate starts off with expression that they are trying to prove and concludes that \({\sin ^2}\theta = (1 - {\cos ^2}\theta )\) award <strong><em>M1A1A1A1A0A0</em></strong>.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 31.0px Times; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;">If the candidate uses two general 3D vectors and explicitly finds the expressions correctly award full marks. Use of 2D vectors gains a maximum of 2 marks.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 31.0px Times; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;">If two specific vectors are used no marks are gained.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 31.0px Times; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;"> </span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 31.0px Times; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[6 marks]</em></strong></span></p>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">Those candidates who chose to use the trigonometric version of Pythagoras’ Theorem were generally successful, although a minority were unconvincing in their reasoning. Some candidates adopted a full component approach, but often seemed to lose track of what they were trying to prove. A few candidates used 2-dimensional vectors or specific rather than general vectors.</span></p>
</div>
<br><hr><br><div class="specification">
<p>The points A and B are given by \({\text{A}}(0,{\text{ }}3,{\text{ }} - 6)\) and \({\text{B}}(6,{\text{ }} - 5,{\text{ }}11)\).</p>
<p>The plane <em>Π</em> is defined by the equation \(4x - 3y + 2z = 20\).</p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Find a vector equation of the line <em>L </em>passing through the points A and B.</p>
<div class="marks">[3]</div>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Find the coordinates of the point of intersection of the line <em>L </em>with the plane <em>Π</em>.</p>
<div class="marks">[3]</div>
<div class="question_part_label">b.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p>\(\overrightarrow {{\text{AB}}} = \left( {\begin{array}{*{20}{c}} 6 \\ { - 8} \\ {17} \end{array}} \right)\) <strong><em>(A1)</em></strong></p>
<p> </p>
<p><strong><em>r </em></strong>= \(\left( {\begin{array}{*{20}{c}} 0 \\ 3 \\ { - 6} \end{array}} \right) + \lambda \left( {\begin{array}{*{20}{c}} 6 \\ { - 8} \\ {17} \end{array}} \right)\) or <strong><em>r</em></strong> = \(\left( {\begin{array}{*{20}{c}} 6 \\ { - 5} \\ {11} \end{array}} \right) + \lambda \left( {\begin{array}{*{20}{c}} 6 \\ { - 8} \\ {17} \end{array}} \right)\) <strong><em>M1A1</em></strong></p>
<p> </p>
<p><strong>Note:</strong> Award <strong><em>M1A0 </em></strong>if <strong><em>r </em></strong>= is not seen (or equivalent).</p>
<p> </p>
<p><strong><em>[3 marks]</em></strong></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p>substitute line <em>L </em>in \(\Pi :4(6\lambda ) - 3(3 - 8\lambda ) + 2( - 6 + 17\lambda ) = 20\) <strong><em>M1</em></strong></p>
<p>\(82\lambda = 41\)</p>
<p>\(\lambda = \frac{1}{2}\) <strong><em>(A1)</em></strong></p>
<p> </p>
<p><strong><em>r</em></strong> = \(\left( {\begin{array}{*{20}{c}} 0 \\ 3 \\ { - 6} \end{array}} \right) + \frac{1}{2}\left( {\begin{array}{*{20}{c}} 6 \\ { - 8} \\ {17} \end{array}} \right) = \left( {\begin{array}{*{20}{c}} 3 \\ { - 1} \\ {\frac{5}{2}} \end{array}} \right)\)</p>
<p>so coordinate is \(\left( {3,{\text{ }} - 1,{\text{ }}\frac{5}{2}} \right)\) <strong><em>A1</em></strong></p>
<p> </p>
<p><strong>Note:</strong> Accept coordinate expressed as position vector \(\left( {\begin{array}{*{20}{c}} 3 \\ { - 1} \\ {\frac{5}{2}} \end{array}} \right)\).</p>
<p> </p>
<p><strong><em>[3 marks]</em></strong></p>
<div class="question_part_label">b.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">b.</div>
</div>
<br><hr><br><div class="question">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 31.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">The three vectors \(\boldsymbol{a}\), \(\boldsymbol{b}\) and \(\boldsymbol{c}\) are given by\[{\boldsymbol{a}} = \left( {\begin{array}{*{20}{c}}<br> {2y} \\ <br> { - 3x} \\ <br> {2x} <br>\end{array}} \right),{\text{ }}{\boldsymbol{b}}{\text{ }} = \left( {\begin{array}{*{20}{c}}<br> {4x} \\ <br> y \\ <br> {3 - x} <br>\end{array}} \right),{\text{ }}{\boldsymbol{c}}{\text{ }} = \left( {\begin{array}{*{20}{c}}<br> 4 \\ <br> { - 7} \\ <br> 6 <br>\end{array}} \right){\text{ where }}x,y \in \mathbb{R}{\text{ }}{\text{.}}\]</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 31.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(a) If <strong><em>a</em></strong> + 2<strong><em>b</em></strong> − <strong><em>c</em></strong> = 0, find the value of <em>x</em> and of <em>y</em>.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 31.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(b) Find the exact value of \(|\)<strong><em>a</em></strong> + 2<strong><em>b</em></strong>\(|\).</span></p>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 29.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(a) \(2y + 8x = 4\) <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 29.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( - 3x + 2y = - 7\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 29.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(2x + 6 - 2x = 6\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 29.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>Note:</strong> Award <strong><em>M1</em></strong> for attempt at components, <strong><em>A1</em></strong> for two correct equations.</span></p>
<p style="margin: 0px 0px 0px 30px; font: 29px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"> No penalty for not checking the third equation.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 29.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"> </span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 29.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">solving : <em>x</em> = 1, <em>y</em> = –2 <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 29.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"> </span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 29.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(b) \(|\)<strong>a</strong> + 2<em><strong>b</strong></em>\(| = \left| {\left( {\begin{array}{*{20}{c}}<br> { - 4} \\ <br> { - 3} \\ <br> 2 <br>\end{array}} \right) + 2\left( {\begin{array}{*{20}{c}}<br> 4 \\ <br> { - 2} \\ <br> 2 <br>\end{array}} \right)} \right|\)<br></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 29.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = \left| {\left( {\begin{array}{*{20}{c}}<br> 4 \\ <br> { - 7} \\ <br> 6 <br>\end{array}} \right)} \right|\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 29.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( \Rightarrow |\)<strong><em>a</em></strong> + 2<strong><em>b</em></strong>\(|\) \( = \sqrt {{4^2} + {{( - 7)}^2} + {6^2}} \) <strong><em>(M1)</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 29.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = \sqrt {101} \) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 29.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em> </em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 29.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[5 marks]</em></strong></span></p>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 24.0px Helvetica;"> </p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">The majority of candidates understood what was required in part (a) of this question and gained the correct answer. Most candidates were able to do part (b) but few realised that they did not have to calculate \(|\)<strong><em>a</em></strong> + 2<strong><em>b</em></strong>\(|\) as this is \(|\)<strong><em>c</em></strong>\(|\). Many candidates lost time on this question.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 24.0px Helvetica;"> </p>
</div>
<br><hr><br><div class="specification">
<p><span style="font-family: times new roman,times; font-size: medium;">The diagram below shows a circle with centre O. The points A, B, C lie on the </span><span style="font-family: times new roman,times; font-size: medium;">circumference of the circle and [AC] is a diameter.</span></p>
<p><br><span style="font-family: times new roman,times; font-size: medium;"><img style="display: block; margin-left: auto; margin-right: auto;" src="data:image/png;base64,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" alt></span><br><span style="font-family: times new roman,times; font-size: medium;">Let \(\overrightarrow {{\text{OA}}} = {\boldsymbol{a}}\) and \(\overrightarrow {{\text{OB}}} = {\boldsymbol{b}}\) .</span></p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">Write down expressions for \(\overrightarrow {{\text{AB}}} \) </span><span style="font-family: times new roman,times; font-size: medium;">and \(\overrightarrow {{\text{CB}}} \) </span><span style="font-family: times new roman,times; font-size: medium;">in terms of the vectors \({\boldsymbol{a}}\) and \({\boldsymbol{b}}\) .</span></p>
<div class="marks">[2]</div>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">Hence prove that angle \({\text{A}}\hat {\rm{B}}{\text{C}}\) is a right angle.</span></p>
<div class="marks">[3]</div>
<div class="question_part_label">b.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">\(\overrightarrow {{\text{AB}}} = {\boldsymbol{b}} - {\boldsymbol{a}}\) <em><strong>A1</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(\overrightarrow {{\text{CB}}} = {\boldsymbol{a}} + {\boldsymbol{b}}\) <em><strong>A1</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"><em><strong>[2 marks]<br></strong></em></span></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">\(\overrightarrow {{\text{AB}}} \cdot \overrightarrow {{\text{CB}}} = \left( {{\boldsymbol{b}} - {\boldsymbol{a}}} \right) \cdot \left( {{\boldsymbol{b}} + {\boldsymbol{a}}} \right)\) </span><em><strong><span style="font-family: times new roman,times; font-size: medium;"> M1</span></strong></em><br><span style="font-family: times new roman,times; font-size: medium;">\( = {\left| {\mathbf{b}} \right|^2} - {\left| {\mathbf{a}} \right|^2}\) </span><em><strong><span style="font-family: times new roman,times; font-size: medium;"> A1</span></strong></em><br><span style="font-family: times new roman,times; font-size: medium;">\( = 0\) since \(\left| {\boldsymbol{b}} \right| = \left| {\boldsymbol{a}} \right|\) <em><strong>R1</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"><strong>Note:</strong> Only award the <em><strong>A1</strong></em> and <em><strong>R1</strong></em> if working indicates that they understand </span><span style="font-family: times new roman,times; font-size: medium;">that they are working with vectors.</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"> </span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">so \(\overrightarrow {{\text{AB}}} \) </span><span style="font-family: times new roman,times; font-size: medium;">is perpendicular to \(\overrightarrow {{\text{CB}}} \) </span><span style="font-family: times new roman,times; font-size: medium;"><em>i.e.</em> \({\text{A}}\hat {\rm{B}}{\text{C}}\) </span><span style="font-family: times new roman,times; font-size: medium;">is a right angle <em><strong>AG</strong></em></span></p>
<p><em><strong><span style="font-family: times new roman,times; font-size: medium;">[3 marks]</span></strong></em></p>
<div class="question_part_label">b.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">This question was poorly done with most candidates having difficulties in using appropriate notation which made unclear the distinction between scalars and vectors. A few candidates scored at least one of the marks in (a) but most candidates had problems in setting up the proof required in (b) with many using a circular argument which resulted in a very poor performance in this part.</span></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">This question was poorly done with most candidates having difficulties in using appropriate notation which made unclear the distinction between scalars and vectors. A few candidates scored at least one of the marks in (a) but most candidates had problems in setting up the proof required in (b) with many using a circular argument which resulted in a very poor performance in this part.</span></p>
<div class="question_part_label">b.</div>
</div>
<br><hr><br><div class="specification">
<p class="p1">Two planes have equations</p>
<p class="p1">\[{\Pi _1}:{\text{ }}4x + y + z = 8{\text{ and }}{\Pi _2}:{\text{ }}4x + 3y - z = 0\]</p>
</div>
<div class="specification">
<p class="p1">Let \(L\) be the line of intersection of the two planes.</p>
</div>
<div class="specification">
<p class="p1"><span class="s1">B </span>is the point on \({\Pi _1}\) <span class="s1">with coordinates \((a,{\text{ }}b,{\text{ }}1)\).</span></p>
</div>
<div class="specification">
<p class="p1">The point P <span class="s1">lies on \(L\) </span>and \({\rm{A\hat BP}} = 45^\circ \).</p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p class="p1">Find the cosine of the angle between the two planes in the form \(\sqrt {\frac{p}{q}} \) where \(p,{\text{ }}q \in \mathbb{Z}\).</p>
<div class="marks">[4]</div>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p class="p1">(i) Show that \(L\) <span class="s1">has direction \(\left( {\begin{array}{*{20}{c}} { - 1} \\ 2 \\ 2 \end{array}} \right)\).</span></p>
<p class="p4">(ii) <span class="Apple-converted-space"> </span>Show that the point \({\text{A }}(1,{\text{ }}0,{\text{ }}4)\) <span class="s2">lies on both planes.</span></p>
<p class="p3">(iii) <span class="Apple-converted-space"> </span>Write down a vector equation of \(L\).</p>
<div class="marks">[6]</div>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p class="p1"><span class="s1">Given the vector \(\overrightarrow {{\text{AB}}} \) </span>is perpendicular to \(L\) find the value of \(a\) and the value of \(b\).</p>
<div class="marks">[5]</div>
<div class="question_part_label">c.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p class="p1">Show that \({\text{AB}} = 3\sqrt 2 \).</p>
<div class="marks">[1]</div>
<div class="question_part_label">d.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p class="p1">Find the coordinates of the two possible positions of \(P\).</p>
<div class="marks">[5]</div>
<div class="question_part_label">e.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p class="p1"><strong>Note: <span class="Apple-converted-space"> </span></strong>Throughout the question condone vectors written horizontally.</p>
<p class="p1">angle between planes is equal to the angles between the normal to the planes <span class="Apple-converted-space"> </span><strong><em>(M1)</em></strong></p>
<p class="p3"><span class="Apple-converted-space">\(\left( {\begin{array}{*{20}{c}} 4 \\ 1 \\ 1 \end{array}} \right) \bullet \left( {\begin{array}{*{20}{c}} 4 \\ 3 \\ { - 1} \end{array}} \right) = 18\) </span><span class="s1"><strong><em>(A1)</em></strong></span></p>
<p class="p1">let \(\theta \) be the angle between the normal to the planes</p>
<p class="p1"><span class="Apple-converted-space">\(\cos \theta = \frac{{18}}{{\sqrt {18} \sqrt {26} }} = \sqrt {\frac{{18}}{{26}}} {\text{ }}\left( {{\text{or equivalent, for example }}\sqrt {\frac{{324}}{{468}}} {\text{ or }}\sqrt {\frac{9}{{13}}} } \right)\) </span><strong><em>M1A1</em></strong></p>
<p class="p1"><strong><em>[4 marks]</em></strong></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1"><strong>Note: <span class="Apple-converted-space"> </span></strong>Throughout the question condone vectors written horizontally.</p>
<p class="p3">(i) <span class="Apple-converted-space"> </span><strong>METHOD 1</strong></p>
<p class="p4"><span class="Apple-converted-space">\(\left( {\begin{array}{*{20}{c}} 4 \\ 1 \\ 1 \end{array}} \right) \times \left( {\begin{array}{*{20}{c}} 4 \\ 3 \\ { - 1} \end{array}} \right) = \left( {\begin{array}{*{20}{c}} { - 4} \\ 8 \\ 8 \end{array}} \right)\) </span><span class="s1"><strong><em>M1A1</em></strong></span></p>
<p class="p4">which is a multiple of \(\left( {\begin{array}{*{20}{c}} { - 1} \\ 2 \\ 2 \end{array}} \right)\) <span class="Apple-converted-space"> </span><span class="s1"><strong><em>R1AG</em></strong></span></p>
<p class="p3"><strong>Note: <span class="Apple-converted-space"> </span></strong>Allow any equivalent wording or \(\left( {\begin{array}{*{20}{c}} { - 4} \\ 8 \\ 8 \end{array}} \right) = 4\left( {\begin{array}{*{20}{c}} { - 1} \\ 2 \\ 2 \end{array}} \right)\), do not allow \(\left( {\begin{array}{*{20}{c}} { - 4} \\ 8 \\ 8 \end{array}} \right) = \left( {\begin{array}{*{20}{c}} { - 1} \\ 2 \\ 2 \end{array}} \right)\).</p>
<p class="p3"><strong>METHOD 2</strong></p>
<p class="p3">let \(z = t\) (or equivalent)</p>
<p class="p3">solve simultaneously to get <span class="Apple-converted-space"> </span><strong><em>M1</em></strong></p>
<p class="p3"><span class="Apple-converted-space">\(y = t - 4,{\text{ }}x = 3 - 0.5t\) </span><strong><em>A1</em></strong></p>
<p class="p4">hence direction vector is \(\left( {\begin{array}{*{20}{c}} { - 0.5} \\ 1 \\ 1 \end{array}} \right)\)</p>
<p class="p4">which is a multiple of \(\left( {\begin{array}{*{20}{c}} { - 1} \\ 2 \\ 2 \end{array}} \right)\) <span class="Apple-converted-space"> </span><span class="s1"><strong><em>R1AG</em></strong></span></p>
<p class="p3"><strong>METHOD 3</strong></p>
<p class="p4"><span class="Apple-converted-space">\(\left( {\begin{array}{*{20}{c}} 4 \\ 1 \\ 1 \end{array}} \right) \bullet \left( {\begin{array}{*{20}{c}} { - 1} \\ 2 \\ 2 \end{array}} \right) = - 4 + 2 + 2 = 0\) </span><span class="s1"><strong><em>M1A1</em></strong></span></p>
<p class="p4"><span class="Apple-converted-space">\(\left( {\begin{array}{*{20}{c}} 4 \\ 3 \\ { - 1} \end{array}} \right) \bullet \left( {\begin{array}{*{20}{c}} { - 1} \\ 2 \\ 2 \end{array}} \right) = - 4 + 6 - 2 = 0\) </span><span class="s1"><strong><em>A1</em></strong></span></p>
<p class="p3"><strong>Note: <span class="Apple-converted-space"> </span></strong>If only one scalar product is found award <strong><em>M0A0A0</em></strong>.</p>
<p class="p3">(ii) <span class="Apple-converted-space"> \({\Pi _1}:{\text{ }}4 + 0 + 4 = 8\)</span> and \({\Pi _2}:{\text{ }}4 + 0 - 4 = 0\) <span class="Apple-converted-space"> </span><strong><em>R1</em></strong></p>
<p class="p6">(iii) <span class="Apple-converted-space"> </span><span class="s2"><strong><em>r</em></strong></span> \( = \left( {\begin{array}{*{20}{c}} 1 \\ 0 \\ 4 \end{array}} \right) + \lambda \left( {\begin{array}{*{20}{c}} { - 1} \\ 2 \\ 2 \end{array}} \right)\) <span class="Apple-converted-space"> </span><span class="s1"><strong><em>A1A1</em></strong></span></p>
<p class="p3"><strong>Note: <span class="Apple-converted-space"> </span><em>A1 </em></strong>for “<span class="s2"><strong><em>r </em></strong>\( = \)</span>” and a correct point on the line, <strong><em>A1 </em></strong>for a parameter and a correct direction vector.</p>
<p class="p3"><strong><em>[6 marks]</em></strong></p>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1"><strong>Note: <span class="Apple-converted-space"> </span></strong>Throughout the question condone vectors written horizontally.</p>
<p class="p2"> </p>
<p class="p3"><span class="Apple-converted-space">\(\overrightarrow {{\text{AB}}} = \left( {\begin{array}{*{20}{c}} a \\ b \\ 1 \end{array}} \right) - \left( {\begin{array}{*{20}{c}} 1 \\ 0 \\ 4 \end{array}} \right) = \left( {\begin{array}{*{20}{c}} {a - 1} \\ b \\ { - 3} \end{array}} \right)\) </span><span class="s1"><strong><em>(A1)</em></strong></span></p>
<p class="p3"><span class="Apple-converted-space">\(\left( {\begin{array}{*{20}{c}} {a - 1} \\ b \\ { - 3} \end{array}} \right) \bullet \left( {\begin{array}{*{20}{c}} { - 1} \\ 2 \\ 2 \end{array}} \right) = 0\) </span><span class="s1"><strong><em>M1</em></strong></span></p>
<p class="p4"> </p>
<p class="p5"><strong>Note: <span class="Apple-converted-space"> </span></strong>Award <strong><em>M0 </em></strong><span class="s2">for \(\left( {\begin{array}{*{20}{c}} a \\ b \\ 1 \end{array}} \right) \bullet \left( {\begin{array}{*{20}{c}} { - 1} \\ 2 \\ 2 \end{array}} \right) = 0\).</span></p>
<p class="p2"> </p>
<p class="p3"><span class="Apple-converted-space">\( - a + 1 + 2b - 6 = 0 \Rightarrow a - 2b = - 5\) </span><span class="s1"><strong><em>A1</em></strong></span></p>
<p class="p5">lies on \({\Pi _1}\) so \(4a + b + 1 = 8 \Rightarrow 4a + b = 7\) <span class="Apple-converted-space"> </span><strong><em>M1</em></strong></p>
<p class="p3"><span class="Apple-converted-space">\(a = 1,{\text{ }}b = 3\) </span><span class="s1"><strong><em>A1</em></strong></span></p>
<p class="p5"><strong><em>[5 marks]</em></strong></p>
<div class="question_part_label">c.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1"><strong>Note: <span class="Apple-converted-space"> </span></strong>Throughout the question condone vectors written horizontally.</p>
<p class="p2"> </p>
<p class="p3"><span class="Apple-converted-space">\({\text{AB}} = \sqrt {{0^2} + {3^2} + {{( - 3)}^2}} = 3\sqrt 2 \) </span><span class="s1"><strong><em>M1AG</em></strong></span></p>
<p class="p4"><strong><em>[1 mark]</em></strong></p>
<div class="question_part_label">d.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1"><strong>Note: <span class="Apple-converted-space"> </span></strong>Throughout the question condone vectors written horizontally.</p>
<p class="p3"><strong>METHOD 1</strong></p>
<p class="p4"><span class="Apple-converted-space">\(\left| {\overrightarrow {{\text{AB}}} } \right| = \left| {\overrightarrow {{\text{AP}}} } \right| = 3\sqrt 2 \) </span><span class="s1"><strong><em>(M1)</em></strong></span></p>
<p class="p4"><span class="Apple-converted-space">\(\overrightarrow {{\text{AP}}} = t\left( {\begin{array}{*{20}{c}} { - 1} \\ 2 \\ 2 \end{array}} \right)\) </span><span class="s1"><strong><em>(A1)</em></strong></span></p>
<p class="p4"><span class="Apple-converted-space">\(\left| {3t} \right| = 3\sqrt 2 \Rightarrow t = \pm \sqrt 2 \) </span><span class="s1"><strong><em>(M1)A1</em></strong></span></p>
<p class="p3"><span class="s2">\({\text{P}}\left( {1 - \sqrt 2 ,{\text{ }}2\sqrt 2 ,{\text{ }}4 + 2\sqrt 2 } \right)\) </span>and \(\left( {1 + \sqrt 2 ,{\text{ }} - 2\sqrt 2 ,{\text{ }}4 - 2\sqrt 2 } \right)\) <span class="Apple-converted-space"> </span><strong><em>A1</em></strong></p>
<p class="p3"><strong><em>[5 marks]</em></strong></p>
<p class="p3"><strong>METHOD 2</strong></p>
<p class="p3"><span class="s2">let P </span>have coordinates \((1 - \lambda ,{\text{ }}2\lambda ,{\text{ }}4 + 2\lambda )\) <span class="Apple-converted-space"> </span><strong><em>M1</em></strong></p>
<p class="p4"><span class="Apple-converted-space">\(\overrightarrow {{\text{BA}}} = \left( {\begin{array}{*{20}{c}} 0 \\ { - 3} \\ 3 \end{array}} \right),{\text{ }}\overrightarrow {{\text{BP}}} = \left( {\begin{array}{*{20}{c}} { - \lambda } \\ {2\lambda - 3} \\ {3 + 2\lambda } \end{array}} \right)\) </span><span class="s1"><strong><em>A1</em></strong></span></p>
<p class="p4"><span class="Apple-converted-space">\(\cos 45^\circ = \frac{{\overrightarrow {{\text{BA}}} \bullet \overrightarrow {{\text{BP}}} }}{{\left| {{\text{BA}}} \right|\left| {{\text{BP}}} \right|}}\) </span><span class="s1"><strong><em>M1</em></strong></span></p>
<p class="p4"><span class="s1"><strong>Note: <span class="Apple-converted-space"> </span></strong>Award <strong><em>M1 </em></strong></span>even if AB rather than BA is used in the scalar product.</p>
<p class="p4">\(\overrightarrow {{\text{BA}}} \bullet \overrightarrow {{\text{BP}}} = 18\)</p>
<p class="p4">\(\frac{1}{{\sqrt 2 }} = \frac{{18}}{{\sqrt {18} \sqrt {9{\lambda ^2} + 18} }}\)</p>
<p class="p4"><span class="Apple-converted-space">\(\lambda = \pm \sqrt 2 \) </span><span class="s1"><strong><em>A1</em></strong></span></p>
<p class="p3"><span class="s2">\({\text{P}}\left( {1 - \sqrt 2 ,{\text{ }}2\sqrt 2 ,{\text{ }}4 + 2\sqrt 2 } \right)\) </span>and \(\left( {1 + \sqrt 2 ,{\text{ }} - 2\sqrt 2 ,{\text{ }}4 - 2\sqrt 2 } \right)\) <span class="Apple-converted-space"> </span><strong><em>A1</em></strong></p>
<p class="p3"><strong>Note: <span class="Apple-converted-space"> </span></strong>Accept answers given as position vectors.</p>
<p class="p3"><strong><em>[5 marks]</em></strong></p>
<div class="question_part_label">e.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
<p class="p1">This was successfully done, though some candidates lost marks unnecessarily by not giving the answer in the form requested in the question.</p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">(b)(i) A variety of techniques were successfully used here. A common error was not to justify why the vector obtained from the vector product was in the same direction as the one given in the question.</p>
<p class="p1">(b)(ii) and (iii) These were well done, though too many candidates still lose a mark unnecessarily by writing the vector equation of a line as \(l = \) rather than \(r = \).</p>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">Weaker candidates found the rest of this question more difficult. Though most obtained one equation for \(a\) and \(b\) they did not take note of the fact that it was also on the given plane, which gave the second equation.</p>
<div class="question_part_label">c.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">This was done successfully by the majority of candidates. Candidates need to be aware that the notation AB means the length of the line segment joining the points A and B (as in the course guide).</p>
<div class="question_part_label">d.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">This proved to be a difficult question for most candidates. Those who were successful were equally split between the two approaches given in the markscheme.</p>
<div class="question_part_label">e.</div>
</div>
<br><hr><br><div class="specification">
<p>In the following diagram, \(\overrightarrow {{\text{OA}}} \) = <strong><em>a</em></strong>, \(\overrightarrow {{\text{OB}}} \) = <strong><em>b</em></strong>. C is the midpoint of [OA] and \(\overrightarrow {{\text{OF}}} = \frac{1}{6}\overrightarrow {{\text{FB}}} \).</p>
<p style="text-align: center;"><img src="images/Schermafbeelding_2018-02-07_om_14.26.10.png" alt="N17/5/MATHL/HP1/ENG/TZ0/09"></p>
</div>
<div class="specification">
<p>It is given also that \(\overrightarrow {{\text{AD}}} = \lambda \overrightarrow {{\text{AF}}} \) and \(\overrightarrow {{\text{CD}}} = \mu \overrightarrow {{\text{CB}}} \), where \(\lambda ,{\text{ }}\mu \in \mathbb{R}\).</p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Find, in terms of <strong><em>a </em></strong>and <strong><em>b </em></strong>\(\overrightarrow {{\text{OF}}} \).</p>
<div class="marks">[1]</div>
<div class="question_part_label">a.i.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Find, in terms of <strong><em>a </em></strong>and <strong><em>b </em></strong>\(\overrightarrow {{\text{AF}}} \).</p>
<div class="marks">[2]</div>
<div class="question_part_label">a.ii.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Find an expression for \(\overrightarrow {{\text{OD}}} \) in terms of <strong><em>a</em></strong>, <strong><em>b </em></strong>and \(\lambda \);</p>
<div class="marks">[2]</div>
<div class="question_part_label">b.i.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Find an expression for \(\overrightarrow {{\text{OD}}} \) in terms of <strong><em>a</em></strong>, <strong><em>b </em></strong>and \(\mu \).</p>
<div class="marks">[2]</div>
<div class="question_part_label">b.ii.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Show that \(\mu = \frac{1}{{13}}\), and find the value of \(\lambda \).</p>
<div class="marks">[4]</div>
<div class="question_part_label">c.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Deduce an expression for \(\overrightarrow {{\text{CD}}} \) in terms of <strong><em>a </em></strong>and <strong><em>b </em></strong>only.</p>
<div class="marks">[2]</div>
<div class="question_part_label">d.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Given that area \(\Delta {\text{OAB}} = k({\text{area }}\Delta {\text{CAD}})\), find the value of \(k\).</p>
<div class="marks">[5]</div>
<div class="question_part_label">e.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p>\(\overrightarrow {{\text{OF}}} = \frac{1}{7}\)<strong><em>b</em></strong> <strong><em>A1</em></strong></p>
<p><strong><em>[1 mark]</em></strong></p>
<div class="question_part_label">a.i.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p>\(\overrightarrow {{\text{AF}}} = \overrightarrow {{\text{OF}}} - \overrightarrow {{\text{OA}}} \) <strong><em>(M1)</em></strong></p>
<p>\( = \frac{1}{7}\)<strong><em>b</em></strong> – <strong><em>a </em></strong><strong><em>A1</em></strong></p>
<p><strong><em>[2 marks]</em></strong></p>
<div class="question_part_label">a.ii.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p>\(\overrightarrow {{\text{OD}}} = \) <strong><em>a</em></strong> \( + \lambda \left( {\frac{1}{7}b -a} \right){\text{ }}\left( { = (1 - \lambda )a + \frac{\lambda }{7}b} \right)\) <strong><em>M1A1</em></strong></p>
<p><strong><em>[2 marks]</em></strong></p>
<div class="question_part_label">b.i.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p>\(\overrightarrow {{\text{OD}}} = \frac{1}{2}\) <strong><em>a</em></strong> \( + \mu \left( { - \frac{1}{2}a + b} \right){\text{ }}\left( { = \left( {\frac{1}{2} - \frac{\mu }{2}} \right)a + \mu b} \right)\) <strong><em>M1A1</em></strong></p>
<p><strong><em>[2 marks]</em></strong></p>
<div class="question_part_label">b.ii.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p>equating coefficients: <strong><em>M1</em></strong></p>
<p>\(\frac{\lambda }{7} = \mu ,{\text{ }}1 - \lambda = \frac{{1 - \mu }}{2}\) <strong><em>A1</em></strong></p>
<p>solving simultaneously: <strong><em>M1</em></strong></p>
<p>\(\lambda = \frac{7}{{13}},{\text{ }}\mu = \frac{1}{{13}}\) <strong><em>A1AG</em></strong></p>
<p><strong><em>[4 marks]</em></strong></p>
<div class="question_part_label">c.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p>\(\overrightarrow {{\text{CD}}} = \frac{1}{{13}}\overrightarrow {{\text{CB}}} \)</p>
<p>\( = \frac{1}{{13}}\left( {b - \frac{1}{2}a} \right){\text{ }}\left( { = - \frac{1}{{26}}a + \frac{1}{{13}}b} \right)\) <strong><em>M1A1</em></strong></p>
<p><strong><em>[2 marks]</em></strong></p>
<div class="question_part_label">d.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><strong>METHOD 1</strong></p>
<p>\({\text{area }}\Delta {\text{ACD}} = \frac{1}{2}{\text{CD}} \times {\text{AC}} \times \sin {\rm{A\hat CB}}\) <strong><em>(M1)</em></strong></p>
<p>\({\text{area }}\Delta {\text{ACB}} = \frac{1}{2}{\text{CB}} \times {\text{AC}} \times \sin {\rm{A\hat CB}}\) <strong><em>(M1)</em></strong></p>
<p>\({\text{ratio }}\frac{{{\text{area }}\Delta {\text{ACD}}}}{{{\text{area }}\Delta {\text{ACB}}}} = \frac{{{\text{CD}}}}{{{\text{CB}}}} = \frac{1}{{13}}\) <strong><em>A1</em></strong></p>
<p>\(k = \frac{{{\text{area }}\Delta {\text{OAB}}}}{{{\text{area }}\Delta {\text{CAD}}}} = \frac{{13}}{{{\text{area }}\Delta {\text{CAB}}}} \times {\text{area }}\Delta {\text{OAB}}\) <strong><em>(M1)</em></strong></p>
<p>\( = 13 \times 2 = 26\) <strong><em>A1</em></strong></p>
<p> </p>
<p><strong>METHOD 2</strong></p>
<p>\({\text{area }}\Delta {\text{OAB}} = \frac{1}{2}\left| {a \times b} \right|\) <strong><em>A1</em></strong></p>
<p>\({\text{area }}\Delta {\text{CAD}} = \frac{1}{2}\left| {\overrightarrow {{\text{CA}}} \times \overrightarrow {{\text{CD}}} } \right|\) or \(\frac{1}{2}\left| {\overrightarrow {{\text{CA}}} \times \overrightarrow {{\text{AD}}} } \right|\) <strong><em>M1</em></strong></p>
<p>\( = \frac{1}{2}\left| {\frac{1}{2}a \times \left( { - \frac{1}{{26}}a + \frac{1}{{13}}b} \right)} \right|\)</p>
<p>\( = \frac{1}{2}\left| {\frac{1}{2}a \times \left( { - \frac{1}{{26}}a} \right) + \frac{1}{2}a \times \frac{1}{{13}}b} \right|\) <strong><em>(M1)</em></strong></p>
<p>\( = \frac{1}{2} \times \frac{1}{2} \times \frac{1}{{13}}\left| {a \times b} \right|{\text{ }}\left( { = \frac{1}{{52}}\left| {a \times b} \right|} \right)\) <strong><em>A1</em></strong></p>
<p>\({\text{area }}\Delta {\text{OAB}} = k({\text{area }}\Delta {\text{CAD}})\)</p>
<p>\(\frac{1}{2}\left| {a \times b} \right| = k\frac{1}{{52}}\left| {a \times b} \right|\)</p>
<p>\(k = 26\) <strong><em>A1</em></strong></p>
<p><strong><em>[5 marks]</em></strong></p>
<div class="question_part_label">e.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">a.i.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">a.ii.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">b.i.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">b.ii.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">c.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">d.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">e.</div>
</div>
<br><hr><br><div class="specification">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-size: medium; font-family: 'times new roman', times;">The vertices of a triangle ABC have coordinates given by A(−1, 2, 3), B(4, 1, 1) and C(3, −2, 2).</span></p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 30.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(i) Find the lengths of the sides of the triangle.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 30.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(ii) Find \(\cos {\rm{B\hat AC}}\).</span></p>
<div class="marks">[6]</div>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 29.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(i) Show that \(\overrightarrow {{\text{BC}}} \times \overrightarrow {{\text{CA}}} = \) −7<strong><em>i</em></strong> − 3<strong><em>j</em></strong> − 16<strong><em>k</em></strong>.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 29.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(ii) Hence, show that the area of the triangle ABC is \(\frac{1}{2}\sqrt {314} \).</span></p>
<div class="marks">[5]</div>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 29.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">Find the Cartesian equation of the plane containing the triangle ABC.</span></p>
<div class="marks">[3]</div>
<div class="question_part_label">c.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 33.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">Find a vector equation of (AB).</span></p>
<div class="marks">[2]</div>
<div class="question_part_label">d.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 25.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">The point D on (AB) is such that \(\overrightarrow {{\text{OD}}} \) is perpendicular to \(\overrightarrow {{\text{BC}}} \) where O is the origin.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 25.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"> </span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 25.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(i) Find the coordinates of D.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 25.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(ii) Show that D does not lie between A and B.</span></p>
<div class="marks">[5]</div>
<div class="question_part_label">e.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(i) \(\overrightarrow {{\text{AB}}} = \overrightarrow {{\text{OB}}} - \overrightarrow {{\text{OA}}} = \) 5<strong><em>i</em></strong> – <strong><em>j</em></strong> – 2<strong><em>k</em></strong> (or in column vector form) <strong><em>(A1)</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><strong style="font-family: 'times new roman', times; font-size: medium;">Note:</strong><span style="font-family: 'times new roman', times; font-size: medium;"> Award </span><strong style="font-family: 'times new roman', times; font-size: medium;"><em>A1</em></strong><span style="font-family: 'times new roman', times; font-size: medium;"> if any one of the vectors, or its negative, representing the sides of the triangle is seen.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"> </span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(\overrightarrow {{\text{AB}}} = \) |5</span><strong style="font-family: 'times new roman', times; font-size: medium;"><em>i</em></strong><span style="font-family: 'times new roman', times; font-size: medium;"> – </span><strong style="font-family: 'times new roman', times; font-size: medium;"><em>j</em></strong><span style="font-family: 'times new roman', times; font-size: medium;"> – 2</span><strong style="font-family: 'times new roman', times; font-size: medium;"><em>k</em></strong><span style="font-family: 'times new roman', times; font-size: medium;">|= \(\sqrt {30} \)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(\overrightarrow {{\text{BC}}} = \) |–<strong><em>i</em></strong> – 3<strong><em>j</em></strong> + <strong><em>k</em></strong>|= \(\sqrt {11} \)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(\overrightarrow {{\text{CA}}} = \) |–4<strong><em>i</em></strong> + 4<strong><em>j</em></strong> + <strong><em>k</em></strong>|= \(\sqrt {33} \) <strong><em>A2</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><strong style="font-family: 'times new roman', times; font-size: medium;">Note:</strong><span style="font-family: 'times new roman', times; font-size: medium;"> Award </span><strong style="font-family: 'times new roman', times; font-size: medium;"><em>A1</em></strong><span style="font-family: 'times new roman', times; font-size: medium;"> for two correct and </span><strong style="font-family: 'times new roman', times; font-size: medium;"><em>A0</em></strong><span style="font-family: 'times new roman', times; font-size: medium;"> for one correct.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"> </span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(ii) </span><strong style="font-family: 'times new roman', times; font-size: medium;">METHOD 1</strong></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(\cos {\text{BAC}} = \frac{{20 + 4 + 2}}{{\sqrt {30} \sqrt {33} }}\) <strong><em>M1A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><strong style="font-family: 'times new roman', times; font-size: medium;">Note:</strong><span style="font-family: 'times new roman', times; font-size: medium;"> Award </span><strong style="font-family: 'times new roman', times; font-size: medium;"><em>M1</em></strong><span style="font-family: 'times new roman', times; font-size: medium;"> for an attempt at the use of the scalar product for two vectors representing the sides AB and AC, or their negatives, </span><strong style="font-family: 'times new roman', times; font-size: medium;"><em>A1</em></strong><span style="font-family: 'times new roman', times; font-size: medium;"> for the correct computation using their vectors.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"> </span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = \frac{{26}}{{\sqrt {990} }}{\text{ }}\left( { = \frac{{26}}{{3\sqrt {110} }}} \right)\) </span><strong style="font-family: 'times new roman', times; font-size: medium;"><em>A1</em></strong></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><strong style="font-family: 'times new roman', times; font-size: medium;">Note:</strong><span style="font-family: 'times new roman', times; font-size: medium;"> Candidates who use the modulus need to justify it – the angle is not stated in the question to be acute.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><strong style="font-family: 'times new roman', times; font-size: medium;"> </strong></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><strong style="font-family: 'times new roman', times; font-size: medium;">METHOD 2</strong></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">using the cosine rule</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(\cos {\text{BAC}} = \frac{{30 + 33 - 11}}{{2\sqrt {30} \sqrt {33} }}\) <strong><em>M1A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = \frac{{26}}{{\sqrt {990} }}{\text{ }}\left( { = \frac{{26}}{{3\sqrt {110} }}} \right)\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[6 marks]</em></strong></span></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(\overrightarrow {{\text{BC}}} \times \overrightarrow {{\text{CA}}} = \left| {\begin{array}{*{20}{c}}<br> i&j&k \\ <br> { - 1}&{ - 3}&1 \\ <br> { - 4}&4&1 <br>\end{array}} \right|\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = \left( {( - 3) \times 1 - 1 \times 4} \right)\)<strong><em>i</em></strong> + \(\left( {1 \times ( - 4) - ( - 1) \times 1} \right)\)<strong><em>j</em></strong> + \(\left( {( - 1) \times 4 - ( - 3) \times ( - 4)} \right)\)<strong><em>k</em></strong> <strong><em>M1A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">= –7<strong><em>i</em></strong> – 3<strong><em>j</em></strong> – 16<strong><em>k</em></strong> <strong><em>AG</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em> </em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(ii) the area of \(\Delta {\text{ABC}} = \frac{1}{2}\left| {\overrightarrow {{\text{BC}}} \times \overrightarrow {{\text{CA}}} } \right|\) <strong><em>(M1)</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(\frac{1}{2}\sqrt {{{( - 7)}^2} + {{( - 3)}^2} + {{( - 16)}^2}} \) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = \frac{1}{2}\sqrt {314} \) <strong><em>AG</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><strong style="font-family: 'times new roman', times; font-size: medium;"><em>[5 marks]</em></strong></p>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 29.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">attempt at the use of “(<strong><em>r</em></strong> – <strong><em>a</em></strong>)\( \cdot \)<strong><em>n</em></strong> = 0” <strong><em>(M1)</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 29.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">using <strong><em>r</em></strong> = <em>x</em><strong><em>i</em></strong> + <em>y</em><strong><em>j</em></strong> + <em>z</em><strong><em>k</em></strong>, <strong><em>a</em></strong> = \(\overrightarrow {{\text{OA}}} \) and <strong><em>n</em></strong> = –7<strong><em>i</em></strong> – 3<strong><em>j</em></strong> – 16<strong><em>k</em></strong> <strong><em>(A1)</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 29.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(7x + 3y + 16z = 47\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 29.0px Helvetica;"><strong style="font-family: 'times new roman', times; font-size: medium;">Note:</strong><span style="font-family: 'times new roman', times; font-size: medium;"> Candidates who adopt a 2-parameter approach should be awarded, </span><strong style="font-family: 'times new roman', times; font-size: medium;"><em>A1</em></strong><span style="font-family: 'times new roman', times; font-size: medium;"> for correct 2-parameter equations for </span><em style="font-family: 'times new roman', times; font-size: medium;">x</em><span style="font-family: 'times new roman', times; font-size: medium;">, </span><em style="font-family: 'times new roman', times; font-size: medium;">y</em><span style="font-family: 'times new roman', times; font-size: medium;"> and </span><em style="font-family: 'times new roman', times; font-size: medium;">z</em><span style="font-family: 'times new roman', times; font-size: medium;">; </span><strong style="font-family: 'times new roman', times; font-size: medium;"><em>M1</em></strong><span style="font-family: 'times new roman', times; font-size: medium;"> for a serious attempt at elimination of the parameters; </span><strong style="font-family: 'times new roman', times; font-size: medium;"><em>A1</em></strong><span style="font-family: 'times new roman', times; font-size: medium;"> for the final Cartesian equation.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 29.0px Helvetica;"><strong style="font-family: 'times new roman', times; font-size: medium;"><em> </em></strong></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 29.0px Helvetica;"><strong style="font-family: 'times new roman', times; font-size: medium;"><em>[3 marks]</em></strong></p>
<div class="question_part_label">c.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 30.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>r</em></strong> = \(\overrightarrow {{\text{OA}}} + t\overrightarrow {{\text{AB}}} \) (or equivalent) <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 30.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>r</em></strong> = (–<strong><em>i</em></strong> + 2<strong><em>j</em></strong> + 3<strong><em>k</em></strong>) + <em>t </em>(5<strong><em>i</em></strong> <em>–</em> <strong><em>j</em></strong> – 2<strong><em>k</em></strong>) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 30.0px Helvetica;"><strong style="font-family: 'times new roman', times; font-size: medium;">Note:</strong><span style="font-family: 'times new roman', times; font-size: medium;"> Award </span><strong style="font-family: 'times new roman', times; font-size: medium;"><em>M1A0</em></strong><span style="font-family: 'times new roman', times; font-size: medium;"> if “</span><strong style="font-family: 'times new roman', times; font-size: medium;"><em>r</em></strong><span style="font-family: 'times new roman', times; font-size: medium;"> =” is missing.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 30.0px Helvetica;"><strong style="font-family: 'times new roman', times; font-size: medium;"> </strong></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 30.0px Helvetica;"><strong style="font-family: 'times new roman', times; font-size: medium;">Note:</strong><span style="font-family: 'times new roman', times; font-size: medium;"> Accept forms of the equation starting with B or with the direction reversed.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 30.0px Helvetica;"><strong style="font-family: 'times new roman', times; font-size: medium;"><em> </em></strong></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 30.0px Helvetica;"><strong style="font-family: 'times new roman', times; font-size: medium;"><em>[2 marks]</em></strong></p>
<div class="question_part_label">d.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(i) \(\overrightarrow {{\text{OD}}} = \) (–<strong><em>i</em></strong> + 2<strong><em>j</em></strong> + 3<strong><em>k</em></strong>) + <em>t</em>(5<strong><em>i</em></strong> – <strong><em>j</em></strong> – 2<strong><em>k</em></strong>)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">statement that \(\overrightarrow {{\text{OD}}} \cdot \overrightarrow {{\text{BC}}} = 0\) <strong><em>(M1)</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(\left( {\begin{array}{*{20}{c}}<br> { - 1 + 5t} \\ <br> {2 - t} \\ <br> {3 - 2t} <br>\end{array}} \right) \cdot \left( {\begin{array}{*{20}{c}}<br> { - 1} \\ <br> { - 3} \\ <br> 1 <br>\end{array}} \right) = 0\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( - 2 - 4t = 0{\text{ or }}t = - \frac{1}{2}\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">coordinates of <em>D</em> are \(\left( { - \frac{7}{2},\frac{5}{2},4} \right)\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>Note:</strong> Different forms of \(\overrightarrow {{\text{OD}}} \) give different values of <em>t</em>, but the same final answer.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"> </span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(ii) \(t < 0 \Rightarrow \) D is not between A and B <strong><em>R1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><strong style="font-family: 'times new roman', times; font-size: medium;"><em>[5 marks]</em></strong></p>
<div class="question_part_label">e.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 19.0px Arial;"><span style="font-family: 'times new roman', times; font-size: medium;">Many candidates confidently tackled most of this many-part question. Part (b)(i) As the answer was given, candidates were required to show they really knew how to work out a vector product in detail, not just by writing down a 3x3 determinant and then the final answer. Part (d) A few candidates failed to realise that the equation of a line is an equation not simply an expression. Part (e) A significant number of candidates did not realise that they could use their result for part (d).</span></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 19.0px Arial;"><span style="font-family: 'times new roman', times; font-size: medium;">Many candidates confidently tackled most of this many-part question. Part (b)(i) As the answer was given, candidates were required to show they really knew how to work out a vector product in detail, not just by writing down a 3x3 determinant and then the final answer. Part (d) A few candidates failed to realise that the equation of a line is an equation not simply an expression. Part (e) A significant number of candidates did not realise that they could use their result for part (d).</span></p>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 19.0px Arial;"><span style="font-family: 'times new roman', times; font-size: medium;">Many candidates confidently tackled most of this many-part question. Part (b)(i) As the answer was given, candidates were required to show they really knew how to work out a vector product in detail, not just by writing down a 3x3 determinant and then the final answer. Part (d) A few candidates failed to realise that the equation of a line is an equation not simply an expression. Part (e) A significant number of candidates did not realise that they could use their result for part (d).</span></p>
<div class="question_part_label">c.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 19.0px Arial;"><span style="font-family: 'times new roman', times; font-size: medium;">Many candidates confidently tackled most of this many-part question. Part (b)(i) As the answer was given, candidates were required to show they really knew how to work out a vector product in detail, not just by writing down a 3x3 determinant and then the final answer. Part (d) A few candidates failed to realise that the equation of a line is an equation not simply an expression. Part (e) A significant number of candidates did not realise that they could use their result for part (d).</span></p>
<div class="question_part_label">d.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 19.0px Arial;"><span style="font-family: 'times new roman', times; font-size: medium;">Many candidates confidently tackled most of this many-part question. Part (b)(i) As the answer was given, candidates were required to show they really knew how to work out a vector product in detail, not just by writing down a 3x3 determinant and then the final answer. Part (d) A few candidates failed to realise that the equation of a line is an equation not simply an expression. Part (e) A significant number of candidates did not realise that they could use their result for part (d).</span></p>
<div class="question_part_label">e.</div>
</div>
<br><hr><br><div class="specification">
<p class="p1">A line \(L\) has equation \(\frac{{x - 2}}{p} = \frac{{y - q}}{2} = z - 1\) where \(p,{\text{ }}q \in \mathbb{R}\).</p>
<p class="p1">A plane \(\Pi \) has equation \(x + y + 3z = 9\).</p>
</div>
<div class="specification">
<p class="p1">Consider the different case where the acute angle between \(L\) and \(\Pi \) is \(\theta \)</p>
<p class="p1">where \(\theta = \arcsin \left( {\frac{1}{{\sqrt {11} }}} \right)\).</p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p class="p1">Show that \(L\) is not perpendicular to \(\Pi \).</p>
<div class="marks">[3]</div>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p class="p1">Given that \(L\) lies in the plane \(\Pi \), find the value of \(p\) and the value of \(q\).</p>
<div class="marks">[4]</div>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p class="p1">(i) <span class="Apple-converted-space"> </span>Show that \(p = - 2\).</p>
<p class="p1">(ii) <span class="Apple-converted-space"> </span>If \(L\) intersects \(\Pi \) at \(z = - 1\), find the value of \(q\).</p>
<div class="marks">[11]</div>
<div class="question_part_label">c.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p class="p1"><strong>EITHER</strong></p>
<p class="p2"><em><strong>n</strong></em> \( = \left( {\begin{array}{*{20}{c}} 1 \\ 1 \\ 3 \end{array}} \right)\) and <em><strong>d</strong></em> \( = \left( {\begin{array}{*{20}{c}} p \\ 2 \\ 1 \end{array}} \right)\) <span class="Apple-converted-space"> </span><span class="s1"><strong><em>A1A1</em></strong></span></p>
<p class="p1">and <em><strong>n</strong></em> \( \ne \) <em>k<strong>d</strong></em> <span class="Apple-converted-space"> </span><strong><em>R1</em></strong></p>
<p class="p1"><strong>OR</strong></p>
<p class="p3"><em><strong>n</strong></em> \( \times \) <em><strong>d</strong></em> \( = \left( {\begin{array}{*{20}{c}} { - 5} \\ {3p - 1} \\ {2 - p} \end{array}} \right)\) <span class="Apple-converted-space"> </span><span class="s1"><strong><em>M1A1</em></strong></span></p>
<p class="p1">the vector product is non-zero for \(p \in \mathbb{R}\) <span class="Apple-converted-space"> </span><strong><em>R1</em></strong></p>
<p class="p1"><strong>THEN</strong></p>
<p class="p1">\(L\) is not perpendicular to \(\Pi \) <span class="Apple-converted-space"> </span><strong><em>AG</em></strong></p>
<p class="p1"><strong><em>[3 marks]</em></strong></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1"><strong>METHOD 1</strong></p>
<p class="p2"><span class="Apple-converted-space">\((2 + p\lambda ) + (q + 2\lambda ) + 3(1 + \lambda ) = 9\) </span><span class="s1"><strong><em>M1</em></strong></span></p>
<p class="p2"><span class="Apple-converted-space">\((q + 5) + (p + 5)\lambda = 9\) </span><span class="s1"><strong><em>(A1)</em></strong></span></p>
<p class="p1">\(p = - 5\) and \(q = 4\) <span class="Apple-converted-space"> </span><strong><em>A1A1</em></strong></p>
<p class="p1"><strong>METHOD 2</strong></p>
<p class="p1">direction vector of line is perpendicular to plane, so</p>
<p class="p1"><span class="Apple-converted-space">\(\left( {\begin{array}{*{20}{c}} p \\ 2 \\ 1 \end{array}} \right) \bullet \left( {\begin{array}{*{20}{c}} 1 \\ 1 \\ 3 \end{array}} \right) = 0\) </span><strong><em>M1</em></strong></p>
<p class="p2"><span class="Apple-converted-space">\(p = - 5\) </span><span class="s1"><strong><em>A1</em></strong></span></p>
<p class="p1"><span class="s2">\((2,{\text{ }}q,{\text{ }}1)\) </span>is common to both \(L\) and \(\Pi \)</p>
<p class="p1"><span class="s2">either \(\left( {\begin{array}{*{20}{c}} 2 \\ q \\ 1 \end{array}} \right) \bullet \left( {\begin{array}{*{20}{c}} 1 \\ 1 \\ 3 \end{array}} \right) = 9\) </span>or by substituting into \(x + y + 3z = 9\) <span class="Apple-converted-space"> </span><strong><em>M1</em></strong></p>
<p class="p2"><span class="Apple-converted-space">\(q = 4\) </span><span class="s1"><strong><em>A1</em></strong></span></p>
<p class="p1"><strong><em>[4 marks]</em></strong></p>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">(i) <span class="Apple-converted-space"> </span><strong>METHOD 1</strong></p>
<p class="p1">\(\alpha \) is the acute angle between <em><strong>n</strong></em> and <em>L</em></p>
<p class="p2">if \(\sin \theta = \frac{1}{{\sqrt {11} }}\) then \(\cos \alpha = \frac{1}{{\sqrt {11} }}\) <span class="Apple-converted-space"> </span><span class="s1"><strong><em>(M1)(A1)</em></strong></span></p>
<p class="p2">attempting to use \(\cos \alpha = \frac{{n \bullet d}}{{\left| n \right|\left| d \right|}}\) or \(\sin \alpha = \frac{{n \bullet d}}{{\left| n \right|\left| d \right|}}\) <span class="Apple-converted-space"> </span><span class="s1"><strong><em>M1</em></strong></span></p>
<p class="p2"><span class="Apple-converted-space">\(\frac{{p + 5}}{{\sqrt {11} \times \sqrt {{p^2} + 5} }} = \frac{1}{{\sqrt {11} }}\) </span><span class="s1"><strong><em>A1A1</em></strong></span></p>
<p class="p2"><span class="Apple-converted-space">\({(p + 5)^2} = {p^2} + 5\) </span><span class="s1"><strong><em>M1</em></strong></span></p>
<p class="p1">\(10p = - 20\) (or equivalent) <span class="Apple-converted-space"> </span><strong><em>A1</em></strong></p>
<p class="p2"><span class="Apple-converted-space">\(p = - 2\) </span><span class="s1"><strong><em>AG</em></strong></span></p>
<p class="p1"><strong>METHOD 2</strong></p>
<p class="p1">\(\alpha \) is the angle between <em><strong>n</strong></em> and <em>L</em></p>
<p class="p1">if \(\sin \theta = \frac{1}{{\sqrt {11} }}\) <span class="s2">then \(\sin \alpha = \frac{{\sqrt {10} }}{{\sqrt {11} }}\) <span class="Apple-converted-space"> </span></span><strong><em>(M1)A1</em></strong></p>
<p class="p1">attempting to use \(\sin \alpha = \frac{{\left| {n \times d} \right|}}{{\left| n \right|\left| d \right|}}\) <span class="Apple-converted-space"> </span><strong><em>M1</em></strong></p>
<p class="p3"><span class="Apple-converted-space">\(\frac{{\sqrt {{{( - 5)}^2} + {{(3p - 1)}^2} + {{(2 - p)}^2}} }}{{\sqrt {11} \times \sqrt {{p^2} + 5} }} = \frac{{\sqrt {10} }}{{\sqrt {11} }}\) </span><span class="s1"><strong><em>A1A1</em></strong></span></p>
<p class="p2"><span class="Apple-converted-space">\({p^2} - p + 3 = {p^2} + 5\) </span><span class="s1"><strong><em>M1</em></strong></span></p>
<p class="p1">\( - p + 3 = 5\) (or equivalent) <span class="Apple-converted-space"> </span><strong><em>A1</em></strong></p>
<p class="p2"><span class="Apple-converted-space">\(p = - 2\) </span><span class="s1"><strong><em>AG</em></strong></span></p>
<p class="p1"><span class="s2">(ii) <span class="Apple-converted-space"> \(p = - 2\)</span> </span>and \(z = - 1 \Rightarrow \frac{{x - 2}}{{ - 2}} = \frac{{y - q}}{2} = - 2\) <span class="Apple-converted-space"> </span><strong><em>(A1)</em></strong></p>
<p class="p1">\(x = 6\) and \(y = q - 4\) <span class="Apple-converted-space"> </span><strong><em>(A1)</em></strong></p>
<p class="p1">this satisfies \(\Pi \) <span class="s2">so \(6 + q - 4 - 3 = 9\) <span class="Apple-converted-space"> </span></span><strong><em>M1</em></strong></p>
<p class="p2"><span class="Apple-converted-space">\(q = 10\) </span><span class="s1"><strong><em>A1</em></strong></span></p>
<p class="p1"><strong><em>[11 marks]</em></strong></p>
<div class="question_part_label">c.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
<p class="p1">Parts (a) and (b) were often well done, though a small number of candidates were clearly puzzled when trying to demonstrate \(\left( {\begin{array}{*{20}{c}} 1 \\ 1 \\ 3 \end{array}} \right) \ne k\left( {\begin{array}{*{20}{c}} p \\ 2 \\ 1 \end{array}} \right)\), with some scripts seen involving needlessly convoluted arguments.</p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p>Parts (a) and (b) were often well done, though a small number of candidates were clearly puzzled when trying to demonstrate \(\left( {\begin{array}{*{20}{c}} 1 \\ 1 \\ 3 \end{array}} \right) \ne k\left( {\begin{array}{*{20}{c}} p \\ 2 \\ 1 \end{array}} \right)\), with some scripts seen involving needlessly convoluted arguments.</p>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">Part (c) often proved problematic, as some candidates unsurprisingly used the sine (or cosine) of an incorrect angle, and few consequent marks were then available. Some good clear solutions were seen, occasionally complete with diagrams in the cases of the thoughtful candidates who were able to ‘work through’ the question rather than just apply a standard vector result.</p>
<div class="question_part_label">c.</div>
</div>
<br><hr><br><div class="question" style="padding-left: 20px; padding-right: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 29.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">For non-zero vectors </span><span style="font-family: 'times new roman', times; font-size: medium;"><span style="font-family: 'times new roman', times; font-size: medium;">\({\boldsymbol{a}}\)</span> and </span><span style="font-family: 'times new roman', times; font-size: medium;"><span style="font-family: 'times new roman', times; font-size: medium;">\({\boldsymbol{b}}\)</span>, show that</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 29.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(i) if \(\left| {{\boldsymbol{a}} - {\boldsymbol{b}}} \right| = \left| {{\boldsymbol{a}} + {\boldsymbol{b}}} \right|\), then \({\boldsymbol{a}}\)</span><span style="font-family: 'times new roman', times; font-size: medium;"> and </span><span style="font-family: 'times new roman', times; font-size: medium;">\({\boldsymbol{b}}\)</span><span style="font-family: 'times new roman', times; font-size: medium;"> are perpendicular;</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 29.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(ii) \({\left| {{\boldsymbol{a}} \times {\boldsymbol{b}}} \right|^2} = {\left| {\boldsymbol{a}} \right|^2}{\left| {\boldsymbol{b}} \right|^2} - {({\boldsymbol{a}} \cdot {\boldsymbol{b}})^2}\).</span></p>
<div class="marks">[8]</div>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">The points A, B and C have position vectors \({\boldsymbol{a}}\), \({\boldsymbol{b}}\) and \({\boldsymbol{c}}\).</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(i) Show that the area of triangle ABC is \(\frac{1}{2}\left| {{\boldsymbol{a}} \times {\boldsymbol{b}} + {\boldsymbol{b}} \times {\boldsymbol{c}} + {\boldsymbol{c}} \times {\boldsymbol{a}}} \right|\).</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(ii) Hence, show that the shortest distance from B to AC is</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\[\frac{{\left| {{\boldsymbol{a}} \times {\boldsymbol{b}} + {\boldsymbol{b}} \times {\boldsymbol{c}} + {\boldsymbol{c}} \times {\boldsymbol{a}}} \right|}}{{\left| {{\boldsymbol{c}} - {\boldsymbol{a}}} \right|}}{\text{.}}\]</span></p>
<div class="marks">[7]</div>
<div class="question_part_label">b.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 31.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(i) </span><span style="font-family: 'times new roman', times; font-size: medium;">\(\left| {{\boldsymbol{a}} - {\boldsymbol{b}}} \right| = \left| {{\boldsymbol{a}} + {\boldsymbol{b}}} \right|\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 31.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( \Rightarrow \left( {{\boldsymbol{a}} - {\boldsymbol{b}}} \right) \cdot \left( {{\boldsymbol{a}} - {\boldsymbol{b}}} \right) = \left( {{\boldsymbol{a}} + {\boldsymbol{b}}} \right) \cdot \left( {{\boldsymbol{a}} + {\boldsymbol{b}}} \right)\)</span><span style="font-family: 'times new roman', times; font-size: medium;"> <strong><em>(M1)</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 31.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( \Rightarrow {\left| {\boldsymbol{a}} \right|^2} - 2{\boldsymbol{a}} \cdot {\boldsymbol{b}} + {\left| {\boldsymbol{b}} \right|^2} = {\left| {\boldsymbol{a}} \right|^2} + 2{\boldsymbol{a}} \cdot {\boldsymbol{b}} + {\left| {\boldsymbol{b}} \right|^2}\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 31.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( \Rightarrow 4{\boldsymbol{a}} \cdot {\boldsymbol{b}} = 0 \Rightarrow {\boldsymbol{a}} \cdot {\boldsymbol{b}} = 0\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 31.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">therefore </span><span style="font-family: 'times new roman', times; font-size: medium;"><span style="font-family: 'times new roman', times; font-size: medium;">\({\boldsymbol{a}}\)</span> and </span><span style="font-family: 'times new roman', times; font-size: medium;"><span style="font-family: 'times new roman', times; font-size: medium;">\({\boldsymbol{b}}\) </span></span><span style="font-family: 'times new roman', times; font-size: medium;">are perpendicular <strong><em>R1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 31.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>Note:</strong> Allow use of 2-d components.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 31.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"> </span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 31.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>Note:</strong> Do not condone sloppy vector notation, so we must see something to the effect that \({\left| {\boldsymbol{c}} \right|^2} = {\boldsymbol{c}} \cdot {\boldsymbol{c}}\) is clearly being used for the <strong><em>M1</em></strong>.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 31.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"> </span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 31.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>Note:</strong> Allow a correct geometric argument, for example that the diagonals of a parallelogram have the same length only if it is a rectangle.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 31.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"> </span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 31.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(ii) \({\left| {{\boldsymbol{a}} \times {\boldsymbol{b}}} \right|^2} = {\left( {\left| {\boldsymbol{a}} \right|\left| {\boldsymbol{b}} \right|\sin \theta } \right)^2} = {\left| {\boldsymbol{a}} \right|^2}{\left| {\boldsymbol{b}} \right|^2}{\sin ^2}\theta \) <strong><em>M1A1</em></strong></span></p>
<p style="margin-top: 0px; margin-right: 0px; margin-bottom: 0px; font: 31px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\({\left| {\boldsymbol{a}} \right|^2}{\left| {\boldsymbol{b}} \right|^2} - {\left( {{\boldsymbol{a}} \cdot {\boldsymbol{b}}} \right)^2} = {\left| {\boldsymbol{a}} \right|^2}{\left| {\boldsymbol{b}} \right|^2} - {\left| {\boldsymbol{a}} \right|^2}{\left| {\boldsymbol{b}} \right|^2}{\cos ^2}\theta \) <strong><em>M1</em></strong></span></p>
<p style="margin-top: 0px; margin-right: 0px; margin-bottom: 0px; font: 31px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = {\left| {\boldsymbol{a}} \right|^2}{\left| {\boldsymbol{b}} \right|^2}\left( {1 - {{\cos }^2}\theta } \right)\) <em><strong>A1</strong></em><em><strong><br></strong></em>\( = {\left| {\boldsymbol{a}} \right|^2}{\left| {\boldsymbol{b}} \right|^2}\left( {{{\sin }^2}\theta } \right)\)</span></p>
<p style="margin-top: 0px; margin-right: 0px; margin-bottom: 0px; font: 31px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( \Rightarrow {\left| {{\boldsymbol{a}} \times {\boldsymbol{b}}} \right|^2} = {\left| {\boldsymbol{a}} \right|^2}{\left| {\boldsymbol{b}} \right|^2} - {\left( {{\boldsymbol{a}} \cdot {\boldsymbol{b}}} \right)^2}\) <strong><em>AG</em></strong></span></p>
<p style="margin-top: 0px; margin-right: 0px; margin-bottom: 0px; font: 31px Helvetica;"><strong style="font-family: 'times new roman', times; font-size: medium;"><em>[8 marks]</em></strong></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(i) area of triangle \( = \frac{1}{2}\left| {\overrightarrow {{\text{AB}}} \times \overrightarrow {{\text{AC}}} } \right|\) <strong><em>(M1)</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = \frac{1}{2}\left| {\left( {{\boldsymbol{b}} - {\boldsymbol{a}}} \right) \times \left( {{\boldsymbol{c}} - {\boldsymbol{a}}} \right)} \right|\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = \frac{1}{2}\left| {{\boldsymbol{b}} \times {\boldsymbol{c}} + {\boldsymbol{b}} \times {\boldsymbol{ - a}} + {\boldsymbol{ - a}} \times {\boldsymbol{c}} + {\boldsymbol{ - a}} \times {\boldsymbol{ - a}}} \right|\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\({\boldsymbol{b}} \times {\boldsymbol{ - a}} = {\boldsymbol{a}} \times {\boldsymbol{b}}\); \({\boldsymbol{c}} \times {\boldsymbol{a}} = {\boldsymbol{ - a}} \times {\boldsymbol{c}}\); \({\boldsymbol{ - a}} \times {\boldsymbol{ - a}} = 0\) <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">hence, area of triangle is \(\frac{1}{2}\left| {{\boldsymbol{a}} \times {\boldsymbol{b}} + {\boldsymbol{b}} \times {\boldsymbol{c}} + {\boldsymbol{c}} \times {\boldsymbol{a}}} \right|\) <strong><em>AG</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"> </span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(ii) D is the foot of the perpendicular from B to AC</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">area of triangle \({\text{ABC}} = \frac{1}{2}\left| {\overrightarrow {{\text{AC}}} } \right|\left| {\overrightarrow {{\text{BD}}} } \right|\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">therefore</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(\frac{1}{2}\left| {\overrightarrow {{\text{AC}}} } \right|\left| {\overrightarrow {{\text{BD}}} } \right| = \frac{1}{2}\left| {\overrightarrow {{\text{AB}}} \times \overrightarrow {{\text{AC}}} } \right|\) <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">hence, \(\left| {\overrightarrow {{\text{BD}}} } \right| = \frac{{\left| {\overrightarrow {{\text{AB}}} \times \overrightarrow {{\text{AC}}} } \right|}}{{\left| {\overrightarrow {{\text{AC}}} } \right|}}\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = \frac{{\left| {{\boldsymbol{a}} \times {\boldsymbol{b}} + {\boldsymbol{b}} \times {\boldsymbol{c}} + {\boldsymbol{c}} \times {\boldsymbol{a}}} \right|}}{{\left| {{\boldsymbol{c}} - {\boldsymbol{a}}} \right|}}\)</span><span style="font-family: 'times new roman', times; font-size: medium;"> <strong><em>AG</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><strong style="font-family: 'times new roman', times; font-size: medium;"><em>[7 marks]</em></strong></p>
<div class="question_part_label">b.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(i) The majority of candidates were very sloppy in their use of vector notation. Some candidates used Cartesian coordinates, which was acceptable. Part (ii) was well done.</span></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">Part (i) was usually well started, but not completed satisfactorily. Many candidates understood the geometry involved in this part.</span></p>
<div class="question_part_label">b.</div>
</div>
<br><hr><br><div class="question">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">Consider the points A(1, −1, 4), B (2, − 2, 5) and O(0, 0, 0).</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(a) Calculate the cosine of the angle between \(\overrightarrow {{\text{OA}}} \) and \(\overrightarrow {{\text{AB}}} \).</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(b) Find a vector equation of the line \({L_1}\) which passes through A and B.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">The line \({L_2}\) has equation <strong><em>r</em></strong> = 2<strong><em>i</em></strong> + 4<strong><em>j</em></strong> + 7<strong><em>k</em></strong> + t(2<strong><em>i</em></strong> + <strong><em>j</em></strong> + 3<strong><em>k</em></strong>), where \(t \in \mathbb{R}\) .</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(c) Show that the lines \({L_1}\) and \({L_2}\) intersect and find the coordinates of their point of intersection.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(d) Find the Cartesian equation of the plane which contains both the line \({L_2}\) and the point A.</span></p>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 16.0px Times; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;">(a) Use of \(\cos \theta = \frac{{\overrightarrow {{\text{OA}}} \cdot \overrightarrow {{\text{AB}}} }}{{\left| {\overrightarrow {{\text{OA}}} } \right|\left| {\overrightarrow {{\text{AB}}} } \right|}}\) <strong><em>(M1)</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 16.0px Times; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;">\({\overrightarrow {{\text{AB}}} }\) = <strong><em>i</em></strong> − <strong><em>j</em></strong> + <strong><em>k</em></strong> <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 16.0px Times; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;">\(\left| {\overrightarrow {{\text{AB}}} } \right| = \sqrt 3 \) <strong>and</strong> \(\left| {\overrightarrow {{\text{OA}}} } \right| = 3\sqrt 2 \) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 16.0px Times; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;">\(\overrightarrow {{\text{OA}}} \cdot \overrightarrow {{\text{AB}}} = 6\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 16.0px Times; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;">substituting gives \(\cos \theta = \frac{2}{{\sqrt 6 }}\,\,\,\,\,\left( { = \frac{{\sqrt 6 }}{3}} \right)\,\,\,\,\,\)or equivalent <strong><em>M1</em></strong> <strong><em>N1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 16.0px Times; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[5 marks]</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 16.0px Times; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em> </em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 16.0px Times; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;">(b) \({L_1}\) : <strong><em>r</em></strong> = \(\overrightarrow {{\text{OA}}} + s\overrightarrow {{\text{AB}}} \,\,\,\,\,\) or equivalent <strong><em>(M1)</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 16.0px Times; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;">\({L_1}\) : <strong><em>r</em></strong> = <strong><em>i</em></strong> − <strong><em>j</em></strong> + 4<strong><em>k</em></strong> + s(<strong><em>i</em></strong> − <strong><em>j</em></strong> + <strong><em>k</em></strong>)</span><span style="font-family: 'times new roman', times; font-size: medium;"><span style="font-family: 'times new roman', times; font-size: medium;">\(\,\,\,\,\,\)</span>or equivalent <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 16.0px Times; color: #3f3f3f;"><strong style="font-family: 'times new roman', times; font-size: medium;">Note:</strong><span style="font-family: 'times new roman', times; font-size: medium;"> Award </span><strong style="font-family: 'times new roman', times; font-size: medium;"><em>(M1)A0</em></strong><span style="font-family: 'times new roman', times; font-size: medium;"> for omitting “ </span><strong style="font-family: 'times new roman', times; font-size: medium;"><em>r</em></strong><span style="font-family: 'times new roman', times; font-size: medium;"> = ” in the final answer.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 16.0px Times; color: #3f3f3f;"><strong style="font-family: 'times new roman', times; font-size: medium;"><em> </em></strong></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 16.0px Times; color: #3f3f3f;"><strong style="font-family: 'times new roman', times; font-size: medium;"><em>[2 marks]</em></strong></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 16.0px Times; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em> </em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 16.0px Times; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;">(c) Equating components and forming equations involving <em>s</em> and <em>t</em> <strong><em>(M1)</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 16.0px Times; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;">1 + <em>s</em> = 2 + 2<em>t</em> , −1 − <em>s</em> = 4 + <em>t</em> , 4 + <em>s</em> = 7 + 3<em>t</em></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 16.0px Times; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;">Having two of the above three equations <strong><em>A1A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 16.0px Times; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;">Attempting to solve for <em>s</em> or <em>t</em> <strong><em>(M1)</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 16.0px Times; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;">Finding either <em>s</em> = −3 or <em>t</em> = −2 <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 16.0px Times; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;">Explicitly showing that these values satisfy the third equation <strong><em>R1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 16.0px Times; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;">Point of intersection is (−2, 2, 1) <strong><em>A1</em></strong> <strong><em>N1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 16.0px Times; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>Note:</strong> Position vector is not acceptable for final <strong><em>A1</em></strong>.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 16.0px Times; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;"> </span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 16.0px Times; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[7 marks]</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 16.0px Times; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em> </em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(d) <strong>METHOD 1</strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(r = \left( {\begin{array}{*{20}{c}}<br> 1 \\ <br> { - 1} \\ <br> 4 <br>\end{array}} \right) + \lambda \left( {\begin{array}{*{20}{c}}<br> 2 \\ <br> 1 \\ <br> 3 <br>\end{array}} \right) + \mu \left( {\begin{array}{*{20}{c}}<br> { - 3} \\ <br> 3 \\ <br> { - 3} <br>\end{array}} \right)\) <strong><em>(A1)</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(x = 1 + 2\lambda - 3\mu {\text{ , }}y = - 1 + \lambda + 3\mu {\text{ and }}z = 4 + 3\lambda - 3\mu \) <strong><em>M1A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">Elimination of the parameters<span style="font: 19.0px Helvetica;"> </span><strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(x + y = 3\lambda {\text{ so }}4(x + y) = 12\lambda {\text{ and }}y + z = 4\lambda + 3{\text{ so }}3(y + z) = 12\lambda + 9\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(3(y + z) = 4(x + y) + 9\)<span style="font: 19.0px Helvetica;"> </span><strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">Cartesian equation of plane is 4<em>x</em> + <em>y</em> − 3<em>z</em> = −9 (or equivalent)<span style="font: 19.0px Helvetica;"> </span><strong><em>A1</em></strong><span style="font: 19.0px Helvetica;"> </span><strong><em>N1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[6 marks]</em></strong></span><strong style="font-family: 'times new roman', times; font-size: medium;"><em> </em></strong></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>METHOD 2</strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>EITHER</strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">The point (2, 4, 7) lies on the plane.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">The vector joining (2, 4, 7) and (1, −1, 4) and 2<strong><em>i</em></strong> + <strong><em>j</em></strong> + 3<strong><em>k</em></strong> are parallel to the plane. So they are perpendicular to the normal to the plane.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(<strong><em>i</em></strong> − <strong><em>j</em></strong> + 4<strong><em>k</em></strong>) − (2<strong><em>i</em></strong> + 4<strong><em>j</em></strong> + 7<strong><em>k</em></strong>) = −<strong><em>i</em></strong> − 5<strong><em>j</em></strong> − 3<strong><em>k</em></strong><span style="font: 19.0px Helvetica;"> </span><strong><em>(A1)</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(n = \left| {\begin{array}{*{20}{c}}<br> i&j&k \\ <br> { - 1}&{ - 5}&{ - 3} \\ <br> 2&1&3 <br>\end{array}} \right|\)<span style="font: 19.0px Helvetica;"> </span><strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">= −12<strong><em>i</em></strong> − 3<strong><em>j</em></strong> + 9<strong><em>k</em></strong>\(\,\,\,\,\,\)or equivalent parallel vector <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>OR</strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\({L_1}\) and \({L_2}\) intersect at D(−2, 2, 1)</span></p>
<p><span style="font-family: times new roman,times;"><span style="font-size: medium;">\(n = \left| {\begin{array}{*{20}{c}}<br> i&j&k \\ <br> 2&1&{ - 3} \\ <br> { - 3}&3&{ - 3} <br>\end{array}} \right|\)</span> <span style="font-size: medium;"><strong><em>M1</em></strong></span></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">= −12<strong><em>i</em></strong> − 3<strong><em>j</em></strong> + 9<strong><em>k</em></strong>\(\,\,\,\,\,\)or equivalent parallel vector <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>THEN</strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>r</em></strong>\( \cdot \)<strong><em>n </em></strong>= (<strong><em>i</em></strong> − <strong><em>j</em></strong> + 4<strong><em>k</em></strong>)\( \cdot \)(−12<strong><em>i</em></strong> − 3<strong><em>j</em></strong> + 9<strong><em>k</em></strong>) <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">= 27 <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">Cartesian equation of plane is 4<em>x</em> + <em>y</em> −3<em>z</em> = −9 (or equivalent) <strong><em>A1 N1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[6 marks]</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>Total [20 marks]</em></strong></span></p>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 24.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">Most candidates scored reasonably well on this question. The most common errors were: Using <strong><em>OB</em></strong> rather than <strong><em>AB</em></strong> in (a); omitting the <em>r</em> = in (b); failure to check that the values of the two parameters satisfied the third equation in (c); the use of an incorrect vector in (d). Even when (d) was correctly answered, there was usually little evidence of why a specific vector had been used.</span></p>
</div>
<br><hr><br><div class="specification">
<p class="p1"><span class="s1">ABCD </span>is a parallelogram, where \(\overrightarrow {{\text{AB}}} \) = –<strong><em>i</em></strong> + 2<strong><em>j</em></strong> + 3<strong><em>k</em></strong> and \(\overrightarrow {{\text{AD}}} \) = 4<strong><em>i</em></strong> – <strong><em>j</em></strong> – 2<strong><em>k</em></strong>.</p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Find the area of the parallelogram ABCD.</p>
<div class="marks">[3]</div>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>By using a suitable scalar product of two vectors, determine whether \({\rm{A\hat BC}}\) is acute or obtuse.</p>
<div class="marks">[4]</div>
<div class="question_part_label">b.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p>\(\overrightarrow {{\text{AB}}} \times \overrightarrow {{\text{AD}}} = - \)<strong><em>i</em></strong> \( + 10\)<strong><em>j</em></strong> – 7<strong><em>k</em></strong> <strong><em>M1A1</em></strong></p>
<p>\({\text{area}} = \left| {\overrightarrow {{\text{AB}}} \times \overrightarrow {{\text{AD}}} } \right|{\text{ = }}\sqrt {{1^2} + {{10}^2} + {7^2}} \)</p>
<p> \( = 5\sqrt 6 \left( {\sqrt {150} } \right)\) <strong><em>A1</em></strong></p>
<p><strong><em>[3 marks]</em></strong></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><strong>METHOD 1</strong></p>
<p>\(\overrightarrow {{\text{AB}}} \bullet \overrightarrow {{\text{AD}}} = - 4 - 2 - 6\) <strong><em>M1A1</em></strong></p>
<p>\( = - 12\)</p>
<p>considering the sign of the answer</p>
<p>\(\overrightarrow {{\text{AB}}} \bullet \overrightarrow {{\text{AD}}} < 0\), therefore angle \({\rm{D\hat AB}}\) is obtuse <strong><em>M1</em></strong></p>
<p>(as it is a parallelogram), \({\rm{A\hat BC}}\) is acute <strong><em>A1</em></strong></p>
<p><strong><em>[4 marks]</em></strong></p>
<p><strong>METHOD 2</strong></p>
<p>\(\overrightarrow {{\text{BA}}} \bullet \overrightarrow {{\text{BC}}} = + 4 + 2 + 6\) <strong><em>M1A1</em></strong></p>
<p>\( = 12\) considering the sign of the answer <strong><em>M1</em></strong></p>
<p>\(\overrightarrow {{\text{BA}}} \bullet \overrightarrow {{\text{BC}}} > 0 \Rightarrow {\rm{A\hat BC}}\) is acute <strong><em>A1</em></strong></p>
<p><strong><em>[4 marks]</em></strong></p>
<div class="question_part_label">b.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">b.</div>
</div>
<br><hr><br><div class="question">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 30.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">The points A, B, C have position vectors <strong><em>i</em></strong> + <strong><em>j</em></strong> + 2<strong><em>k</em></strong> , <strong><em>i</em></strong> + 2<strong><em>j</em></strong> + 3<strong><em>k</em></strong> , 3<strong><em>i</em></strong> + <strong><em>k</em></strong> respectively and lie in the plane \(\pi \) .</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 30.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(a) Find</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 30.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(i) the area of the triangle ABC;</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 30.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(ii) the shortest distance from C to the line AB;</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 30.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(iii) the cartesian equation of the plane \(\pi \) . </span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 30.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">The line <em>L</em> passes through the origin and is normal to the plane \(\pi \) , it intersects \(\pi \) at the</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 30.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">point D.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 30.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(b) Find</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 30.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(i) the coordinates of the point D;</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 30.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(ii) the distance of \(\pi \) from the origin.</span></p>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question">
<p><span style="font-family: times new roman,times; font-size: medium;">(a) (i) <strong>METHOD 1</strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(\overrightarrow {{\text{AB}}} = \boldsymbol{b} - \boldsymbol{a} = \left( {\begin{array}{*{20}{c}}<br> 1 \\ <br> 2 \\ <br> 3 <br>\end{array}} \right) - \left( {\begin{array}{*{20}{c}}<br> 1 \\ <br> 1 \\ <br> 2 <br>\end{array}} \right) = \left( {\begin{array}{*{20}{c}}<br> 0 \\ <br> 1 \\ <br> 1 <br>\end{array}} \right)\) <strong><em>(A1)</em></strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(\overrightarrow {{\text{AC}}} = \boldsymbol{c} - \boldsymbol{a} = \left( {\begin{array}{*{20}{c}}<br> 3 \\ <br> 0 \\ <br> 1 <br>\end{array}} \right) - \left( {\begin{array}{*{20}{c}}<br> 1 \\ <br> 1 \\ <br> 2 <br>\end{array}} \right) = \left( {\begin{array}{*{20}{c}}<br> 2 \\ <br> { - 1} \\ <br> { - 1} <br>\end{array}} \right)\) <strong><em>(A1)</em></strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(\overrightarrow {{\text{AB}}} \times \overrightarrow {{\text{AC}}} = \left| {\begin{array}{*{20}{c}}<br> \boldsymbol{i}&\boldsymbol{j}&\boldsymbol{k} \\ <br> 0&1&1 \\ <br> 2&{ - 1}&{ - 1} <br>\end{array}} \right|\) <strong><em>M1</em></strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">= <strong><em>i</em></strong>(−1 + 1) − <strong><em>j</em></strong>(0 − 2) + <strong><em>k</em></strong> (0 − 2) <strong><em>(A1)</em></strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">= 2<strong><em>j</em></strong> − 2<strong><em>k</em></strong> <strong><em>A1</em></strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">Area of triangle ABC}} \( = \frac{1}{2}\left| {2{\mathbf{j}} - 2{\mathbf{k}}} \right| = {\text{ }}\frac{1}{2}\sqrt 8 \) \(( = \sqrt 2 )\) sq. units <strong><em>M1A1</em></strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"><strong>Note:</strong> Allow <strong><em>FT</em></strong> on final <strong><em>A1</em></strong>.</span></p>
<p><strong style="font-family: 'times new roman', times; font-size: medium;"> </strong></p>
<p><strong style="font-family: 'times new roman', times; font-size: medium;">METHOD 2</strong></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(\left| {{\text{AB}}} \right| = \sqrt 2 ,{\text{ }}\left| {{\text{BC}}} \right| = \sqrt {12} ,{\text{ }}\left| {{\text{AC}}} \right| = \sqrt 6 \) <strong><em>A1A1A1</em></strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">Using cosine rule, e.g. on \({\hat C}\) <strong><em>M1</em></strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(\cos C = \frac{{6 + 12 - 2}}{{2\sqrt {72} }} = \frac{{2\sqrt 2 }}{3}\) <strong><em>A1</em></strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(\therefore {\text{Area }}\Delta {\text{ABC}} = \frac{1}{2}ab\sin C\) <strong><em>M1</em></strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\( = \frac{1}{2}\sqrt {12} \sqrt 6 \sin \left( {\arccos \frac{{2\sqrt 2 }}{3}} \right)\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\( = 3\sqrt 2 \sin \left( {\arccos \frac{{2\sqrt 2 }}{3}} \right){\text{ }}\left( { = \sqrt 2 } \right)\) <strong><em>A1</em></strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"><strong>Note:</strong> Allow <strong><em>FT</em></strong> on final <strong><em>A1</em></strong>.</span></p>
<p><span style="font-family: 'times new roman', times; font-size: medium;"> </span></p>
<p><span style="font-family: 'times new roman', times; font-size: medium;">(ii) \({\text{AB}} = \sqrt 2 \) </span><strong style="font-family: 'times new roman', times; font-size: medium;"><em>A1</em></strong></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(\sqrt 2 = \frac{1}{2}{\text{AB}} \times h = \frac{1}{2}\sqrt 2 \times h{\text{ , }}h\) equals the shortest distance <strong><em>(M1)</em></strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\( \Rightarrow h = 2\) <strong><em>A1</em></strong></span></p>
<p><span style="font-family: 'times new roman', times; font-size: medium;"> </span></p>
<p><span style="font-family: 'times new roman', times; font-size: medium;">(iii) </span><strong style="font-family: 'times new roman', times; font-size: medium;">METHOD 1</strong></p>
<p><span style="font-family: times new roman,times; font-size: medium;">(\pi \) has form \(r \cdot \left( {\begin{array}{*{20}{c}}<br> 0 \\ <br> 2 \\ <br> { - 2} <br>\end{array}} \right) = d\) <strong><em>(M1)</em></strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">Since (1, 1, 2) is on the plane</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(d = \left( {\begin{array}{*{20}{c}}<br> 1 \\ <br> 1 \\ <br> 2 <br>\end{array}} \right) \cdot \left( {\begin{array}{*{20}{c}}<br> 0 \\ <br> 2 \\ <br> { - 2} <br>\end{array}} \right) = 2 - 4 = -2\) <strong><em>M1A1</em></strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">Hence \(r \cdot \left( {\begin{array}{*{20}{c}}<br> 0 \\ <br> 2 \\ <br> { - 2} <br>\end{array}} \right) = - 2\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(2y - 2z = - 2{\text{ (or }}y - z = - 1)\) <strong><em>A1</em></strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"><strong>METHOD 2</strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(r = \left( {\begin{array}{*{20}{c}}<br> 1 \\ <br> 1 \\ <br> 2 <br>\end{array}} \right) + \lambda \left( {\begin{array}{*{20}{c}}<br> 0 \\ <br> 1 \\ <br> 1 <br>\end{array}} \right) + \mu \left( {\begin{array}{*{20}{c}}<br> 2 \\ <br> { - 1} \\ <br> { - 1} <br>\end{array}} \right)\) <strong><em>(M1)</em></strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(x = 1 + 2\mu \) (i)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(y = 1 + \lambda - \mu \) (ii)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(z = 2 + \lambda - \mu \) (iii) <strong><em>A1</em></strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"><strong>Note:</strong> Award <strong><em>A1</em></strong> for all three correct, <strong><em>A0</em></strong> otherwise.</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"> </span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">From (i) \(\mu = \frac{{x - 1}}{2}\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">substitute in (ii) \(y = 1 + \lambda - \left( {\frac{{x - 1}}{2}} \right)\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\( \Rightarrow \lambda = y - 1 + \left( {\frac{{x - 1}}{2}} \right)\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">substitute \(\lambda \) and \(\mu \) in (iii) <strong><em>M1</em></strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\( \Rightarrow z = 2 + y - 1 + \left( {\frac{{x - 1}}{2}} \right) - \left( {\frac{{x - 1}}{2}} \right)\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\( \Rightarrow y - z = - 1\) <strong><em>A1</em></strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"><strong><em>[14 marks]</em></strong></span></p>
<p><span style="font-family: 'times new roman', times; font-size: medium;"> </span></p>
<p><span style="font-family: 'times new roman', times; font-size: medium;">(b) (i) The equation of OD is</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(r = \lambda \left( {\begin{array}{*{20}{c}}<br> 0 \\ <br> 2 \\ <br> { - 2} <br>\end{array}} \right)\), \(\left( {{\text{or }}r = \lambda \left( {\begin{array}{*{20}{c}}<br> 0 \\ <br> 1 \\ <br> { - 1} <br>\end{array}} \right)} \right)\) <strong><em>M1</em></strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">This meets \(\pi \) where</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(2\lambda + 2\lambda = - 1\) <strong><em>(M1)</em></strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(\lambda = - \frac{1}{4}\) <strong><em>A1</em></strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">Coordinates of D are \(\left( {0, - \frac{1}{2},\frac{1}{2}} \right)\) <strong><em>A1</em></strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"><strong><em> </em></strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">(ii) \(\left| {\overrightarrow {{\text{OD}}} } \right| = \sqrt {0 + {{\left( { - \frac{1}{2}} \right)}^2} + {{\left( {\frac{1}{2}} \right)}^2}} = \frac{1}{{\sqrt 2 }}\) <strong><em>(M1)A1</em></strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"><strong><em>[6 marks]</em></strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"><strong><em>Total [20 marks]</em></strong></span></p>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question">
<p style="margin: 0.0px 0.0px 10.0px 0.0px; font: 22.0px Times; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;">It was disappointing to see that a number of candidates did not appear to be well prepared for this question and made no progress at all. There were a number of schools where no candidate made any appreciable progress with the question. A good number of students, however, were successful with part (a) (i). A good number of candidates were also successful with part a (ii) but few realised that the shortest distance was the height of the triangle. Candidates used a variety of methods to answer (a) (iii) but again a reasonable number of correct answers were seen. Candidates also had a reasonable degree of success with part (b), with a respectable number of correct answers seen.</span></p>
</div>
<br><hr><br><div class="specification">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 25.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">Consider the points A(1, 2, 3), B(1, 0, 5) and C(2, −1, 4).</span></p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 29.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">Find \(\overrightarrow {{\text{AB}}} \times \overrightarrow {{\text{AC}}} \).</span></p>
<div class="marks">[4]</div>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 23.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">Hence find the area of the triangle ABC.</span></p>
<div class="marks">[2]</div>
<div class="question_part_label">b.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(\overrightarrow {{\text{AB}}} = \left( {\begin{array}{*{20}{c}}<br> 1 \\ <br> 0 \\ <br> 5 <br>\end{array}} \right) - \left( {\begin{array}{*{20}{c}}<br> 1 \\ <br> 2 \\ <br> 3 <br>\end{array}} \right) = \left( {\begin{array}{*{20}{c}}<br> 0 \\ <br> { - 2} \\ <br> 2 <br>\end{array}} \right)\), \(\overrightarrow {{\text{AC}}} = \left( {\begin{array}{*{20}{c}}<br> 2 \\ <br> { - 1} \\ <br> 4 <br>\end{array}} \right) - \left( {\begin{array}{*{20}{c}}<br> 1 \\ <br> 2 \\ <br> 3 <br>\end{array}} \right) = \left( {\begin{array}{*{20}{c}}<br> 1 \\ <br> { - 3} \\ <br> 1 <br>\end{array}} \right)\) <strong><em>A1A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>Note:</strong> Award the above marks if the components are seen in the line below.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"> </span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(\overrightarrow {{\text{AB}}} \times \overrightarrow {{\text{AC}}} = \left| {\begin{array}{*{20}{c}}<br> i&j&k \\ <br> 0&{ - 2}&2 \\ <br> 1&{ - 3}&1 <br>\end{array}} \right| = \left( {\begin{array}{*{20}{c}}<br> 4 \\ <br> 2 \\ <br> 2 <br>\end{array}} \right)\) <strong><em>(M1)A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[4 marks]</em></strong></span></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 29.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">area \( = \frac{1}{2}\left| {\left( {\overrightarrow {{\text{AB}}} \times \overrightarrow {{\text{AC}}} } \right)} \right|\) <strong><em>(M1)</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 29.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = \frac{1}{2}\sqrt {{4^2} + {2^2} + {2^2}} = \frac{1}{2}\sqrt {24} {\text{ }}\left( { = \sqrt 6 } \right)\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 29.0px Helvetica; min-height: 35.0px;"><strong style="font-family: 'times new roman', times; font-size: medium;">Note:</strong><span style="font-family: 'times new roman', times; font-size: medium;"> Award </span><strong style="font-family: 'times new roman', times; font-size: medium;"><em>M0A0</em></strong><span style="font-family: 'times new roman', times; font-size: medium;"> for attempts that do not involve the answer to (a).</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 29.0px Helvetica; min-height: 35.0px;"><strong style="font-family: 'times new roman', times; font-size: medium;"><em>[2 marks]</em></strong></p>
<div class="question_part_label">b.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">Candidates showed a good understanding of the vector techniques required in this question.</span></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">Candidates showed a good understanding of the vector techniques required in this question.</span></p>
<div class="question_part_label">b.</div>
</div>
<br><hr><br><div class="specification">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">In the diagram below, [AB] is a diameter of the circle with centre O. Point C is on the circumference of the circle. Let \(\overrightarrow {{\text{OB}}} = \boldsymbol{b} \) and \(\overrightarrow {{\text{OC}}} = \boldsymbol{c}\) .</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"> </span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><img style="display: block; margin-left: auto; margin-right: auto;" 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" alt></p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">Find an expression for \(\overrightarrow {{\text{CB}}} \) and for \(\overrightarrow {{\text{AC}}} \) in terms of \(\boldsymbol{b}\) and \(\boldsymbol{c}\) .</span></p>
<div class="marks">[2]</div>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">Hence prove that \({\rm{A\hat CB}}\) is a right angle.</span></p>
<div class="marks">[3]</div>
<div class="question_part_label">b.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 23.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(\overrightarrow {{\text{CB}}} = \boldsymbol{b} - \boldsymbol{c}\) , \(\overrightarrow {{\text{AC}}} = \boldsymbol{b} + \boldsymbol{c}\) <strong><em>A1A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 23.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>Note:</strong> Condone absence of vector notation in (a).</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 23.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><em><strong>[2 marks]</strong></em><br></span></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 24.0px Times; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;">\(\overrightarrow {{\text{AC}}} \cdot \overrightarrow {{\text{CB}}} = \)(<strong><em>b</em></strong> + <strong><em>c</em></strong>)\( \cdot \)(<strong><em>b</em></strong> – <strong><em>c</em></strong>) <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 24.0px Times; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;">= \(|\)<strong><em>b</em></strong>\({|^2}\) – \(|\)<strong><em>c</em></strong>\({|^2}\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 24.0px Times; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;">= 0 since \(|\)<strong><em>b</em></strong>\(|\) = \(|\)<strong><em>c</em></strong>\(|\) <strong><em>R1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 24.0px Times; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>Note:</strong> Only award the <strong><em>A1</em></strong> and <strong><em>R1</em></strong> if working indicates that they understand that they are working with vectors.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 24.0px Times; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;"> </span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 24.0px Times; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;">so \(\overrightarrow {{\text{AC}}} \) is perpendicular to \(\overrightarrow {{\text{CB}}} \) <em>i.e.</em> \({\rm{A\hat CB}}\) is a right angle <strong><em>AG</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 24.0px Times; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[3 marks]</em></strong></span></p>
<div class="question_part_label">b.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 22.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">Most candidates were able to find the expressions for the two vectors although a number were not able to do this. Most then tried to use Pythagoras’ theorem and confused scalars and vectors. There were few correct responses to the second part. Candidates did not seem to be able to use the algebra of vectors comfortably. </span></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 22.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">Most candidates were able to find the expressions for the two vectors although a number were not able to do this. Most then tried to use Pythagoras’ theorem and confused scalars and vectors. There were few correct responses to the second part. Candidates did not seem to be able to use the algebra of vectors comfortably. </span></p>
<div class="question_part_label">b.</div>
</div>
<br><hr><br><div class="question">
<p class="p1">Find the coordinates of the point of intersection of the planes defined by the equations \(x + y + z = 3,{\text{ }}x - y + z = 5\) and \(x + y + 2z = 6\).</p>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question">
<p class="p1"><strong>METHOD 1</strong></p>
<p class="p1">for eliminating one variable from two equations <span class="Apple-converted-space"> </span><strong><em>(M1)</em></strong></p>
<p class="p1"><em>eg</em>, \(\left\{ {\begin{array}{*{20}{l}} {(x + y + z = 3)} \\ {2x + 2z = 8} \\ {2x + 3z = 11} \end{array}} \right.\) <span class="Apple-converted-space"> </span><strong><em>A1A1</em></strong></p>
<p class="p1">for finding correctly one coordinate</p>
<p class="p1"><em>eg</em>, \(\left\{ {\begin{array}{*{20}{l}} {(x + y + z = 3)} \\ {(2x + 2z = 8)} \\ {z = 3} \end{array}} \right.\) <span class="Apple-converted-space"> </span><strong><em>A1</em></strong></p>
<p class="p1">for finding correctly the other two coordinates <span class="Apple-converted-space"> </span><strong><em>A1</em></strong></p>
<p class="p2">\( \Rightarrow \left\{ {\begin{array}{*{20}{l}} {x = 1} \\ {y = - 1} \\ {z = 3} \end{array}} \right.\)</p>
<p class="p2">the intersection point has coordinates \((1,{\text{ }} - 1,{\text{ }}3)\)</p>
<p class="p1"><strong>METHOD 2</strong></p>
<p class="p1">for eliminating two variables from two equations or using row reduction <span class="Apple-converted-space"> </span><strong><em>(M1)</em></strong></p>
<p class="p1"><em>eg</em>, \(\left\{ {\begin{array}{*{20}{l}} {(x + y + z = 3)} \\ { - 2 = 2} \\ {z = 3} \end{array}} \right.\) <strong>or</strong> \(\left( {\begin{array}{*{20}{c}} 1&1&1 \\ 0&{ - 2}&0 \\ 0&0&1 \end{array}\left| {\begin{array}{*{20}{c}} 3 \\ 2 \\ 3 \end{array}} \right.} \right)\) <span class="Apple-converted-space"> </span><strong><em>A1A1</em></strong></p>
<p class="p1">for finding correctly the other coordinates <span class="Apple-converted-space"> </span><strong><em>A1A1</em></strong></p>
<p class="p1"><span class="s1">\( \Rightarrow \left\{ {\begin{array}{*{20}{l}} {x = 1} \\ {y = - 1} \\ {(z = 3)} \end{array}} \right.\) </span><strong>or</strong> \(\left( {\begin{array}{*{20}{c}} 1&0&0 \\ 0&1&0 \\ 0&0&1 \end{array}\left| {\begin{array}{*{20}{c}} 1 \\ { - 1} \\ 3 \end{array}} \right.} \right)\)</p>
<p class="p2">the intersection point has coordinates \((1,{\text{ }} - 1,{\text{ }}3)\)</p>
<p class="p1"><strong>METHOD 3</strong></p>
<p class="p2"><span class="Apple-converted-space">\(\left| {\begin{array}{*{20}{c}} 1&1&1 \\ 1&{ - 1}&1 \\ 1&1&2 \end{array}} \right| = - 2\) </span><span class="s2"><strong><em>(A1)</em></strong></span></p>
<p class="p1">attempt to use Cramer’s rule <span class="Apple-converted-space"> </span><strong><em>M1</em></strong></p>
<p class="p2"><span class="Apple-converted-space">\(x = \frac{{\left| {\begin{array}{*{20}{c}} 3&1&1 \\ 5&{ - 1}&1 \\ 6&1&2 \end{array}} \right|}}{{ - 2}} = \frac{{ - 2}}{{ - 2}} = 1\) </span><span class="s2"><strong><em>A1</em></strong></span></p>
<p class="p2"><span class="Apple-converted-space">\(y = \frac{{\left| {\begin{array}{*{20}{c}} 1&3&1 \\ 1&5&1 \\ 1&6&2 \end{array}} \right|}}{{ - 2}} = \frac{2}{{ - 2}} = - 1\) </span><span class="s2"><strong><em>A1</em></strong></span></p>
<p class="p2"><span class="Apple-converted-space">\(z = \frac{{\left| {\begin{array}{*{20}{c}} 1&1&3 \\ 1&{ - 1}&5 \\ 1&1&6 \end{array}} \right|}}{{ - 2}} = \frac{{ - 6}}{{ - 2}} = 3\) </span><span class="s2"><strong><em>A1</em></strong></span></p>
<p class="p3"> </p>
<p class="p1"><strong>Note: <span class="Apple-converted-space"> </span></strong>Award <strong><em>M1 </em></strong>only if candidate attempts to determine at least one of the variables using this method.</p>
<p class="p3"> </p>
<p class="p1"><strong><em>[5 marks]</em></strong></p>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question">
[N/A]
</div>
<br><hr><br><div class="specification">
<p>The following figure shows a square based pyramid with vertices at O(0, 0, 0), A(1, 0, 0), B(1, 1, 0), C(0, 1, 0) and D(0, 0, 1).</p>
<p style="text-align: center;"><img src="data:image/png;base64,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"></p>
</div>
<div class="specification">
<p>The Cartesian equation of the plane \({\Pi _2}\), passing through the points B , C and D , is \(y + z = 1\).</p>
</div>
<div class="specification">
<p>The plane \({\Pi _3}\) passes through O and is normal to the line BD.</p>
</div>
<div class="specification">
<p>\({\Pi _3}\) cuts AD and BD at the points P and Q respectively.</p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Find the Cartesian equation of the plane \({\Pi _1}\), passing through the points A , B and D.</p>
<div class="marks">[3]</div>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Find the angle between the faces ABD and BCD.</p>
<div class="marks">[4]</div>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Find the Cartesian equation of \({\Pi _3}\).</p>
<div class="marks">[3]</div>
<div class="question_part_label">c.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Show that P is the midpoint of AD.</p>
<div class="marks">[4]</div>
<div class="question_part_label">d.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Find the area of the triangle OPQ.</p>
<div class="marks">[5]</div>
<div class="question_part_label">e.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p>recognising normal to plane or attempting to find cross product of two vectors lying in the plane <em><strong>(M1)</strong></em></p>
<p>for example, \(\mathop {{\text{AB}}}\limits^ \to \,\, \times \mathop {{\text{AD}}}\limits^ \to = \left( \begin{gathered}<br> 0 \hfill \\<br> 1 \hfill \\<br> 0 \hfill \\ <br>\end{gathered} \right) \times \left( \begin{gathered}<br> - 1 \hfill \\<br> \,0 \hfill \\<br> \,1 \hfill \\ <br>\end{gathered} \right) = \left( \begin{gathered}<br> 1 \hfill \\<br> 0 \hfill \\<br> 1 \hfill \\ <br>\end{gathered} \right)\) <em><strong>(A1)</strong></em></p>
<p>\({\Pi _1}\,{\text{:}}\,\,x + z = 1\) <em><strong>A1</strong></em></p>
<p><em><strong>[3 marks]</strong></em></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><strong>EITHER</strong></p>
<p>\(\left( \begin{gathered}<br> 1 \hfill \\<br> 0 \hfill \\<br> 1 \hfill \\ <br>\end{gathered} \right) \bullet \left( \begin{gathered}<br> 0 \hfill \\<br> 1 \hfill \\<br> 1 \hfill \\ <br>\end{gathered} \right) = 1 = \sqrt 2 \sqrt 2 \,{\text{cos}}\,\theta \) <em><strong> M1A1</strong></em></p>
<p><strong>OR</strong></p>
<p>\(\left| {\left( \begin{gathered}<br> 1 \hfill \\<br> 0 \hfill \\<br> 1 \hfill \\ <br>\end{gathered} \right) \times \left( \begin{gathered}<br> 0 \hfill \\<br> 1 \hfill \\<br> 1 \hfill \\ <br>\end{gathered} \right)} \right| = \sqrt 3 = \sqrt 2 \sqrt 2 \,{\text{sin}}\,\theta \) <em><strong>M1A1</strong></em></p>
<p><strong>Note:</strong> <em><strong>M1</strong> </em>is for an attempt to find the scalar or vector product of the two normal vectors.</p>
<p>\( \Rightarrow \theta = 60^\circ \left( { = \frac{\pi }{3}} \right)\) <em><strong>A1</strong></em></p>
<p>angle between faces is \(20^\circ \left( { = \frac{{2\pi }}{3}} \right)\) <em><strong>A1</strong></em></p>
<p><em><strong>[4 marks]</strong></em></p>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p>\(\mathop {{\text{DB}}}\limits^ \to = \left( \begin{gathered}<br> \,1 \hfill \\<br> \,1 \hfill \\<br> - 1 \hfill \\ <br>\end{gathered} \right)\) or \(\mathop {{\text{BD}}}\limits^ \to = \left( \begin{gathered}<br> - 1 \hfill \\<br> - 1 \hfill \\<br> \,1 \hfill \\ <br>\end{gathered} \right)\) <em><strong>(A1)</strong></em></p>
<p>\({\Pi _3}\,{\text{:}}\,\,x + y - z = k\) <em><strong>(M1)</strong></em></p>
<p>\({\Pi _3}\,{\text{:}}\,\,x + y - z = 0\) <em><strong>A1</strong></em></p>
<p><em><strong>[3 marks]</strong></em></p>
<div class="question_part_label">c.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><strong>METHOD 1</strong></p>
<p>line AD : (<strong>r</strong> =)\(\left( \begin{gathered}<br> 0 \hfill \\<br> 0 \hfill \\<br> 1 \hfill \\ <br>\end{gathered} \right) + \lambda \left( \begin{gathered}<br> \,1 \hfill \\<br> \,0 \hfill \\<br> - 1 \hfill \\ <br>\end{gathered} \right)\) <em><strong>M1A1</strong></em></p>
<p>intersects \({\Pi _3}\) when \(\lambda - \left( {1 - \lambda } \right) = 0\) <em><strong>M1</strong></em></p>
<p>so \(\lambda = \frac{1}{2}\) <em><strong>A1</strong></em></p>
<p>hence P is the midpoint of AD <em><strong>AG</strong></em></p>
<p> </p>
<p><strong>METHOD 2</strong></p>
<p>midpoint of AD is (0.5, 0, 0.5) <em><strong>(M1)A1</strong></em></p>
<p>substitute into \(x + y - z = 0\) <em><strong> M1</strong></em></p>
<p>0.5 + 0.5 − 0.5 = 0 <em><strong>A1</strong></em></p>
<p>hence P is the midpoint of AD <em><strong> AG</strong></em></p>
<p><em><strong>[4 marks]</strong></em></p>
<div class="question_part_label">d.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><strong>METHOD 1</strong></p>
<p>\({\text{OP}} = \frac{1}{{\sqrt 2 }},\,\,{\text{O}}\mathop {\text{P}}\limits^ \wedge {\text{Q}} = 90^\circ ,\,\,{\text{O}}\mathop {\text{Q}}\limits^ \wedge {\text{P}} = 60^\circ \) <em><strong>A1A1A1</strong></em></p>
<p>\({\text{PQ}} = \frac{1}{{\sqrt 6 }}\) <em><strong> A1</strong></em></p>
<p>area \( = \frac{1}{{2\sqrt {12} }} = \frac{1}{{4\sqrt 3 }} = \frac{{\sqrt 3 }}{{12}}\) <em><strong>A1</strong></em></p>
<p> </p>
<p><strong>METHOD 2</strong></p>
<p>line BD : (<strong>r </strong> =)\(\left( \begin{gathered}<br> 1 \hfill \\<br> 1 \hfill \\<br> 0 \hfill \\ <br>\end{gathered} \right) + \lambda \left( \begin{gathered}<br> - 1 \hfill \\<br> - 1 \hfill \\<br> \,1 \hfill \\ <br>\end{gathered} \right)\)</p>
<p>\( \Rightarrow \lambda = \frac{2}{3}\) <em><strong>(A1)</strong></em></p>
<p>\(\mathop {{\text{OQ}}}\limits^ \to = \left( \begin{gathered}<br> \frac{1}{3} \hfill \\<br> \frac{1}{3} \hfill \\<br> \frac{2}{3} \hfill \\ <br>\end{gathered} \right)\) <em><strong>A1</strong></em></p>
<p>area = \(\frac{1}{2}\left| {\mathop {{\text{OP}}}\limits^ \to \, \times \mathop {{\text{OQ}}}\limits^ \to } \right|\) <em><strong>M1</strong></em></p>
<p>\(\mathop {{\text{OP}}}\limits^ \to = \left( \begin{gathered}<br> \frac{1}{2} \hfill \\<br> 0 \hfill \\<br> \frac{1}{2} \hfill \\ <br>\end{gathered} \right)\) <em><strong>A1</strong></em></p>
<p><strong>Note</strong>: This <em><strong>A1</strong> </em>is dependent on <em><strong>M1</strong></em>.</p>
<p>area = \(\frac{{\sqrt 3 }}{{12}}\) <em><strong>A1</strong></em></p>
<p><em><strong>[5 marks]</strong></em></p>
<div class="question_part_label">e.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">c.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">d.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">e.</div>
</div>
<br><hr><br><div class="question">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">PQRS is a rhombus. Given that \(\overrightarrow {{\text{PQ}}} = \) \(\boldsymbol{a}\) and \(\overrightarrow {{\text{QR}}} = \) \(\boldsymbol{b}\),</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">(a) express the vectors \(\overrightarrow {{\text{PR}}} \) and \(\overrightarrow {{\text{QS}}} \) in terms of \(\boldsymbol{a}\) and \(\boldsymbol{b}\);</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">(b) hence show that the diagonals in a rhombus intersect at right angles.</span></p>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(a) \(\overrightarrow {{\text{PR}}} = \) <strong>a</strong> + <em><strong>b A1</strong></em></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(\overrightarrow {{\text{QS}}} = \) <strong><em>b</em></strong> − <strong><em>a</em></strong> <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[2 marks]</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em> </em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(b) \(\overrightarrow {{\text{PR}}} \cdot \overrightarrow {{\text{QS}}} = \) (<strong><em>a</em></strong> + <strong><em>b</em></strong>) \( \cdot \) (<strong><em>b </em></strong>− <strong><em>a</em></strong>) <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.5px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = |\)<strong><em>b</em></strong>\({|^2} - |\)<strong><em>a</em></strong>\({|^2}\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.5px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">for a rhombus \(|\)</span><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>a</em></strong>\(| = |\)<strong><em>b</em></strong>\(|\) <strong><em>R1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">hence \(|\)<em><strong>b</strong></em>\({|^2} - |\)<em><strong>a</strong></em>\({|^2} = 0\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica; min-height: 25.0px;"> </p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>Note:</strong> Do not award the final <strong><em>A1 </em></strong>unless <strong><em>R1 </em></strong>is awarded.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman'; min-height: 25.0px;"> </p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">hence the diagonals intersect at right angles <strong><em>AG</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[4 marks]</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em> </em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>Total [6 marks]</em></strong></span></p>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question">
[N/A]
</div>
<br><hr><br><div class="question">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">The vectors <strong><em>a</em></strong> , <strong><em>b</em></strong> , <strong><em>c</em></strong> satisfy the equation <strong><em>a</em></strong> + <strong><em>b</em></strong> + <strong><em>c</em></strong> = <strong>0</strong> . Show that <strong><em>a</em></strong> \( \times \) <strong><em>b</em></strong> = <strong><em>b</em></strong> \( \times \) <strong><em>c</em></strong> = <strong><em>c</em></strong> \( \times \) <strong><em>a</em></strong> .</span></p>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 31.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">taking cross products with <strong><em>a</em></strong>, <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 31.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>a</em></strong> \( \times \) (<strong><em>a</em></strong> + <strong><em>b</em></strong> + <strong><em>c</em></strong>) = <strong><em>a</em></strong> \( \times \) <strong><em>0</em></strong> = <strong><em>0</em></strong> <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 31.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">using the algebraic properties of vectors and the fact that <strong><em>a</em></strong> \( \times \) <strong><em>a</em></strong> = <strong><em>0</em></strong> , <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 31.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>a</em></strong> \( \times \) <strong><em>b</em></strong> + <strong><em>a</em></strong> \( \times \) <strong><em>c</em></strong> = <strong><em>0</em></strong> <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 31.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>a</em></strong> \( \times \) <strong><em>b</em></strong> = <strong><em>c</em></strong> \( \times \) <strong><em>a</em></strong> <strong><em>AG</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 31.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">taking cross products with <strong><em>b</em></strong>, <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 31.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>b</em></strong> \( \times \) (<strong><em>a</em></strong> + <strong><em>b</em></strong> + <strong><em>c</em></strong>) = <strong><em>0</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 31.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>b</em></strong> \( \times \) <strong><em>a</em></strong> + <strong><em>b</em></strong> \( \times \) <strong><em>c</em></strong> = <strong><em>0</em></strong> <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 31.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>a</em></strong> \( \times \) <strong><em>b</em></strong> = <strong><em>b</em></strong> \( \times \) <strong><em>c</em></strong> <strong><em>AG</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 31.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">this completes the proof</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 31.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[6 marks]</em></strong></span></p>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question">
[N/A]
</div>
<br><hr><br><div class="question" style="padding-left: 20px; padding-right: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">Consider the vectors <strong><em>a</em></strong> = 6<strong><em>i</em></strong> + 3<strong><em>j</em></strong> + 2<strong><em>k</em></strong>, <strong><em>b</em></strong> = −3<strong><em>j</em></strong> + 4<strong><em>k</em></strong>.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(i) Find the cosine of the angle between vectors <strong><em>a</em></strong> and <strong><em>b</em></strong>.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(ii) Find <strong><em>a</em></strong> \( \times \) <strong><em>b</em></strong>.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(iii) <strong>Hence</strong> find the Cartesian equation of the plane \(\prod \) containing the vectors <strong><em>a</em></strong> and <strong><em>b</em></strong> and passing through the point (1, 1, −1).</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(iv) The plane \(\prod \) intersects the <em>x-y</em> plane in the line <em>l</em>. Find the area of the finite triangular region enclosed by <em>l</em>, the <em>x</em>-axis and the <em>y</em>-axis.</span></p>
<div class="marks">[11]</div>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 29.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">Given two vectors <strong><em>p</em></strong> and <strong><em>q</em></strong>,</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 29.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(i) show that <strong><em>p</em></strong>\( \cdot \)<strong><em>p</em></strong> = \(|\)<strong><em>p</em></strong>\({|^2}\);</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 29.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(ii) hence, or otherwise, show that \(|\)<strong><em>p</em></strong> + <strong><em>q</em></strong>\({|^2}\) = \(|\)<strong><em>p</em></strong>\({|^2}\) + 2<strong><em>p</em></strong>\( \cdot \)<strong><em>q</em></strong> + \(|\)<strong><em>q</em></strong>\({|^2}\);</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 29.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(iii) deduce that \(|\)<strong><em>p</em></strong> + <strong><em>q</em></strong>\(|\) ≤ \(|\)<strong><em>p</em></strong>\(|\) + \(|\)<strong><em>q</em></strong>\(|\).</span></p>
<div class="marks">[8]</div>
<div class="question_part_label">b.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 32.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(i) use of <strong><em>a</em></strong>\( \cdot \)<strong><em>b</em></strong> = \(|\)<strong><em>a</em></strong>\(|\)\(|\)<strong><em>b</em></strong>\(|\cos \theta \) <strong><em>(M1)</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 32.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>a</em></strong>\( \cdot \)<strong><em>b</em></strong> = –1 <strong><em>(A1)</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 32.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(|\)<strong><em>a</em></strong>\(|\) = 7, \(|\)<strong><em>b</em></strong>\(|\) = 5 <strong><em>(A1)</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 32.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(\cos \theta = - \frac{1}{{35}}\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 32.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em> </em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 32.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(ii) the required cross product is</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 32.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(\left| {\begin{array}{*{20}{c}}<br> i&j&k \\ <br> 6&3&2 \\ <br> 0&{ - 3}&4 <br>\end{array}} \right| = \) 18<em><strong>i</strong></em> - 24<em><strong>j</strong></em> -18<em><strong>k</strong></em> <strong><em>M1A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 32.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em> </em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 32.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(iii) using <strong><em>r</em></strong>\( \cdot \)<strong><em>n</em></strong> = <strong><em>p</em></strong>\( \cdot \)<strong><em>n</em></strong> the equation of the plane is <strong><em>(M1)</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 32.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(18x - 24y - 18z = 12\,\,\,\,\,(3x - 4y - 3z = 2)\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 32.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em> </em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 32.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(iv) recognizing that <em>z</em> = 0 <strong><em>(M1)</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 32.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><em>x</em>-intercept \( = \frac{2}{3}\), <em>y</em>-intercept \( = - \frac{1}{2}\) <strong><em>(A1)</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 32.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">area \( = \left( {\frac{2}{3}} \right)\left( {\frac{1}{2}} \right)\left( {\frac{1}{2}} \right) = \frac{1}{6}\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 32.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[11 marks]</em></strong></span></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(i) <strong><em>p</em></strong>\( \cdot \)<strong><em>p</em></strong> = \(|\)<strong><em>p</em></strong>\(|\)\(|\)<strong><em>p</em></strong>\(|\cos 0\) <strong><em>M1A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">= \(|\)<strong><em>p</em></strong>\({|^2}\) <strong><em>AG</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em> </em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(ii) consider the LHS, and use of result from part (i)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(|\)<strong><em>p</em></strong> + <strong><em>q</em></strong>\({|^2}\) = (<strong><em>p</em></strong> + <strong><em>q</em></strong>)\( \cdot \)(<strong><em>p</em></strong> + <strong><em>q</em></strong>) <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">= <strong><em>p</em></strong>\( \cdot \)<strong><em>p</em></strong> + <strong><em>p</em></strong>\( \cdot \)<strong><em>q</em></strong> + <strong><em>q</em></strong>\( \cdot \)<strong><em>p</em></strong> + <strong><em>q</em></strong>\( \cdot \)<strong><em>q</em></strong> <strong><em>(A1)</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">= <strong><em>p</em></strong>\( \cdot \)<strong><em>p</em></strong> + 2<strong><em>p</em></strong>\( \cdot \)<strong><em>q</em></strong> + <strong><em>q</em></strong>\( \cdot \)<strong><em>q</em></strong> <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">= \(|\)<strong><em>p</em></strong>\({|^2}\) + 2<strong><em>p</em></strong>\( \cdot \)<strong><em>q</em></strong> + \(|\)<strong><em>q</em></strong>\({|^2}\) <strong><em>AG</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em> </em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(iii) <strong>EITHER</strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">use of <strong><em>p</em></strong>\( \cdot \)<strong><em>q</em></strong> \( \leqslant \) \(|\)<strong><em>p</em></strong>\(|\)\(|\)<strong><em>q</em></strong>\(|\) <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">so 0 \( \leqslant \) \(|\)<strong><em>p</em></strong> + <strong><em>q</em></strong>\({|^2}\) = \(|\)<strong><em>p</em></strong>\({|^2}\) + 2<strong><em>p</em></strong>\( \cdot \)<strong><em>q</em></strong> + \(|\)<strong><em>q</em></strong>\({|^2}\) \( \leqslant \) \(|\)<strong><em>p</em></strong>\({|^2}\) + 2 \(|\)<strong><em>p</em></strong>\(|\)\(|\)<strong><em>q</em></strong>\(|\) + \(|\)q\({|^2}\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">take square root (of these positive quantities) to establish <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(|\)</span><strong style="font-family: 'times new roman', times; font-size: medium;"><em>p</em></strong><span style="font-family: 'times new roman', times; font-size: medium;"> + </span><strong style="font-family: 'times new roman', times; font-size: medium;"><em>q</em></strong><span style="font-family: 'times new roman', times; font-size: medium;">\(|\) \( \leqslant \) \(|\)</span><strong style="font-family: 'times new roman', times; font-size: medium;"><em>p</em></strong><span style="font-family: 'times new roman', times; font-size: medium;">\(|\) + \(|\)</span><strong style="font-family: 'times new roman', times; font-size: medium;"><em>q</em></strong><span style="font-family: 'times new roman', times; font-size: medium;">\(|\) </span><strong style="font-family: 'times new roman', times; font-size: medium;"><em>AG</em></strong></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>OR</strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><img src="data:image/png;base64,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" alt> <strong><em>M1M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>Note:</strong> Award <strong><em>M1</em></strong> for correct diagram and <strong><em>M1</em></strong> for correct labelling of vectors including arrows.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"> </span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">since the sum of any two sides of a triangle is greater than the third side,</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(|\)<strong><em>p</em></strong>\(|\) + \(|\)<strong><em>q</em></strong>\(|\) > \(|\)<strong><em>p</em></strong> + <strong><em>q</em></strong>\(|\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">when <strong><em>p</em></strong> and <strong><em>q</em></strong> are collinear \(|\)<strong><em>p</em></strong>\(|\) + \(|\)<strong><em>q</em></strong>\(|\) = \(|\)<strong><em>p</em></strong> + <strong><em>q</em></strong>\(|\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( \Rightarrow |\)<strong><em>p</em></strong> + <strong><em>q</em></strong>\(|\) \( \leqslant \) \(|\)<strong><em>p</em></strong>\(|\) + \(|\)<strong><em>q</em></strong>\(|\) <strong><em>AG</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em> </em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[8 marks]</em></strong></span></p>
<div class="question_part_label">b.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">This was the most accessible question in section B for the candidates. The majority of candidates produced partially correct answers to part (a), with nearly all candidates being able to use the scalar and vector product. Candidates found part (iv) harder and often did not appreciate the significance of letting <em>z</em> = 0. Candidates clearly found part (b) harder and again this was a point where candidates lost time. Many candidates attempted this using components, which was fine in part (i), fine, but time consuming in part (ii), and extremely complicated in part (iii). A number of candidates lost marks because they were careless in showing their working in part (ii) which required them to “show that”.</span></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 25.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">This was the most accessible question in section B for the candidates. The majority of candidates produced partially correct answers to part (a), with nearly all candidates being able to use the scalar and vector product. Candidates found part (iv) harder and often did not appreciate the significance of letting <em>z</em> = 0. Candidates clearly found part (b) harder and again this was a point where candidates lost time. Many candidates attempted this using components, which was fine in part (i), fine, but time consuming in part (ii), and extremely complicated in part (iii). A number of candidates lost marks because they were careless in showing their working in part (ii) which required them to “show that”. </span></p>
<div class="question_part_label">b.</div>
</div>
<br><hr><br><div class="question">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 25.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(a) Show that a Cartesian equation of the line, \({l_1}\), containing points A(1, −1, 2) and B(3, 0, 3) has the form \(\frac{{x - 1}}{2} = \frac{{y + 1}}{1} = \frac{{z - 2}}{1}\).</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 25.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(b) An equation of a second line, \({l_2}\), has the form \(\frac{{x - 1}}{1} = \frac{{y - 2}}{2} = \frac{{z - 3}}{1}\). Show that the lines \({l_1}\) and \({l_2}\) intersect, and find the coordinates of their point of intersection.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 25.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(c) Given that direction vectors of \({l_1}\) and \({l_2}\) are <strong><em>d</em></strong>\(_1\) and <strong><em>d</em></strong>\(_2\) respectively, determine <strong><em>d</em></strong>\(_1 \times \) <strong><em>d</em></strong>\(_2\).</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 25.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(d) Show that a Cartesian equation of the plane, \(\prod \), that contains \({l_1}\) and \({l_2}\) is \( - x - y + 3z = 6\).</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 25.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(e) Find a vector equation of the line \({l_3}\) which is perpendicular to the plane \(\prod \) and passes through the point T(3, 1, −4).</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 25.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(f) (i) Find the point of intersection of the line \({l_3}\) and the plane \(\prod \).</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 25.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"> (ii) Find the coordinates of \({{\text{T}}}'\), the reflection of the point T in the plane \(\prod \).</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 25.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"> (iii) Hence find the magnitude of the vector \(\overrightarrow {{\text{TT}}'} \).</span></p>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 24.0px Times; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;">(a) identifies a direction vector <em>e.g.</em> \(\overrightarrow {{\text{AB}}} = \left( {\begin{array}{*{20}{c}}<br> 2 \\ <br> 1 \\ <br> 1 <br>\end{array}} \right)\) or \(\overrightarrow {{\text{BA}}} = \left( {\begin{array}{*{20}{c}}<br> { - 2} \\ <br> { - 1} \\ <br> { - 1} <br>\end{array}} \right)\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 24.0px Times; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;">identifies the point (1, –1, 2) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 24.0px Times; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;">line \({l_1}:{\text{ }}\frac{{x - 1}}{2} = \frac{{y + 1}}{1} = \frac{{z - 2}}{1}\) <strong><em>AG</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 24.0px Times; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[2 marks]</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 24.0px Times; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;"> </span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 24.0px Times; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;">(b) \(r = \left( {\begin{array}{*{20}{c}}<br> 1 \\ <br> { - 1} \\ <br> 2 <br>\end{array}} \right) + \lambda \left( {\begin{array}{*{20}{c}}<br> 2 \\ <br> 1 \\ <br> 1 <br>\end{array}} \right)\) \(r = \left( {\begin{array}{*{20}{c}}<br> 1 \\ <br> 2 \\ <br> 3 <br>\end{array}} \right) + \mu \left( {\begin{array}{*{20}{c}}<br> 1 \\ <br> 2 \\ <br> 1 <br>\end{array}} \right)\)<br></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 24.0px Times; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;">\(1 + 2\lambda = 1 + \mu ,{\text{ }} - 1 + \lambda = 2 + 2\mu ,{\text{ }}2 + \lambda = 3 + \mu \) <strong><em>(M1)</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 24.0px Times; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;">equating two of the three equations gives \(\lambda = - 1\) and \(\mu = - 2\) <strong><em>A1A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 24.0px Times; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;">check in the third equation</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 24.0px Times; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;">satisfies third equation therefore the lines intersect <strong><em>R1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 24.0px Times; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;">therefore coordinates of intersection are (–1, –2, 1) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 24.0px Times; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[5 marks]</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 24.0px Times; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;"> </span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 24.0px Times; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;">(c) <strong><em>d</em></strong>\(_1\) = 2<strong><em>i</em></strong> + <strong><em>j</em></strong> + <strong><em>k</em></strong>, <strong><em>d</em></strong>\(_2\) = <strong><em>i</em></strong> + 2<strong><em>j</em></strong> + <strong><em>k</em></strong> <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 24.0px Times; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;"><em><strong>d</strong></em>\(_1\) \( \times \) <em><strong>d</strong></em>\(_2\) \( = \left| {\begin{array}{*{20}{c}}<br> \boldsymbol{i}&\boldsymbol{j}&\boldsymbol{k} \\ <br> 2&1&1 \\ <br> 1&2&1 <br>\end{array}} \right| = \) </span><span style="font-family: 'times new roman', times; font-size: medium;">–<em><strong>i</strong></em> \(-\) <em><strong>j</strong></em> \(+\) 3<em><strong>k</strong></em> <strong><em>M1A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 24.0px Times; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>Note:</strong> Accept scalar multiples of above vectors.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 24.0px Times; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;"> </span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 24.0px Times; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[3 marks]</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 24.0px Times; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;"> </span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 24.0px Times; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;">(d) equation of plane is \( - x - y + 3z = k\) <strong><em>M1A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 24.0px Times; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;">contains (1, 2, 3) \(\left( {{\text{or }}( - 1,{\text{ }} - 2,{\text{ }}1){\text{ or }}(1,{\text{ }} - 1,{\text{ }}2)} \right){\text{ }}\therefore k = - 1 - 2 + 3 \times 3 = 6\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 24.0px Times; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;">\( - x - y + 3z = 6\) <strong><em>AG</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 24.0px Times; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[3 marks]</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 24.0px Times; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;"> </span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 24.0px Times; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;">(e) direction vector of the perpendicular line is \(\left( {\begin{array}{*{20}{c}}<br> { - 1} \\ <br> { - 1} \\ <br> 3 <br>\end{array}} \right)\) <strong><em>(M1)</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 24.0px Times; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;">\(r = \left( {\begin{array}{*{20}{c}}<br> 3 \\ <br> 1 \\ <br> { - 4} <br>\end{array}} \right) + m\left( {\begin{array}{*{20}{c}}<br> { - 1} \\ <br> { - 1} \\ <br> 3 <br>\end{array}} \right)\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 24.0px Times; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>Note:</strong> Award <strong><em>A0</em></strong> if <strong><em>r</em></strong> omitted.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 24.0px Times; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;"> </span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 24.0px Times; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[2 marks]</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 24.0px Times; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;"> </span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 24.0px Times; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;">(f) (i) find point where line meets plane</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 24.0px Times; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;">\( - (3 - m) - (1 - m) + 3( - 4 + 3m) = 6\) <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 24.0px Times; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;"><em>m</em> = 2 <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 24.0px Times; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;">point of intersection is (1, –1, 2) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 24.0px Times; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em> </em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 24.0px Times; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;">(ii) for \({\text{T}}', m = 4\) <strong><em>(M1)</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 24.0px Times; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;">so \({\text{T}}' = \) (\( - 1\), \( - 3\), \(8\)) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 24.0px Times; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em> </em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 24.0px Times; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;">(iii) \(\overrightarrow {{\text{TT}'}} = \sqrt {{{(3 + 1)}^2} + {{(1 + 3)}^2} + {{( - 4 - 8)}^2}} \) <strong><em>(M1)</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 24.0px Times; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = \sqrt {176} \,\,\,\,\,( = 4\sqrt {11} )\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 24.0px Times; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[7 marks]</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 24.0px Times; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em> </em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 24.0px Times; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>Total [22 marks]</em></strong></span></p>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 29.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">This question was done very well by many students. The common errors were using the same variable for line 2 and in stating the vectors in b) were not parallel and therefore the lines did intersect. Many students did not check the solution in order to establish this.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 29.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">When required to give the equation of the line in e) many did not state it as an equation, let alone a vector equation. </span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 29.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">The difference between position vectors and coordinates was not clear on many papers.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 29.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">In f) many used inefficient techniques that were time consuming to find the point of reflection.</span></p>
</div>
<br><hr><br><div class="question">
<p>The acute angle between the vectors 3<em><strong>i</strong></em> − 4<em><strong>j</strong></em> − 5<em><strong>k</strong></em> and 5<em><strong>i</strong></em> − 4<em><strong>j</strong></em> + 3<em><strong>k</strong></em> is denoted by <em>θ</em>.</p>
<p>Find cos <em>θ</em>.</p>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question">
<p>cos <em>θ</em> = \(\frac{{\left( {3i - 4j - 5k} \right) \bullet \left( {5i - 4j + 3k} \right)}}{{\left| {3i - 4j - 5k} \right|\left| {5i - 4j + 3k} \right|}}\) <em><strong>(M1)</strong></em></p>
<p>\( = \frac{{16}}{{\sqrt {50} \sqrt {50} }}\) <em><strong>A1A1</strong></em></p>
<p><strong>Note:</strong> <em><strong>A1</strong></em> for correct numerator and <em><strong>A1</strong> </em>for correct denominator.</p>
<p>\( = \frac{8}{{25}}\left( { = \frac{{16}}{{50}} = 0.32} \right)\) <em><strong> A1</strong></em></p>
<p><em><strong>[4 marks]</strong></em></p>
<p> </p>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question">
[N/A]
</div>
<br><hr><br><div class="question">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 31.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">Let \(\alpha \) be the angle between the unit vectors <strong><em>a</em></strong> and <strong><em>b</em></strong>, where \(0 \leqslant \alpha \leqslant \pi \).</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 31.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(a) Express \(|\)<strong><em>a</em></strong> − <strong><em>b</em></strong>\(|\) and \(|\)<strong><em>a</em></strong> + <strong><em>b</em></strong>\(|\) in terms of \(\alpha \).</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 31.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(b) Hence determine the value of \(\cos \alpha \) for which \(|\)<strong><em>a</em></strong> + <strong><em>b</em></strong>\(|\) = 3 \(|\)<strong><em>a</em></strong> − <strong><em>b</em></strong>\(|\).</span></p>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 23.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>METHOD 1</strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 23.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(a) \(|\)<strong><em>a</em></strong> – <strong><em>b</em></strong>\(|\) = \(\sqrt {{{\left| a \right|}^2} + {{\left| b \right|}^2} - 2\left| a \right|\left| b \right|\cos \alpha } \) <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 23.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = \sqrt {2 - 2\cos \alpha } \) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 23.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(|\)<strong><em>a</em></strong> + <strong><em>b</em></strong>\(|\) = \(\sqrt {{{\left| a \right|}^2} + {{\left| b \right|}^2} - 2\left| a \right|\left| b \right|\cos (\pi - \alpha )} \)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 23.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = \sqrt {2 + 2\cos \alpha } \) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 23.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>Note:</strong> Accept the use of <em>a</em>, <em>b</em> for \(|\)<strong><em>a</em></strong>\(|\), \(|\)<strong><em>b</em></strong>\(|\).</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 23.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"> </span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 23.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(b) \( = \sqrt {2 + 2\cos \alpha } = 3\sqrt {2 - 2\cos \alpha } \) <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 23.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(\cos \alpha = \frac{4}{5}\) <strong><em>A1</em></strong></span><strong style="font-family: 'times new roman', times; font-size: medium;"><em> </em></strong></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 23.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>METHOD 2</strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 23.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(a) \(|\)<strong><em>a</em></strong> – <strong><em>b</em></strong>\(|\) = \(2\sin \frac{\alpha}{2}\) <strong><em>M1A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 23.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(|\)<strong><em>a</em></strong> + <strong><em>b</em></strong>\(|\) = \(2\sin \left( {\frac{\pi }{2} - \frac{\alpha}{2}} \right) = 2\cos \frac{\alpha}{2}\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 23.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>Note:</strong> Accept the use of <em>a</em>, <em>b</em> for \(|\)<strong><em>a</em></strong> \(|\), \(|\)<strong><em>b</em></strong>\(|\).</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 23.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"> </span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 23.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(b) \(2\cos \frac{\alpha}{2} = 6\sin \frac{\alpha }{2}\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 23.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(\tan \frac{\alpha }{2} = \frac{1}{3} \Rightarrow {\cos ^2}\frac{\alpha }{2} = \frac{9}{{10}}\) <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 23.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(\cos \alpha = 2{\cos ^2}\frac{\alpha }{2} - 1 = \frac{4}{5}\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 23.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[5 marks]</em></strong></span></p>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 25.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">To solve this problem, candidates had to know either that (<strong><em>a</em></strong> + <strong><em>b</em></strong>)(<strong><em>a</em></strong> + <strong><em>b</em></strong>) = \(|\)<strong><em>a</em></strong> + <strong><em>b</em></strong>\({|^2}\) or that the diagonals of a parallelogram whose sides are <strong><em>a</em></strong> and <strong><em>b</em></strong> represent the vectors <strong><em>a</em></strong> + <strong><em>b</em></strong> and <strong><em>a</em></strong> – <strong><em>b</em></strong>. It was clear from the scripts that many candidates were unaware of either result and were therefore unable to make any progress in this question.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 25.0px Helvetica;"> </p>
</div>
<br><hr><br><div class="question">
<p>A triangle has vertices A(1, −1, 1), B(1, 1, 0) and C(−1, 1, −1) .</p>
<p>Show that the area of the triangle is \(\sqrt 6 \) .</p>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question">
<p><strong>METHOD 1</strong></p>
<p>for finding two of the following three vectors (or their negatives)</p>
<p>\(\overrightarrow {{\text{AB}}} = \left( {\begin{array}{*{20}{c}}<br> 0 \\ <br> 2 \\ <br> { - 1} <br>\end{array}} \right)\), \(\overrightarrow {{\text{AC}}} = \left( {\begin{array}{*{20}{c}}<br> { - 2} \\ <br> 2 \\ <br> { - 2} <br>\end{array}} \right)\), \(\overrightarrow {{\text{BC}}} = \left( {\begin{array}{*{20}{c}}<br> { - 2} \\ <br> 0 \\ <br> { - 1} <br>\end{array}} \right)\) <strong><em>(A1)(A1)</em></strong></p>
<p>and calculating </p>
<p><strong>EITHER</strong></p>
<p>\(\overrightarrow {{\text{AB}}} \times \overrightarrow {{\text{AC}}} = \left| {\begin{array}{*{20}{c}}<br> i&j&k \\ <br> 0&2&{ - 1} \\ <br> { - 2}&2&{ - 2} <br>\end{array}} \right| = \left( {\begin{array}{*{20}{c}}<br> { - 2} \\ <br> 2 \\ <br> 4 <br>\end{array}} \right)\) <strong><em>M1A1</em></strong></p>
<p>area \(\Delta {\text{ABC}} = \frac{1}{2}\left| {\overrightarrow {{\text{AB}}} \times \overrightarrow {{\text{AC}}} } \right|\) <strong><em>M1</em></strong></p>
<p><strong>OR</strong></p>
<p>\(\overrightarrow {{\text{BA}}} \times \overrightarrow {{\text{BC}}} = \left| {\begin{array}{*{20}{c}}<br> i&j&k \\ <br> 0&{ - 2}&1 \\ <br> { - 2}&0&{ - 1} <br>\end{array}} \right| = \left( {\begin{array}{*{20}{c}}<br> 2 \\ <br> { - 2} \\ <br> { - 4} <br>\end{array}} \right)\) <strong><em>M1A1</em></strong></p>
<p>area \(\Delta {\text{ABC}} = \frac{1}{2}\left| {\overrightarrow {{\text{BA}}} \times \overrightarrow {{\text{BC}}} } \right|\) <strong><em>M1</em></strong></p>
<p><strong>OR</strong></p>
<p>\(\overrightarrow {{\text{CA}}} \times \overrightarrow {{\text{CB}}} = \left| {\begin{array}{*{20}{c}}<br> i&j&k \\ <br> 2&{ - 2}&2 \\ <br> 2&0&1 <br>\end{array}} \right| = \left( {\begin{array}{*{20}{c}}<br> { - 2} \\ <br> 2 \\ <br> 4 <br>\end{array}} \right)\) <strong><em>M1A1</em></strong></p>
<p>area \(\Delta {\text{ABC}} = \frac{1}{2}\left| {\overrightarrow {{\text{CA}}} \times \overrightarrow {{\text{CB}}} } \right|\) <strong><em>M1</em></strong></p>
<p><strong>THEN</strong></p>
<p>area \(\Delta {\text{ABC}} = \frac{{\sqrt {24} }}{2}\) <strong><em>A1</em></strong></p>
<p>\( = \sqrt 6 \) <strong><em>AG N0</em></strong></p>
<p><strong>METHOD 2</strong></p>
<p>for finding two of the following three vectors (or their negatives)</p>
<p>\(\overrightarrow {{\text{AB}}} = \left( {\begin{array}{*{20}{c}}<br> 0 \\ <br> 2 \\ <br> { - 1} <br>\end{array}} \right)\), \(\overrightarrow {{\text{AC}}} = \left( {\begin{array}{*{20}{c}}<br> { - 2} \\ <br> 2 \\ <br> { - 2} <br>\end{array}} \right)\), \(\overrightarrow {{\text{BC}}} = \left( {\begin{array}{*{20}{c}}<br> { - 2} \\ <br> 0 \\ <br> { - 1} <br>\end{array}} \right)\) <strong><em>(A1)(A1)</em></strong></p>
<p><strong>EITHER</strong></p>
<p>\(\cos A = \frac{{\overrightarrow {{\text{AB}}} \cdot \overrightarrow {{\text{AC}}} }}{{\left| {\overrightarrow {{\text{AB}}} } \right|\left| {\overrightarrow {{\text{AC}}} } \right|}}\) <strong><em>M1</em></strong></p>
<p>\( = \frac{6}{{\sqrt 5 \sqrt {12} }} = \frac{6}{{\sqrt {60} }}{\text{ }}\left( {{\text{or }}\frac{3}{{\sqrt {15} }}} \right)\)</p>
<p>\(\sin A = \sqrt {\frac{2}{5}} \) <strong><em>A1</em></strong></p>
<p>area \(\Delta {\text{ABC}} = \frac{1}{2}\left| {\overrightarrow {{\text{AB}}} } \right|\left| {\overrightarrow {{\text{AC}}} } \right|\sin A\) <strong><em>M1</em></strong></p>
<p>\( = \frac{1}{2}\sqrt 5 \sqrt {12} \sqrt {\frac{2}{5}} \)</p>
<p>\( = \frac{1}{2}\sqrt {24} \) <strong><em>A1</em></strong></p>
<p>\( = \sqrt 6 \) <strong><em>AG N0</em></strong></p>
<p><strong>OR</strong></p>
<p>\(\cos B = \frac{{\overrightarrow {{\text{BA}}} \cdot \overrightarrow {{\text{BC}}} }}{{\left| {\overrightarrow {{\text{BA}}} } \right|\left| {\overrightarrow {{\text{BC}}} } \right|}}\) <strong><em>M1</em></strong></p>
<p>\( = - \frac{1}{{\sqrt 5 \sqrt 5 }} = - \frac{1}{5}\)</p>
<p>\(\sin B = \sqrt {\frac{{24}}{{25}}} {\text{ }}\left( {{\text{or }}\frac{{\sqrt {24} }}{5}} \right)\) <strong><em>A1</em></strong></p>
<p>area \(\Delta {\text{ABC}} = \frac{1}{2}\left| {\overrightarrow {{\text{BA}}} } \right|\left| {\overrightarrow {{\text{BC}}} } \right|\sin B\) <strong><em>M1</em></strong></p>
<p>\( = \frac{1}{2}\sqrt 5 \sqrt 5 \sqrt {\frac{{24}}{{25}}} \)</p>
<p>\( = \frac{1}{2}\sqrt {24} \) <strong><em>A1</em></strong></p>
<p>\( = \sqrt 6 \) <strong><em>AG N0</em></strong></p>
<p><strong>OR</strong></p>
<p>\(\cos C = \frac{{\overrightarrow {{\text{CA}}} \cdot \overrightarrow {{\text{CB}}} }}{{\left| {\overrightarrow {{\text{CA}}} } \right|\left| {\overrightarrow {{\text{CB}}} } \right|}}\) <strong><em>M1</em></strong></p>
<p>\( = \frac{6}{{\sqrt {12} \sqrt 5 }} = \frac{6}{{\sqrt {60} }}{\text{ }}\left( {{\text{or }}\frac{3}{{\sqrt {15} }}} \right)\)</p>
<p>\(\sin C = \sqrt {\frac{2}{5}} \) <strong><em>A1</em></strong></p>
<p>area \(\Delta {\text{ABC}} = \frac{1}{2}\left| {\overrightarrow {{\text{CA}}} } \right|\left| {\overrightarrow {{\text{CB}}} } \right|\sin C\) <strong><em>M1</em></strong></p>
<p>\( = \frac{1}{2}\sqrt {12} \sqrt 5 \sqrt {\frac{2}{5}} \)</p>
<p>\( = \frac{1}{2}\sqrt {24} \) <strong><em>A1</em></strong></p>
<p>\( = \sqrt 6 \) <strong><em>AG N0</em></strong></p>
<p><strong>METHOD 3</strong></p>
<p>for finding two of the following three vectors (or their negatives)</p>
<p>\(\overrightarrow {{\text{AB}}} = \left( {\begin{array}{*{20}{c}}<br> 0 \\ <br> 2 \\ <br> { - 1} <br>\end{array}} \right)\), \(\overrightarrow {{\text{AC}}} = \left( {\begin{array}{*{20}{c}}<br> { - 2} \\ <br> 2 \\ <br> { - 2} <br>\end{array}} \right)\), \(\overrightarrow {{\text{BC}}} = \left( {\begin{array}{*{20}{c}}<br> { - 2} \\ <br> 0 \\ <br> { - 1} <br>\end{array}} \right)\) <strong><em>(A1)(A1)</em></strong></p>
<p>\({\text{AB}} = \sqrt 5 = c\) , \({\text{AC}} = \sqrt {12} = 2\sqrt 3 = b\) , \({\text{BC}} = \sqrt 5 = a\) <strong><em>M1A1</em></strong></p>
<p>\(s = \frac{{\sqrt 5 + 2\sqrt 3 + \sqrt 5 }}{2} = \sqrt 3 + \sqrt 5 \) <strong><em>M1</em></strong></p>
<p>area \(\Delta {\text{ABC}} = \sqrt {s(s - a)(s - b)(s - c)} \)</p>
<p>\( = \sqrt {(\sqrt 3 + \sqrt 5 )(\sqrt 3 )(\sqrt 5 - \sqrt 3 )(\sqrt 3 )} \)</p>
<p>\( = \sqrt {3(5 - 3)} \) <strong><em>A1</em></strong></p>
<p>\( = \sqrt 6 \) <strong><em>AG N0</em></strong></p>
<p><strong>METHOD 4</strong></p>
<p>for finding two of the following three vectors (or their negatives)</p>
<p>\(\overrightarrow {{\text{AB}}} = \left( {\begin{array}{*{20}{c}}<br> 0 \\ <br> 2 \\ <br> { - 1} <br>\end{array}} \right)\), \(\overrightarrow {{\text{AC}}} = \left( {\begin{array}{*{20}{c}}<br> { - 2} \\ <br> 2 \\ <br> { - 2} <br>\end{array}} \right)\), \(\overrightarrow {{\text{BC}}} = \left( {\begin{array}{*{20}{c}}<br> { - 2} \\ <br> 0 \\ <br> { - 1} <br>\end{array}} \right)\) <strong><em>(A1)(A1)</em></strong></p>
<p>\({\text{AB}} = {\text{BC}} = \sqrt 5 \) and \({\text{AC}} = \sqrt {12} = 2\sqrt 3 \) <strong><em>M1A1</em></strong></p>
<p>\(\Delta {\text{ABC}}\) is isosceles</p>
<p><img 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" alt></p>
<p>let <em>M</em> be the midpoint of [AC] , the height \({\text{BM}} = \sqrt {5 - 3} = \sqrt 2 \) <strong><em>M1</em></strong></p>
<p>area \(\Delta {\text{ABC}} = \frac{{2\sqrt 3 \times \sqrt 2 }}{2}\) <strong><em>A1</em></strong></p>
<p>\( = \sqrt 6 \) <strong><em>AG N0</em></strong></p>
<p><strong><em>[6 marks]</em></strong></p>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">This question was answered fairly well by most candidates using a diversity of approaches.</span></p>
</div>
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