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</div><h2>SL Paper 2</h2><div class="specification">
<p class="p1">Let \(u = 6i + 3j + 6k\) and \(v = 2i + 2j + k\).</p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p class="p1">Find</p>
<p class="p1">(i) <span class="Apple-converted-space"> \(u \bullet v\)</span>;</p>
<p class="p1">(ii) <span class="Apple-converted-space"> </span>\(\left| {{u}} \right|\);</p>
<p class="p1">(iii) <span class="Apple-converted-space"> </span>\(\left| {{v}} \right|\).</p>
<div class="marks">[5]</div>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p class="p1">Find the angle between \({{u}}\) and \({{v}}\).</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">(i) <span class="Apple-converted-space"> </span>correct substitution <span class="Apple-converted-space"> </span><strong><em>(A1)</em></strong></p>
<p class="p1"><em>eg</em>\(\;\;\;6 \times 2 + 3 \times 2 + 6 \times 1\)</p>
<p class="p1">\(u \bullet v = 24\) <span class="Apple-converted-space"> </span><strong><em>A1 <span class="Apple-converted-space"> </span>N2</em></strong></p>
<p class="p1">(ii) <span class="Apple-converted-space"> </span>correct substitution into magnitude formula for \({{u}}\) or \({{v}}\) <span class="Apple-converted-space"> </span><strong><em>(A1)</em></strong></p>
<p class="p1"><em>eg</em>\(\;\;\;\sqrt {{6^2} + {3^2} + {6^2}} ,{\text{ }}\sqrt {{2^2} + {2^2} + {1^2}} \), correct value for \(\left| {{v}} \right|\)</p>
<p class="p1">\(\left| {{u}} \right| = 9\) <span class="Apple-converted-space"> </span><strong><em>A1 <span class="Apple-converted-space"> </span>N2</em></strong></p>
<p class="p1">(iii) <span class="Apple-converted-space"> </span>\(\left| {{v}} \right| = 3\) <span class="Apple-converted-space"> </span><strong><em>A1 <span class="Apple-converted-space"> </span>N1</em></strong></p>
<p class="p1"><strong><em>[5 marks]</em></strong></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">correct substitution into angle formula <span class="Apple-converted-space"> </span><strong><em>(A1)</em></strong></p>
<p class="p1"><em>eg</em>\(\;\;\;\frac{{24}}{{9 \times 3}},{\text{ }}0.\bar 8\)</p>
<p class="p1">\(0.475882,{\text{ }}27.26604^\circ \) <span class="Apple-converted-space"> </span><strong><em>A1 <span class="Apple-converted-space"> </span>N2</em></strong></p>
<p class="p1">\(0.476,{\text{ }}27.3^\circ \)</p>
<p class="p1"><em><strong>[2 marks]</strong></em></p>
<p class="p1"><em><strong>Total [7 marks]</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 class="p1">Many candidates performed well in this question. Some candidates were unfamiliar with the basis vector notation and wrongly substituted the <strong><em>i-j-k </em></strong>into the formulas. Others occasionally assumed that the magnitude could be negative.</p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">Many candidates performed well in this question. Some candidates were unfamiliar with the basis vector notation and wrongly substituted the <strong><em>i-j-k </em></strong>into the formulas. Others occasionally assumed that the magnitude could be negative.</p>
<div class="question_part_label">b.</div>
</div>
<br><hr><br><div class="question">
<p><span style="font-family: times new roman,times; font-size: medium;">Two lines with equations \({{\boldsymbol{r}}_1} = \left( {\begin{array}{*{20}{c}}<br>2\\<br>3\\<br>{ - 1}<br>\end{array}} \right) + s\left( {\begin{array}{*{20}{c}}<br>5\\<br>{ - 3}\\<br>2<br>\end{array}} \right)\) and \({{\boldsymbol{r}}_2} = \left( {\begin{array}{*{20}{c}}<br>9\\<br>2\\<br>2<br>\end{array}} \right) + t\left( {\begin{array}{*{20}{c}}<br>{ - 3}\\<br>5\\<br>{ - 1}<br>\end{array}} \right)\) intersect at the point P. Find the coordinates of P.</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;">evidence of appropriate approach <em><strong>(M1)</strong> </em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g. \(\left( {\begin{array}{*{20}{c}}<br>2\\<br>3\\<br>{ - 1}<br>\end{array}} \right) + s\left( {\begin{array}{*{20}{c}}<br>5\\<br>{ - 3}\\<br>2<br>\end{array}} \right) = \left( {\begin{array}{*{20}{c}}<br>9\\<br>2\\<br>2<br>\end{array}} \right) + t\left( {\begin{array}{*{20}{c}}<br>{ - 3}\\<br>5\\<br>{ - 1}<br>\end{array}} \right)\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">two correct equations <em><strong>A1A1 </strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g. \(2 + 5s = 9 - 3t\) , \(3 - 3s = 2 + 5t\) , \( - 1 + 2s = 2 - t\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">attempting to solve the equations <em><strong>(M1)</strong> </em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">one correct parameter \(s = 2\) , \(t = - 1\) <em><strong>A1</strong> </em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">P is \((12, - 3,3)\) (accept \(\left( {\begin{array}{*{20}{c}}<br>{12}\\<br>{ - 3}\\<br>3<br>\end{array}} \right)\)) <em><strong>A1 N3</strong> </em></span></p>
<p><strong><em><span style="font-family: times new roman,times; font-size: medium;"> [6 marks] </span></em></strong></p>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question">
<p><span style="font-family: times new roman,times; font-size: medium;">If this topic had been taught well then the candidates scored highly. The question was either well answered or not at all. Many candidates did not understand what was needed and tried to find the length of vectors or mid-points of lines. The other most common mistake was to use the values of the parameters to write the coordinates as \({\text{P}}(2{\text{, }} - 1)\). </span></p>
</div>
<br><hr><br><div class="question">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">Let \({\boldsymbol{v}} = \left( {\begin{array}{*{20}{c}}<br>2\\<br>{ - 3}\\<br>6<br>\end{array}} \right)\) and \({\boldsymbol{w}} = \left( {\begin{array}{*{20}{c}}<br>k\\<br>{ - 2}\\<br>4<br>\end{array}} \right)\) , for \(k > 0\) . The angle between <strong><em>v</em></strong> and <strong><em>w</em></strong> is \(\frac{\pi }{3}\) </span><span style="font-family: times new roman,times; font-size: medium;">.</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">Find the value of \(k\) .<br></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;">correct substitutions for \({\boldsymbol{v}} \bullet {\boldsymbol{w}}\) ; \(\left| {\boldsymbol{v}} \right|\) ; \(\left| {\boldsymbol{w}} \right|\) <em><strong>(A1)(A1)(A1)</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g. \(2k + ( - 3) \times ( - 2) + 6 \times 4\) , \(2k + 30\) ; \(\sqrt {{2^2} + {{( - 3)}^2} + {6^2}} \) , \(\sqrt {49} \) ; \(\sqrt {{k^2} + {{( - 2)}^2} + {4^2}} \) , \(\sqrt {{k^2} + 20} \)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">evidence of substituting into the formula for scalar product <em><strong>(M1)</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g. </span><span style="font-family: times new roman,times; font-size: medium;">\(\frac{{2k + 30}}{{7 \times \sqrt {{k^2} + 20} }}\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">correct substitution <em><strong>A1</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g. </span><span style="font-family: times new roman,times; font-size: medium;">\(\cos \frac{\pi }{3} = \frac{{2k + 30}}{{7 \times \sqrt {{k^2} + 20} }}\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(k = 18.8\) <em><strong>A2 N5</strong></em></span></p>
<p><em><strong><span style="font-family: times new roman,times; font-size: medium;"> [7 marks]</span></strong></em></p>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question">
<p><span style="font-family: times new roman,times; font-size: medium;">For the most part, this question was well done and candidates had little difficulty finding the scalar product, the appropriate magnitudes and then correctly substituting into the formula for the angle between vectors. However, few candidates were able to solve the resulting equation using their GDCs to obtain the correct answer. Problems arose when candidates attempted to solve the resulting equation analytically. </span></p>
</div>
<br><hr><br><div class="specification">
<p>Let \(\overrightarrow {{\text{AB}}} = \left( {\begin{array}{*{20}{c}} 4 \\ 1 \\ 2 \end{array}} \right)\).</p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Find \(\left| {\overrightarrow {{\text{AB}}} } \right|\).</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>Let \(\overrightarrow {{\text{AC}}} = \left( {\begin{array}{*{20}{c}} 3 \\ 0 \\ 0 \end{array}} \right)\). Find \({\rm{B\hat AC}}\).</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>correct substitution <strong><em>(A1)</em></strong></p>
<p><em>eg</em>\(\,\,\,\,\,\)\(\sqrt {{4^2} + {1^2} + {2^2}} \)</p>
<p>4.58257</p>
<p>\(\left| {\overrightarrow {{\text{AB}}} } \right| = \sqrt {21} \) (exact), 4.58 <strong><em>A1 N2</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>finding scalar product and \(\left| {\overrightarrow {{\text{AC}}} } \right|\) <strong><em>(A1)(A1)</em></strong></p>
<p>scalar product \( = (4 \times 3) + (1 \times 0) + (2 \times 0){\text{ }}( = 12)\)</p>
<p>\(\left| {\overrightarrow {{\text{AC}}} } \right| = \sqrt {{3^2} + 0 + 0} {\text{ }}( = 3)\)</p>
<p>substituting <strong>their</strong> values into cosine formula <strong><em>(M1)</em></strong></p>
<p><em>eg </em>cos B\(\hat A\)C\({\text{ = }}\frac{{4 \times 3 + 0 + 0}}{{\sqrt {{3^2}} \times \sqrt {21} }},{\text{ }}\frac{4}{{\sqrt {21} }},{\text{ }}\cos \theta = 0.873\)</p>
<p>0.509739 (29.2059°)</p>
<p>\({\rm{B\hat AC}} = 0.510\) (29.2°) <strong><em>A1 N2</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="specification">
<p align="LEFT"><em><span style="font-family: times new roman,times; font-size: medium;">In this question, distance is in metres.</span></em></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">Toy airplanes fly in a straight line at a constant speed. Airplane 1 passes through a point A.</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">Its position, <em>p</em> seconds after it has passed through A, is given by \(\left( {\begin{array}{*{20}{c}}<br>x\\<br>y\\<br>z<br>\end{array}} \right) = \left( {\begin{array}{*{20}{c}}<br>3\\<br>{ - 4}\\<br>0<br>\end{array}} \right) + p\left( {\begin{array}{*{20}{c}}<br>{ - 2}\\<br>3\\<br>1<br>\end{array}} \right)\) .</span></p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">(i) Write down the coordinates of A.</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">(ii) Find the speed of the airplane in \({\text{m}}{{\text{s}}^{ - 1}}\).</span></p>
<div class="marks">[4]</div>
<div class="question_part_label">a(i) and (ii).</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">After seven seconds the airplane passes through a point B.</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">(i) Find the coordinates of B.</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">(ii) Find the distance the airplane has travelled during the seven seconds.</span></p>
<div class="marks">[5]</div>
<div class="question_part_label">b(i) and (ii).</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">Airplane 2 passes through a point C. Its position <em>q</em> seconds after it passes </span><span style="font-family: times new roman,times; font-size: medium;">through C is given by \(\left( {\begin{array}{*{20}{c}}<br>x\\<br>y\\<br>z<br>\end{array}} \right) = \left( {\begin{array}{*{20}{c}}<br>2\\<br>{ - 5}\\<br>8<br>\end{array}} \right) + q\left( {\begin{array}{*{20}{c}}<br>{ - 1}\\<br>2\\<br>a<br>\end{array}} \right),a \in \mathbb{R}\) .</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">The angle between the flight paths of Airplane 1 and Airplane 2 is \({40^ \circ }\) . Find the </span><span style="font-family: times new roman,times; font-size: medium;">two values of <em>a</em>.</span></p>
<div class="marks">[7]</div>
<div class="question_part_label">c.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">(i) (3, \( - 4\), 0) <em><strong>A1 N1</strong></em></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">(ii) choosing velocity vector \(\left( {\begin{array}{*{20}{c}}<br>{ - 2}\\<br>3\\<br>1<br>\end{array}} \right)\) <em><strong>(M1)</strong></em><br></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">finding magnitude of velocity vector <em><strong>(A1)</strong></em></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">e.g. \(\sqrt {{{( - 2)}^2} + {3^2} + {1^2}} \) , \(\sqrt {4 + 9 + 1} \)</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">speed \(= 3.74\) \(\left( {\sqrt {14} } \right)\) <em><strong>A1 N2</strong></em></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">a(i) and (ii).</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">(i) substituting \(p = 7\) <em><strong>(M1)</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\({\text{B}} = ( - 11{\text{, }}17{\text{, }}7)\) <em><strong>A1 N2</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">(ii) <strong>METHOD 1</strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">appropriate method to find \(\overrightarrow {{\rm{AB}}} \) or \(\overrightarrow {{\rm{BA}}} \) </span><em><strong><span style="font-family: times new roman,times; font-size: medium;">(M1)</span></strong></em></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g. \(\overrightarrow {{\rm{AO}}} + \overrightarrow {{\rm{OB}}} \) , \({\rm{A}} - {\rm{B}}\)</span></p>
<p><span style="font-family: Times New Roman; font-size: medium;">\(\overrightarrow {{\rm{AB}}} = \left( {\begin{array}{*{20}{c}}<br>{ - 14}\\<br>{21}\\<br>7<br>\end{array}} \right)\) or \(\overrightarrow {{\rm{BA}}} = \left( {\begin{array}{*{20}{c}}<br>{14}\\<br>{ - 21}\\<br>{ - 7}<br>\end{array}} \right)\) </span><em><strong><span style="font-family: times new roman,times; font-size: medium;"> (A1)</span></strong></em></p>
<p><span style="font-family: times new roman,times; font-size: medium;">distance \(= 26.2\) \(\left( {7\sqrt {14} } \right)\) <em><strong>A1 N3</strong></em></span></p>
<p><strong> <span style="font-family: times new roman,times; font-size: medium;">METHOD 2</span></strong></p>
<p><span style="font-family: times new roman,times; font-size: medium;">evidence of applying distance is speed × time <em><strong>(M2)</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g. \(3.74 \times 7\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">distance \(= 26.2\) \(\left( {7\sqrt {14} } \right)\) <em><strong>A1 N3</strong></em></span></p>
<p><strong> <span style="font-family: times new roman,times; font-size: medium;">METHOD 3</span></strong></p>
<p><span style="font-family: times new roman,times; font-size: medium;">attempt to find AB<sup>2</sup> , AB <em><strong>(M1)</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g. \({(3 - ( - 11))^2} + {( - 4 - 17)^2} + (0 - 7){)^2}\) , \(\sqrt {{{(3 - ( - 11))}^2} + {{( - 4 - 17)}^2} + (0 - 7){)^2}} \)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">AB<sup>2</sup> \(= 686\), AB \(= \sqrt {686} \) <em><strong>(A1)</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">distance AB \(= 26.2\) \(\left( {7\sqrt {14} } \right)\) <em><strong>A1 N3</strong></em></span></p>
<p><em><strong><span style="font-family: times new roman,times; font-size: medium;"> [5 marks]</span></strong></em></p>
<div class="question_part_label">b(i) and (ii).</div>
</div>
<div class="question" style="padding-left: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">correct direction vectors \(\left( {\begin{array}{*{20}{c}}<br>{ - 2}\\<br>3\\<br>1<br>\end{array}} \right)\) and \(\left( {\begin{array}{*{20}{c}}<br>{ - 1}\\<br>2\\<br>a<br>\end{array}} \right)\) </span><em><strong><span style="font-family: times new roman,times; font-size: medium;">(A1)(A1)</span></strong></em></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\(\left| {\begin{array}{*{20}{c}}<br>{ - 1}\\<br>2\\<br>a<br>\end{array}} \right| = \sqrt {{a^2} + 5} \) , \(\left( {\begin{array}{*{20}{c}}<br>{ - 2}\\<br>3\\<br>1<br>\end{array}} \right) \bullet \left( {\begin{array}{*{20}{c}}<br>{ - 1}\\<br>2\\<br>a<br>\end{array}} \right) = a + 8\) <em><strong>(A1)(A1)</strong></em></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">substituting <em><strong>M1</strong></em></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">e.g. \(\cos {40^ \circ } = \frac{{a + 8}}{{\sqrt {14} \sqrt {{a^2} + 5} }}\)</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\(a = 3.21\) , \(a = - 0.990\) <em><strong>A1A1 N3</strong></em></span></p>
<p><strong><em><span style="font-family: times new roman,times; font-size: medium;">[7 marks]</span></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 align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">Many candidates demonstrated a good understanding of the vector equation of a line and its </span><span style="font-family: times new roman,times; font-size: medium;">application to a kinematics problem by correctly answering the first two parts of this question.</span></p>
<div class="question_part_label">a(i) and (ii).</div>
</div>
<div class="question" style="padding-left: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">Many candidates demonstrated a good understanding of the vector equation of a line and its </span><span style="font-family: times new roman,times; font-size: medium;">application to a kinematics problem by correctly answering the first two parts of this question.</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">Some knew that speed and distance were magnitudes of vectors but chose the wrong vectors </span><span style="font-family: times new roman,times; font-size: medium;">to calculate magnitudes.</span></p>
<div class="question_part_label">b(i) and (ii).</div>
</div>
<div class="question" style="padding-left: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">Very few candidates were able to get the two correct answers in (c) even if they set up the </span><span style="font-family: times new roman,times; font-size: medium;">equation correctly. Much contorted algebra was seen and little evidence of using the GDC to </span><span style="font-family: times new roman,times; font-size: medium;">solve the equation. Many made simple algebraic errors by combining unlike terms in working </span><span style="font-family: times new roman,times; font-size: medium;">with the scalar product (often writing \(8a\) rather than \(8 + a\) ) or the magnitude (often writing </span><span style="font-family: times new roman,times; font-size: medium;">\(5{a^2}\) rather than \(5 + {a^2}\) ).</span></p>
<div class="question_part_label">c.</div>
</div>
<br><hr><br><div class="specification">
<p><span style="font-family: times new roman,times; font-size: medium;">Consider the points P(2, −1, 5) and Q(3, − 3, 8). Let \({L_1}\) be the line through P and Q.</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;">Show that \(\overrightarrow {{\rm{PQ}}} = \left( {\begin{array}{*{20}{c}}<br>1\\<br>{ - 2}\\<br>3<br>\end{array}} \right)\) .</span></p>
<div class="marks">[1]</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;">The line \({L_1}\) may be represented by \({\boldsymbol{r}} = \left( {\begin{array}{*{20}{c}}<br> 3 \\ <br> { - 3} \\ <br> 8 <br>\end{array}} \right) + s\left( {\begin{array}{*{20}{c}}<br> 1 \\ <br> { - 2} \\ <br> 3 <br>\end{array}} \right)\) .</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">(i) What information does the vector \(\left( {\begin{array}{*{20}{c}}<br>3\\<br>{ - 3}\\<br>8<br>\end{array}} \right)\) give about \({L_1}\) ?</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"> (ii) Write down another vector representation for \({L_1}\) using \(\left( {\begin{array}{*{20}{c}}<br>3\\<br>{ - 3}\\<br>8<br>\end{array}} \right)\) .</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><span style="font-family: times new roman,times; font-size: medium;">The point \({\text{T}}( - 1{\text{, }}5{\text{, }}p)\) lies on \({L_1}\) .</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">Find the value of \(p\) .</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><span style="font-family: times new roman,times; font-size: medium;">The point T also lies on \({L_2}\) with equation \(\left( {\begin{array}{*{20}{c}}<br>x\\<br>y\\<br>z<br>\end{array}} \right) = \left( {\begin{array}{*{20}{c}}<br>{ - 3}\\<br>9\\<br>2<br>\end{array}} \right) + t\left( {\begin{array}{*{20}{c}}<br>1\\<br>{ - 2}\\<br>q<br>\end{array}} \right)\) .</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">Show that \(q = - 3\) .</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><span style="font-family: times new roman,times; font-size: medium;">Let \(\theta \) be the <strong>obtuse</strong> angle between \({L_1}\) and \({L_2}\) . Calculate the size of \(\theta \) .</span></p>
<div class="marks">[7]</div>
<div class="question_part_label">e.</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;">evidence of correct approach <em><strong>A1</strong> </em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g. \(\overrightarrow {{\rm{PQ}}} = \overrightarrow {{\rm{OQ}}} - \overrightarrow {{\rm{OP}}} \) , \(\left( {\begin{array}{*{20}{c}}<br>3\\<br>{ - 3}\\<br>8<br>\end{array}} \right) - \left( {\begin{array}{*{20}{c}}<br>2\\<br>{ - 1}\\<br>5<br>\end{array}} \right)\)</span></p>
<p><span style="font-family: Times New Roman; font-size: medium;">\(\overrightarrow {{\rm{PQ}}} = \left( {\begin{array}{*{20}{c}}<br>1\\<br>{ - 2}\\<br>3<br>\end{array}} \right)\) </span><em><span style="font-family: times new roman,times; font-size: medium;"><strong>AG N0</strong> </span></em></p>
<p><em><strong><span style="font-family: times new roman,times; font-size: medium;"> [1 mark]</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;"> (i) correct description <em><strong>R1 N1</strong> </em></span></p>
<p><span style="font-family: times new roman,times;"><span style="font-size: medium;"> e.g. reference to \(\left( {\begin{array}{*{20}{c}}<br>3\\<br>{ - 3}\\<br>8<br>\end{array}} \right)\) being the position vector of a point on the line, </span><span style="font-size: medium;">a vector to the line, a point on the line. </span></span></p>
<p><span style="font-family: times new roman,times;"><span style="font-size: medium;"> (ii) <strong>any</strong> correct expression in the form </span><span style="font-size: medium;">\({\boldsymbol{r}} = {\boldsymbol{a}} + t{\boldsymbol{b}}\)</span><span style="font-size: medium;"> <em><strong>A2 N2</strong> </em></span></span></p>
<p><span style="font-family: times new roman,times;"><span style="font-size: medium;"> where </span><span style="font-size: medium;">\({\boldsymbol{a}}\) is \(\left( {\begin{array}{*{20}{c}}<br> 3 \\ <br> { - 3} \\ <br> 8 <br>\end{array}} \right)\) , and \({\boldsymbol{b}}\)</span><span style="font-size: medium;"> is a scalar multiple of \(\left( {\begin{array}{*{20}{c}}<br> 1 \\<br> { - 2} \\<br> 3<br>\end{array}} \right)\)</span></span></p>
<p><span style="font-family: times new roman,times;"><span style="font-size: medium;"> e.g. </span><span style="font-size: medium;">\({\boldsymbol{r}} = \left( {\begin{array}{*{20}{c}}<br> 3 \\ <br> { - 3} \\ <br> 8 <br>\end{array}} \right) + t\left( {\begin{array}{*{20}{c}}<br> { - 1} \\ <br> 2 \\ <br> { - 3} <br>\end{array}} \right)\)</span><span style="font-size: medium;"> , \({\boldsymbol{r}} = \left( {\begin{array}{*{20}{c}}<br> {3 + 2s} \\ <br> { - 3 - 4s} \\ <br> {8 + 6s} <br>\end{array}} \right)\)</span></span></p>
<p><span style="font-family: times new roman,times;"><em><strong><span style="font-size: medium;"> [3 marks]</span></strong></em></span></p>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;"><strong>one</strong> correct equation <em><strong>(A1)</strong> </em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g. \(3 + s = - 1\) , \( - 3 - 2s = 5\) </span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(s = - 4\) <em><strong>A1</strong> </em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(p = - 4\) <em><strong>A1 N2</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">c.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">one correct equation <em><strong>A1</strong></em></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">e.g. \( - 3 + t = - 1\) , \(9 - 2t = 5\)</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\(t = 2\) <em><strong>A1</strong></em></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">substituting \(t = 2\)</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">e.g. \(2 + 2q = - 4\) , \(2q = - 6\) <em><strong>A1</strong></em></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\(q = - 3\) <em><strong>AG N0</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">d.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">choosing correct direction vectors \(\left( {\begin{array}{*{20}{c}}<br>1\\<br>{ - 2}\\<br>3<br>\end{array}} \right)\) and \(\left( {\begin{array}{*{20}{c}}<br>1\\<br>{ - 2}\\<br>{ - 3}<br>\end{array}} \right)\) <em><strong>(A1)(A1)</strong> </em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">finding correct scalar product and magnitudes <em><strong>(A1)(A1)(A1)</strong> </em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">scalar product \((1)(1) + ( - 2)( - 2) + ( - 3)(3)\) \(( = - 4)\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">magnitudes \(\sqrt {{1^2} + {{( - 2)}^2} + {3^2}} \) \( = \sqrt {14} \) , \(\sqrt {{1^2} + {{( - 2)}^2} + {{( - 3)}^2}} \) \( = \sqrt {14} \)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">evidence of substituting into scalar product <em><strong>M1</strong> </em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g. \(\cos \theta = \frac{{ - 4}}{{3.741 \ldots \times 3.741 \ldots }}\)</span></p>
<p><span style="font-family: Times New Roman; font-size: medium;">\(\theta = 1.86\) radians (or \(107^\circ \)) </span><em><span style="font-family: times new roman,times; font-size: medium;"><strong>A1 N4</strong> </span></em></p>
<p><em><strong><span style="font-family: times new roman,times; font-size: medium;"> [7 marks]</span></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;">
<p><span style="font-family: times new roman,times; font-size: medium;">Most candidates answered part (a) easily. </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;">For part (b), a number of candidates stated that the vector was a "starting point," which misses the idea that it is a position vector to some point on the line.</span></p>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">Parts (c) and (d) proved accessible to many.</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;">Parts (c) and (d) proved accessible to many. </span></p>
<div class="question_part_label">d.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">For part (e), a surprising number of candidates chose incorrect vectors. Few candidates seemed to have a good conceptual understanding of the vector equation of a line. </span></p>
<div class="question_part_label">e.</div>
</div>
<br><hr><br><div class="specification">
<p class="p1">Consider the points \({\text{A }}(1,{\text{ }}5,{\text{ }} - 7)\) and \({\text{B }}( - 9,{\text{ }}9,{\text{ }} - 6)\).</p>
</div>
<div class="specification">
<p class="p1">Let <span class="s1">C </span>be a point such that \(\overrightarrow {{\text{AC}}} = \left( {\begin{array}{*{20}{c}} 6 \\ { - 4} \\ 0 \end{array}} \right)\).</p>
</div>
<div class="specification">
<p class="p1"><span class="s1">The line \(L\) </span>passes through B and is parallel to (AC)<span class="s1">.</span></p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p class="p1">Find \(\overrightarrow {{\text{AB}}} \).</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">Find the coordinates of C.</p>
<div class="marks">[2]</div>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p class="p1">Write down a vector equation for \(L\).</p>
<div class="marks">[2]</div>
<div class="question_part_label">c.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p class="p1">Given that \(\left| {\overrightarrow {{\text{AB}}} } \right| = k\left| {\overrightarrow {{\text{AC}}} } \right|\), <span class="s1">find \(k\).</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 class="p1">The point D <span class="s1">lies on \(L\) </span>such that \(\left| {\overrightarrow {{\text{AB}}} } \right| = \left| {\overrightarrow {{\text{BD}}} } \right|\). Find the possible coordinates of D<span class="s1">.</span></p>
<div class="marks">[6]</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">valid approach <span class="Apple-converted-space"> </span><strong><em>(M1)</em></strong></p>
<p class="p1"><em>eg</em>\(\,\,\,\,\,\)\({\text{B}} - {\text{A}},{\text{ AO}} + {\text{OB}},{\text{ }}\left( {\begin{array}{*{20}{c}} { - 9} \\ 9 \\ { - 6} \end{array}} \right) - \left( {\begin{array}{*{20}{c}} 1 \\ 5 \\ { - 7} \end{array}} \right)\)</p>
<p class="p1"><span class="s1">\(\overrightarrow {{\text{AB}}} = \left( {\begin{array}{*{20}{c}} { - 10} \\ 4 \\ 1 \end{array}} \right)\) <span class="Apple-converted-space"> </span></span><strong><em>A1 <span class="Apple-converted-space"> </span>N2</em></strong></p>
<p class="p1"><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">valid approach <span class="Apple-converted-space"> </span><strong><em>(M1)</em></strong></p>
<p class="p1"><em>eg</em>\(\,\,\,\,\,\)\({\text{OC}} = {\text{OA}} + {\text{AC}},{\text{ }}\left( {\begin{array}{*{20}{c}} {1 + 6} \\ {5 - 4} \\ { - 7 + 0} \end{array}} \right)\)</p>
<p class="p1"><span class="Apple-converted-space">\({\text{C}}(7,{\text{ }}1,{\text{ }} - 7)\) </span><strong><em>A1 <span class="Apple-converted-space"> </span>N2</em></strong></p>
<p class="p1"><strong><em>[2 marks]</em></strong></p>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">any correct equation in the form <span class="s1"><strong><em>r</em></strong> \( = \) <strong><em>a</em></strong> \( + t\)<strong><em>b</em></strong> </span>(accept any parameter for \(t\)</p>
<p class="p2"><span class="s2">where </span><strong><em>a</em></strong> is \(\left( {\begin{array}{*{20}{c}} { - 9} \\ 9 \\ { - 6} \end{array}} \right)\), <span class="s2">and </span><strong><em>b</em></strong> is a scalar multiple of \(\left( {\begin{array}{*{20}{c}} 6 \\ { - 4} \\ 0 \end{array}} \right)\) <span class="Apple-converted-space"> </span><span class="s2"><strong><em>A2 <span class="Apple-converted-space"> </span>N2</em></strong></span></p>
<p class="p2"><span class="s2"><em>eg</em>\(\,\,\,\,\,\)</span><strong><em>r</em></strong> \( = \left( {\begin{array}{*{20}{c}} { - 9} \\ 9 \\ { - 6} \end{array}} \right) + t\left( {\begin{array}{*{20}{c}} 6 \\ { - 4} \\ 0 \end{array}} \right)\), <strong><em>r</em></strong> \( = - 9\)<strong><em>i</em></strong> \( + 9\)<strong><em>j</em></strong> \( - 6\)<strong><em>k</em></strong> \( + s(6\)<strong><em>i</em></strong> \( - 4\)<strong><em>j</em></strong> \( + 0\)<strong><em>k</em></strong>\()\)</p>
<p class="p1"><strong><em>[2 marks]</em></strong></p>
<div class="question_part_label">c.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">correct magnitudes <span class="Apple-converted-space"> </span><strong><em>(A1)(A1)</em></strong></p>
<p class="p1"><em>eg</em>\(\,\,\,\,\,\)\(\sqrt {{{( - 10)}^2} + {{( - 4)}^2} + {1^2}} ,{\text{ }}\sqrt {{6^2} + {{( - 4)}^2} + {{(0)}^2}} ,{\text{ }}\sqrt {{{10}^2} + {4^2} + 1} ,{\text{ }}\sqrt {{6^2} + {4^2}} \)</p>
<p class="p1"><span class="Apple-converted-space">\(k = \frac{{\sqrt {117} }}{{\sqrt {52} }}{\text{ }}( = 1.5){\text{ }}({\text{exact}})\) </span><strong><em>A1 <span class="Apple-converted-space"> </span>N3</em></strong></p>
<p class="p1"><strong><em>[3 marks]</em></strong></p>
<div class="question_part_label">d.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">correct interpretation of relationship between magnitudes <span class="Apple-converted-space"> </span><strong><em>(A1)</em></strong></p>
<p class="p1"><em>eg</em>\(\,\,\,\,\,\)\({\text{AB}} = 1.5{\text{AC}},{\text{ BD}} = 1.5{\text{AC}},{\text{ }}\sqrt {117} = \sqrt {52{t^2}} \)</p>
<p class="p2">recognizing D can have two positions (may be seen in w<span class="s1">orking) <span class="Apple-converted-space"> </span><strong><em>R1</em></strong></span></p>
<p class="p1"><em>eg</em>\(\,\,\,\,\,\)\(\overrightarrow {{\text{BD}}} = 1.5\overrightarrow {{\text{AC}}} \) and \(\overrightarrow {{\text{BD}}} = - 1.5\overrightarrow {{\text{AC}}} ,{\text{ }}t = \pm 1.5\)<span class="s2">, diagram, t</span>wo answers</p>
<p class="p1">valid approach (seen anywhere) <span class="Apple-converted-space"> </span><strong><em>(M1)</em></strong></p>
<p class="p1"><em>eg</em>\(\,\,\,\,\,\)\(\overrightarrow {{\text{OD}}} = \overrightarrow {{\text{OB}}} + \overrightarrow {{\text{BD}}} {\text{, }}\left( {\begin{array}{*{20}{c}} { - 9} \\ 9 \\ { - 6} \end{array}} \right) + t\left( {\begin{array}{*{20}{c}} 6 \\ { - 4} \\ 0 \end{array}} \right),{\text{ }}\overrightarrow {{\text{BD}}} = k\left( {\begin{array}{*{20}{c}} 6 \\ { - 4} \\ 0 \end{array}} \right)\)</p>
<p class="p1">one correct expression for \(\overrightarrow {{\text{OD}}} \) <span class="Apple-converted-space"> </span><strong><em>(A1)</em></strong></p>
<p class="p1"><em>eg</em>\(\,\,\,\,\,\)\(\overrightarrow {{\text{OD}}} = \left( {\begin{array}{*{20}{c}} { - 9} \\ 9 \\ { - 6} \end{array}} \right) + 1.5\left( {\begin{array}{*{20}{c}} 6 \\ { - 4} \\ 0 \end{array}} \right),{\text{ }}\left( {\begin{array}{*{20}{c}} { - 9} \\ 9 \\ { - 6} \end{array}} \right) - 1.5\left( {\begin{array}{*{20}{c}} 6 \\ { - 4} \\ 0 \end{array}} \right)\)</p>
<p class="p2"><span class="s1">\({\text{D}} = (0,{\text{ }}3,{\text{ }} - 6),{\text{ D}} = ( - 18,{\text{ }}15,{\text{ }} - 6)\) </span>(accept position vectors) <span class="Apple-converted-space"> </span><span class="s1"><strong><em>A1A1 <span class="Apple-converted-space"> </span>N3</em></strong></span></p>
<p class="p1"><strong><em>[6 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">Parts (a) and (b) were attempted by the great majority of the candidates and appropriate approaches were seen, earning at least the method marks.</p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">Parts (a) and (b) were attempted by the great majority of the candidates and appropriate approaches were seen, earning at least the method marks.</p>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">Part (c) was generally well done, with many candidates writing the equation of the line as \(L = \) <em><strong>a</strong></em> \( + t\)<em><strong>b</strong></em>, losing one mark.</p>
<div class="question_part_label">c.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">Part (d) was also well answered by a great majority of students. Even those candidates who had part (a) incorrect, could gain all the marks here.</p>
<div class="question_part_label">d.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">Part (e) was the most challenging of the paper. For many a major problem was to set up the equation \(\sqrt {117} = \sqrt {52{t^2}} \) and, hence, realize that D could have two positions.</p>
<div class="question_part_label">e.</div>
</div>
<br><hr><br><div class="specification">
<p><span style="font-family: times new roman,times; font-size: medium;">The point O has coordinates (0 , 0 , 0) , point A has coordinates (1 , – 2 , 3) and point B has coordinates (– 3 , 4 , 2) .</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;">(i) Show that \(\overrightarrow {{\rm{AB}}} = \left( {\begin{array}{*{20}{c}}<br>{ - 4}\\<br>6\\<br>{ - 1}<br>\end{array}} \right)\) .</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">(ii) Find \({\rm{B}}\widehat {\rm{A}}{\rm{O}}\) .</span></p>
<div class="marks">[8]</div>
<div class="question_part_label">a(i) and (ii).</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">The line \({L_1}\) has equation \(\left( {\begin{array}{*{20}{c}}<br>x\\<br>y\\<br>z<br>\end{array}} \right) = \left( {\begin{array}{*{20}{c}}<br>{ - 3}\\<br>4\\<br>2<br>\end{array}} \right) + s\left( {\begin{array}{*{20}{c}}<br>{ - 4}\\<br>6\\<br>{ - 1}<br>\end{array}} \right)\) .</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"> Write down the coordinates of two points on \({L_1}\) .</span></p>
<div class="marks">[2]</div>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">The line \({L_2}\) passes through A and is parallel to \(\overrightarrow {{\rm{OB}}} \) . </span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">(i) Find a vector equation for \({L_2}\) , giving your answer in the form \({\boldsymbol{r}} = {\boldsymbol{a}} + t{\boldsymbol{b}}\) . </span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"> (ii) Point \(C(k, - k,5)\) is on \({L_2}\) . Find the coordinates of C.</span></p>
<div class="marks">[6]</div>
<div class="question_part_label">c(i) and (ii).</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">The line \({L_3}\) has equation \(\left( {\begin{array}{*{20}{c}}<br>x\\<br>y\\<br>z<br>\end{array}} \right) = \left( {\begin{array}{*{20}{c}}<br>3\\<br>{ - 8}\\<br>0<br>\end{array}} \right) + p\left( {\begin{array}{*{20}{c}}<br>1\\<br>{ - 2}\\<br>{ - 1}<br>\end{array}} \right)\) and passes through the point C. </span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"> Find the value of <em>p</em> at C. </span></p>
<div class="marks">[2]</div>
<div class="question_part_label">d.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">(i) evidence of approach <em><strong>M1</strong></em></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">e.g. \(\overrightarrow {{\rm{AO}}} + \overrightarrow {{\rm{OB}}} = \overrightarrow {{\rm{AB}}} \)</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\(\overrightarrow {{\rm{AB}}} = \left( {\begin{array}{*{20}{c}}<br>{ - 4}\\<br>6\\<br>{ - 1}<br>\end{array}} \right)\) <em><strong>AG N0</strong></em></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">(ii) for choosing <strong>correct</strong> vectors, (\(\overrightarrow {{\rm{AO}}} \) with \(\overrightarrow {{\rm{AB}}} \) or \(\overrightarrow {{\rm{OA}}} \) with \(\overrightarrow {{\rm{BA}}} \) </span><span style="font-family: times new roman,times; font-size: medium;">) <em><strong>(A1)(A1)</strong></em></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;"><strong>Note</strong>: Using \(\overrightarrow {{\rm{AO}}} \) with \(\overrightarrow {{\rm{BA}}} \) </span><span style="font-family: times new roman,times; font-size: medium;">will lead to \(\pi - 0.799\) . If they then say </span><span style="font-family: times new roman,times; font-size: medium;">\({\rm{B}}\widehat {\rm{A}}{\rm{O}} = 0.799\) , this is a correct solution.</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;"> </span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">calculating \(\overrightarrow {{\rm{AO}}} \bullet \overrightarrow {{\rm{AB}}} \) , \(\left| {\overrightarrow {{\rm{AO}}} } \right|\) , \(\left| {\overrightarrow {{\rm{AB}}} } \right|\) </span><em><strong><span style="font-family: times new roman,times; font-size: medium;">(A1)(A1)(A1)</span></strong></em></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">e.g. \({d_1} \bullet {d_2} = ( - 1)( - 4) + (2)(6) + ( - 3)( - 1)( = 19)\)</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\(\left| {{d_1}} \right| = \sqrt {{{( - 1)}^2} + {2^2} + {{( - 3)}^2}} ( = \sqrt {14} )\) , \(\left| {{d_2}} \right| = \sqrt {{{( - 4)}^2} + {6^2} + {{( - 1)}^2}} ( = \sqrt {53} )\)</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">evidence of using the formula to find the angle <em><strong>M1</strong></em></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">e.g. \(\cos \theta = \frac{{( - 1)( - 4) + (2)(6) + ( - 3)( - 1)}}{{\sqrt {{{( - 1)}^2} + {2^2} + {{( - 3)}^2}} \sqrt {{{( - 4)}^2} + {6^2} + {{( - 1)}^2}} }}\) , \(\frac{{19}}{{\sqrt {14} \sqrt {53} }}\) , \(0.69751 \ldots \)</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\({\rm{B}}\widehat {\rm{A}}{\rm{O}} = 0.799\) radians (accept \(45.8^\circ \)) <strong><em>A1 N3</em></strong></span></p>
<p><em><strong><span style="font-family: times new roman,times; font-size: medium;">[8 marks]</span></strong></em></p>
<div class="question_part_label">a(i) and (ii).</div>
</div>
<div class="question" style="padding-left: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">two correct answers <em><strong>A1A1</strong></em></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">e.g. (1, \( - 2\), 3) , (\( - 3\), 4, 2) , (\( - 7\), 10, 1), (\( - 11\), 16, 0) <em><strong>N2</strong></em></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">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">(i) \({\boldsymbol{r}} = \left( {\begin{array}{*{20}{c}}<br>1\\<br>{ - 2}\\<br>3<br>\end{array}} \right) + t\left( {\begin{array}{*{20}{c}}<br>{ - 3}\\<br>4\\<br>2<br>\end{array}} \right)\) </span><em><strong><span style="font-family: times new roman,times; font-size: medium;">A2 N2</span></strong></em></p>
<p> </p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">(ii) C on \({L_2}\) , so \(\left( {\begin{array}{*{20}{c}}<br>k\\<br>{ - k}\\<br>5<br>\end{array}} \right) = \left( {\begin{array}{*{20}{c}}<br>1\\<br>{ - 2}\\<br>3<br>\end{array}} \right) + t\left( {\begin{array}{*{20}{c}}<br>{ - 3}\\<br>4\\<br>2<br>\end{array}} \right)\) </span><em><strong><span style="font-family: times new roman,times; font-size: medium;">(M1)</span></strong></em></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">evidence of equating components <em><strong>(A1)</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g. \(1 - 3t = k\) , \( - 2 + 4t = - k\) , \(5 = 3 + 2t\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">one correct value \(t = 1\) , \(k = - 2\) (seen anywhere) <em><strong>(A1)</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">coordinates of C are \(( - 2{\text{, }}2{\text{, }}5)\) <em><strong>A1 N3</strong></em></span></p>
<p><strong><em><span style="font-family: times new roman,times; font-size: medium;">[6 marks]</span></em></strong></p>
<div class="question_part_label">c(i) and (ii).</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">for setting up one (or more) correct equation using \(\left( {\begin{array}{*{20}{c}}<br>{ - 2}\\<br>2\\<br>5<br>\end{array}} \right) = \left( {\begin{array}{*{20}{c}}<br>3\\<br>{ - 8}\\<br>0<br>\end{array}} \right) + p\left( {\begin{array}{*{20}{c}}<br>1\\<br>{ - 2}\\<br>{ - 1}<br>\end{array}} \right)\) <em><strong>(M1)</strong></em></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">e.g. \(3 + p = - 2\) , \( - 8 - 2p = 2\) , \( - p = 5\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(p = - 5\) <em><strong>A1 N2</strong></em></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">d.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">Part (ai) was done well by most students. Most knew how to approach finding the angle in </span><span style="font-family: times new roman,times; font-size: medium;">part (aii). The problems occurred when the incorrect vectors were chosen. If the vectors being </span><span style="font-family: times new roman,times; font-size: medium;">used were stated, then follow through marks could be given.</span></p>
<div class="question_part_label">a(i) and (ii).</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">Part (b) was well done.</span></p>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">In part </span><span style="font-family: times new roman,times; font-size: medium;">(ci), the error that occurred most often was the incorrect choice for the direction vector.</span></p>
<div class="question_part_label">c(i) and (ii).</div>
</div>
<div class="question" style="padding-left: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">Those </span><span style="font-family: times new roman,times; font-size: medium;">that were able to find the coordinates in part (cii) were also able to be successful in part (d).</span></p>
<div class="question_part_label">d.</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><span style="font-family: 'times new roman', times; font-size: medium;"><span style="line-height: normal;">Consider the lines \({L_1}\) and \({L_2}\) with equations \({L_1}\) : </span></span><span style="font-family: 'times new roman', times; font-size: medium;"><span style="line-height: normal;">\(\boldsymbol{r}=\left( \begin{array}{c}11\\8\\2\end{array} \right) + s\left( \begin{array}{c}4\\3\\ - 1\end{array} \right)\) and \({L_2}\) : </span></span></span><span style="background-color: #f7f7f7; line-height: normal;"><span style="font-family: 'times new roman', times; font-size: medium;">\(\boldsymbol{r} = \left( \begin{array}{c}1\\1\\ - 7\end{array} \right) + t\left( \begin{array}{c}2\\1\\11\end{array} \right)\).</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 lines intersect at point \(\rm{P}\).</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;">Find the coordinates of \({\text{P}}\).</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: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">Show that the lines are perpendicular.</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;">The point \({\text{Q}}(7, 5, 3)\) lies on \({L_1}\). The point \({\text{R}}\) is the reflection of \({\text{Q}}\) in the line \({L_2}\).</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;">Find the coordinates of \({\text{R}}\).</span></p>
<div class="marks">[6]</div>
<div class="question_part_label">c.</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: 20.5px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">appropriate approach <strong><em>(M1)</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 20.5px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;"><em>eg</em> \(\left( \begin{array}{c}11\\8\\2\end{array} \right) + s\left( \begin{array}{c}4\\3\\ - 1\end{array} \right) = \left( \begin{array}{c}1\\1\\ - 7\end{array} \right) + t\left( \begin{array}{c}2\\1\\11\end{array} \right)\), \({L_1} = {L_2}\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 20.5px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">any <strong>two </strong>correct equations <strong><em>A1A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 20.5px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;"><em>eg</em> \(11 + 4s = 1 + 2t,{\text{ }}8 + 3s = 1 + t,{\text{ }}2 - s = - 7 + 11t\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 20.5px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">attempt to solve system of equations <strong><em>(M1)</em></strong></span></p>
<p><span style="font-family: 'times new roman', times; font-size: medium;"><em>eg </em>\(10 + 4s = 2(7 + 3s), \left\{ {\begin{array}{*{20}{c}} {4s - 2t = - 10} \\ {3s - t = - 7} \end{array}} \right.\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 20.5px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">one correct parameter <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 20.5px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;"><em>eg</em> \(s = - 2,{\text{ }}t = 1\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 20.5px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">\({\text{P}}(3, 2, 4)\) (accept position vector) <strong><em>A1 N3</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 20.5px 'Times New Roman';"><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: 20.5px 'Times New Roman';"> </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: 20.5px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">choosing correct direction vectors for \({L_1}\) and \({L_2}\) <strong><em>(A1)(A1)</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 20.5px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;"><em>eg</em> </span><span style="font-family: 'times new roman', times; font-size: medium;">\(\left( {\begin{array}{*{20}{c}} 4 \\ 3 \\ { - 1} \end{array}} \right),\left( {\begin{array}{*{20}{c}} 2 \\ 1 \\ {11} \end{array}} \right)\) (or any scalar multiple)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 20.5px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">evidence of scalar product (with any vectors) <strong><em>(M1)</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 20.5px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;"><em>eg</em> </span><span style="font-family: 'times new roman', times; font-size: medium;">\(a \cdot b\), \(\left( \begin{array}{c}4\\3\\ - 1\end{array} \right) \bullet \left( \begin{array}{c}2\\1\\11\end{array} \right)\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 20.5px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">correct substitution <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 20.5px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;"><em>eg</em> \(4(2) + 3(1) + ( - 1)(11),{\text{ }}8 + 3 - 11\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 20.5px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">calculating \(a \cdot b = 0\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 20.5px 'Times New Roman'; min-height: 23.0px;"> </p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 20.5px '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>without evidence of calculation.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 20.5px 'Times New Roman'; min-height: 23.0px;"> </p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 20.5px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">vectors are perpendicular <strong><em>AG N0</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 20.5px '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: 20.5px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>Note: </strong>Candidates may take different approaches, which do not necessarily involve vectors.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 20.5px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">In particular, most of the working could be done on a diagram. Award marks in line with the markscheme.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 20.5px 'Times New Roman'; min-height: 23.0px;"> </p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 20.5px 'Times New Roman';"><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: 20.5px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">attempt to find \(\overrightarrow {{\text{QP}}} \) or \(\overrightarrow {{\text{PQ}}} \) <strong><em>(M1)</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 20.5px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">correct working (may be seen on diagram) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 20.5px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;"><em>eg</em> \(\overrightarrow {{\text{QP}}} \) = \(\left( \begin{array}{c} - 4\\ - 3\\1\end{array} \right)\), \(\overrightarrow {{\text{PQ}}} \) = \(\left( \begin{array}{c}7\\5\\3\end{array} \right) - \left( \begin{array}{c}3\\2\\4\end{array} \right)\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 20.5px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">recognizing \({\text{R}}\) is on \({L_1}\) (seen anywhere) <strong><em>(R1)</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 20.5px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;"><em>eg</em> on diagram</span></p>
<p><span style="font-family: 'times new roman', times; font-size: medium;">\({\text{Q}}\) and \({\text{R}}\) are equidistant from \({\text{P}}\) (seen anywhere) <em><strong>(R1)</strong></em></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 20.5px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;"><em>eg</em> \(\overrightarrow {{\text{QP}}} = \overrightarrow {{\text{PR}}} \), marked on diagram</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 20.5px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">correct working <strong><em>(A1)</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 20.5px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;"><em>eg</em> \(\left( \begin{array}{c}3\\2\\4\end{array} \right) - \left( \begin{array}{c}7\\5\\3\end{array} \right) = \left( \begin{array}{c}x\\y\\z\end{array} \right) - \left( \begin{array}{c}3\\2\\4\end{array} \right),\left( \begin{array}{c} - 4\\ - 3\\1\end{array} \right) + \left( \begin{array}{c}3\\2\\4\end{array} \right)\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 20.5px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">\({\text{R}}(–1, –1, 5)\) (accept position vector) <strong><em>A1 N3</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 20.5px '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: 20.5px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">recognizing \({\text{R}}\) is on \({L_1}\) (seen anywhere) <strong><em>(R1)</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 20.5px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;"><em>eg</em> on diagram</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 20.5px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">\({\text{Q}}\) and \({\text{R}}\) are equidistant from \({\text{P}}\) (seen anywhere) <strong><em>(R1)</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 20.5px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;"><em>eg</em> \({\text{P}}\) midpoint of \({\text{QR}}\), marked on diagram</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 20.5px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">valid approach to find <strong>one </strong>coordinate of mid-point <strong><em>(M1)</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 20.5px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;"><em>eg</em> \({x_p} = \frac{{{x_Q} + {x_R}}}{2},{\text{ }}2{y_p} = {y_Q} + {y_R},{\text{ }}\frac{1}{2}\left( {{z_Q} + {z_R}} \right)\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 20.5px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>one </strong>correct substitution <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 20.5px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;"><em>eg</em> \({x_R} = 3 + (3 - 7),{\text{ }}2 = \frac{{5 + {y_R}}}{2},{\text{ }}4 = \frac{1}{2}(z + 3)\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 20.5px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">correct working for one coordinate <strong><em>(A1)</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 20.5px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;"><em>eg</em> \({x_R} = 3 - 4,{\text{ }}4 - 5 = {y_R},{\text{ }}8 = (z + 3)\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 20.5px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">\({\text{R}} (-1, -1, 5)\) (accept position vector) <strong><em>A1 N3</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 20.5px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[6 marks]</em></strong></span></p>
<div class="question_part_label">c.</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>
<br><hr><br><div class="specification">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">The following diagram shows the cuboid (rectangular solid) OABCDEFG, where O is the origin, and \(\overrightarrow {{\rm{OA}}} = 4\boldsymbol{i}\) , \(\overrightarrow {{\rm{OC}}} = 3\boldsymbol{j}\) , \(\overrightarrow {{\rm{OD}}} = 2\boldsymbol{k}\) .</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;"><br><img src="images/Lars.png" alt></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;">(i) Find \(\overrightarrow {{\rm{OB}}} \) .</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">(ii) Find \(\overrightarrow {{\rm{OF}}} \) .</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">(iii) Show that \(\overrightarrow {{\rm{AG}}} = - 4{\boldsymbol{i}} + 3{\boldsymbol{j}} + 2{\boldsymbol{k}}\) .</span></p>
<div class="marks">[5]</div>
<div class="question_part_label">a(i), (ii) and (iii).</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">Write down a vector equation for</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">(i) the line OF;</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">(ii) the line AG.</span></p>
<div class="marks">[4]</div>
<div class="question_part_label">b(i) and (ii).</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 obtuse angle between the lines OF and AG.</span></p>
<div class="marks">[7]</div>
<div class="question_part_label">c.</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;">(i) valid approach <em><strong>(M1)</strong> </em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g. \({\rm{OA + OB}}\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(\overrightarrow {{\rm{OB}}} = 4{\boldsymbol{i}} + 3{\boldsymbol{j}}\) <em><strong>A1 N2</strong> </em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">(ii) valid approach <em><strong>(M1) </strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g. \(\overrightarrow {{\rm{OA}}} {\rm{ + }}\overrightarrow {{\rm{AB}}} {\rm{ + }}\overrightarrow {{\rm{BF}}} \) ; \(\overrightarrow {{\rm{OB}}} {\rm{ + }}\overrightarrow {{\rm{BF}}} \) ; \(\overrightarrow {{\rm{OC}}} {\rm{ + }}\overrightarrow {{\rm{CG}}} {\rm{ + }}\overrightarrow {{\rm{GF}}} \)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(\overrightarrow {{\rm{OF}}} = 4{\boldsymbol{i}} + 3{\boldsymbol{j}} + 2{\boldsymbol{k}}\) <em><strong>A1 N2</strong> </em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">(iii) correct approach <em><strong>A1</strong> </em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g. \(\overrightarrow {{\rm{AO}}} {\rm{ + }}\overrightarrow {{\rm{OC}}} {\rm{ + }}\overrightarrow {{\rm{CG}}} \) ; \(\overrightarrow {{\rm{AB}}} {\rm{ + }}\overrightarrow {{\rm{BF}}} {\rm{ + }}\overrightarrow {{\rm{FG}}} \) ; \(\overrightarrow {{\rm{AB}}} {\rm{ + }}\overrightarrow {{\rm{BC}}} {\rm{ + }}\overrightarrow {{\rm{CG}}} \)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(\overrightarrow {{\rm{AG}}} = - 4{\boldsymbol{i}} + 3{\boldsymbol{j}} + 2{\boldsymbol{k}}\) <em><strong>AG N0</strong> </em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"><em> <strong>[5 marks]</strong> </em></span></p>
<div class="question_part_label">a(i), (ii) and (iii).</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">(i) any correct equation for (OF) in the form \({\boldsymbol{r}} = {\boldsymbol{a}} + t{\boldsymbol{b}}\) <em><strong>A2 N2</strong> </em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">where \({\boldsymbol{a}}\) is 0 or \(4{\boldsymbol{i}} + 3{\boldsymbol{j}} + 2{\boldsymbol{k}}\) , and \({\boldsymbol{b}}\) is a scalar multiple of \(4{\boldsymbol{i}} + 3{\boldsymbol{j}} + 2{\boldsymbol{k}}\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g. \({\boldsymbol{r}} = t(4{\text{, }}3{\text{, }}2)\) , \({\boldsymbol{r}} = \left( {\begin{array}{*{20}{c}}<br>{4t}\\<br>{3t}\\<br>{2t}<br>\end{array}} \right)\) , \({\boldsymbol{r}} = 4{\boldsymbol{i}} + 3{\boldsymbol{j}} + 2{\boldsymbol{k}} + t(4{\boldsymbol{i}} + 3{\boldsymbol{j}} + 2{\boldsymbol{k}})\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">(ii) any correct equation for (AG) in the form \({\boldsymbol{r}} = {\boldsymbol{a}} + s{\boldsymbol{b}}\) <em><strong>A2 N2</strong> </em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">where \({\boldsymbol{a}}\) is \(4{\boldsymbol{i}}\) or \(3{\boldsymbol{j}} + 2{\boldsymbol{k}}\) and \({\boldsymbol{b}}\) is a scalar multiple of \( - 4{\boldsymbol{i}} + 3{\boldsymbol{j}} + 2{\boldsymbol{k}}\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g. \({\boldsymbol{r}} = (4{\text{, }}0{\text{, }}0) + s( - 4{\text{, }}3{\text{, }}2)\) , \({\boldsymbol{r}} = \left( {\begin{array}{*{20}{c}}<br>{4 - 4s}\\<br>{3s}\\<br>{2s}<br>\end{array}} \right)\) , \({\boldsymbol{r}} = 3{\boldsymbol{j}} + 2{\boldsymbol{k}} + s( - 4{\boldsymbol{i}} + 3{\boldsymbol{j}} + 2{\boldsymbol{k}})\)</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(i) and (ii).</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">choosing correct direction vectors, \(\overrightarrow {{\rm{OF}}} \) and \(\overrightarrow {{\rm{AG}}} \) <em><strong>(A1)(A1)</strong> </em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">scalar product \( = - 16 + 9 + 4\) \(( = - 3)\) <em><strong> (A1)</strong> </em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">magnitudes \(\sqrt {{4^2} + {3^2} + {2^2}} \) , \(\sqrt {{{( - 4)}^2} + {3^2} + {2^2}} \) , \(\left( {\sqrt {29} ,\sqrt {29} } \right)\) <em><strong>(A1)(A1)</strong></em> </span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">substitution into formula <em><strong>M1</strong> </em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g. \(\cos \theta = \frac{{ - 16 + 9 + 4}}{{\left( {\sqrt {{4^2} + {3^2} + {2^2}} } \right) \times \sqrt {{{( - 4)}^2} + {3^2} + {2^2}} }} = \left( { - \frac{3}{{29}}} \right)\) </span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(95.93777^\circ \) , \(1.67443{\text{ radians}}\) </span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(\theta = 95.9^\circ \) or \(1.67\) <em><strong>A1 N4</strong> </em></span></p>
<p><em><strong><span style="font-family: times new roman,times; font-size: medium;">[7 marks] </span></strong></em></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><span style="font-family: times new roman,times; font-size: medium;">Although a large proportion of candidates managed to answer this question, their biggest challenge was the use of a proper notation to represent the vectors and vector equations of lines. </span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">In part (a), finding \(\overrightarrow {{\rm{OB}}} \) and \(\overrightarrow {{\rm{OF}}} \) was generally well done, although many lost the mark for (iii) due to poor working or not clearly showing the result. </span></p>
<div class="question_part_label">a(i), (ii) and (iii).</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">Part (b) was very poorly done. Not all the students recognized which correct position vectors they had to use to write the equations of the lines. It was seen that they frequently failed to present the equations in the required format, which prevented these candidates from achieving full marks. The notations generally seen were \({\text{AG}} = {\boldsymbol{a}} + {\boldsymbol{b}}t\) , \({\boldsymbol{r}} = 4 + t(4{\text{, }}3{\text{, }}2)\) or \(L = {\boldsymbol{a}} + {\boldsymbol{b}}t\) . </span></p>
<div class="question_part_label">b(i) and (ii).</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">Most achieved the correct result in part (c) with many others gaining most of the marks as follow through from choosing incorrect vectors. Some students did not state which vectors had been used, another cause for losing marks. A few showed poor notation, including <strong><em>i</em></strong>, <strong><em>j</em></strong> and <strong><em>k</em></strong> in the working. </span></p>
<div class="question_part_label">c.</div>
</div>
<br><hr><br><div class="specification">
<p><span style="font-family: times new roman,times; font-size: medium;">Consider the points A(\(5\), \(2\), \(1\)) , B(\(6\), \(5\), \(3\)) , and C(\(7\), \(6\), \(a + 1\)) , \(a \in{\mathbb{R}}\) .</span></p>
</div>
<div class="specification">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">Let \({\rm{q}}\) be the angle between \(\overrightarrow {{\rm{AB}}} \) </span><span style="font-family: times new roman,times; font-size: medium;">and \(\overrightarrow {{\rm{AC}}} \) </span><span style="font-family: times new roman,times; font-size: medium;">.</span></p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">Find</span></p>
<p style="margin-left: 30px;" align="LEFT"><span style="font-family: times new roman,times; font-size: medium;"> (i) \(\overrightarrow {{\rm{AB}}} \) </span><span style="font-family: times new roman,times; font-size: medium;">;</span></p>
<p style="margin-left: 30px;" align="LEFT"><span style="font-family: times new roman,times; font-size: medium;"> (ii) \(\overrightarrow {{\rm{AC}}} \) </span><span style="font-family: times new roman,times; font-size: medium;">.</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 align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">Find the value of \(a\) for which \({\rm{q}} = \frac{\pi }{2}\)</span><span style="font-family: times new roman,times; font-size: medium;"> .</span></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 align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">i. Show that \(\cos q = \frac{{2a + 14}}{{\sqrt {14{a^2} + 280} }}\) </span><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. Hence, find the value of a for which \({\rm{q}} = 1.2\) .</span></p>
<div class="marks">[8]</div>
<div class="question_part_label">c.</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, find the value of a for which \({\rm{q}} = 1.2\) .</span></p>
<div class="marks">[4]</div>
<div class="question_part_label">c.ii.</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;">(i) appropriate approach <em><strong>(M1)</strong> </em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"><em>eg </em> \(\overrightarrow {{\rm{AO}}} {\rm{ + }}\overrightarrow {{\rm{OB}}} \) , \({\rm{B}} - {\rm{A}}\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(\overrightarrow {{\rm{AB}}} = \left( \begin{array}{l}<br>1\\<br>3\\<br>2<br>\end{array} \right)\) <em><strong>A1 N2</strong> </em></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) \(\overrightarrow {{\rm{AC}}} = \left( \begin{array}{l}<br>2\\<br>4\\<br>a<br>\end{array} \right)\) <em><strong>A1 N1 </strong></em></span></p>
<p><em><strong><span style="font-family: times new roman,times; font-size: medium;"> </span></strong></em></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">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">valid reasoning (seen anywhere) <em><strong>R1 </strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"><em>eg </em> scalar product is zero, \(\cos \frac{\pi }{2} = \frac{{\boldsymbol{u} \bullet \boldsymbol{v}}}{{\left| \boldsymbol{u} \right|\left| \boldsymbol{v} \right|}}\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">correct scalar product of <strong>their</strong> \(\overrightarrow {{\rm{AB}}} \) and \(\overrightarrow {{\rm{AC}}} \) (may be seen in part (c)) <em><strong> (A1)</strong> </em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"><em>eg </em> \(1(2) + 3(4) + 2(a)\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">correct working for<strong> their</strong> \(\overrightarrow {{\rm{AB}}} \) and \(\overrightarrow {{\rm{AC}}} \) <strong><em>(A1)</em> </strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"><em>eg </em> \(2a + 14\) , \(2a = - 14\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(a = - 7\) <em><strong>A1 N3 </strong></em></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>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">correct magnitudes (may be seen in (b)) <em><strong>(A1)(A1)</strong> </em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"> \(\sqrt {{1^2} + {3^2} + {2^2}} \left( { = \sqrt {14} } \right)\) , \(\sqrt {{2^2} + {4^2} + {a^2}} \left( { = \sqrt {20 + {a^2}} } \right)\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">substitution into formula <em><strong>(M1)</strong> </em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"><em>eg </em> \(\cos \theta \frac{{1 \times 2 + 3 \times 4 + 2 \times a}}{{\sqrt {{1^2} + {3^2} + {2^2}} \sqrt {{2^2} + {4^2} + {a^2}} }}\) , \(\frac{{14 + 2a}}{{\sqrt {14} \sqrt {4 + 16 + {a^2}} }}\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">simplification leading to required answer <em><strong>A1</strong> </em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"><em>eg </em> \(\cos \theta = \frac{{14 + 2a}}{{\sqrt {14} \sqrt {20 + {a^2}} }}\)</span></p>
<p><span style="font-family: Times New Roman; font-size: medium;">\(\cos \theta = \frac{{2a + 14}}{{\sqrt {14{a^2} + 280} }}\) </span><strong><span style="font-family: times new roman,times; font-size: medium;"><em>AG N0</em> </span></strong></p>
<p><em><strong><span style="font-family: times new roman,times; font-size: medium;"> [4 marks]</span></strong></em></p>
<p> </p>
<p><span style="font-family: times new roman,times; font-size: medium;">correct setup <em><strong>(A1)</strong> </em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"><em>eg </em> \(\cos 1.2 = \frac{{2a + 14}}{{\sqrt {14{a^2} + 280} }}\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">valid attempt to solve <em><strong>(M1)</strong> </em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"><em>eg</em> sketch, \(\frac{{2a + 14}}{{\sqrt {14{a^2} + 280} }} - \cos 1.2 = 0\) , attempt to square </span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(a = - 3.25\) </span><em><strong><span style="font-family: times new roman,times; font-size: medium;">A2 N3 </span></strong></em></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">c.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">correct setup <em><strong>(A1)</strong> </em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"><em>eg </em> \(\cos 1.2 = \frac{{2a + 14}}{{\sqrt {14{a^2} + 280} }}\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">valid attempt to solve <em><strong>(M1)</strong> </em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"><em>eg</em> sketch, \(\frac{{2a + 14}}{{\sqrt {14{a^2} + 280} }} - \cos 1.2 = 0\) , attempt to square </span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(a = - 3.25\) </span><em><strong><span style="font-family: times new roman,times; font-size: medium;">A2 N3 </span></strong></em></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">c.ii.</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;">The majority of candidates successfully found the vectors between the given points in part (a).</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;">In part (b), while most candidates correctly found the value of \(a\), many unnecessarily worked with the magnitudes of the vectors, sometimes leading to algebra errors.</span></p>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">Some candidates showed a minimum of working in part (c)(i); in a “show that” question, candidates need to ensure that their working clearly leads to the answer given. A common error was simplifying the magnitude of vector AC to \(\sqrt {20{a^2}} \) instead of \(\sqrt {20 + {a^2}} \) .</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">In part (c)(ii), a disappointing number of candidates embarked on a usually fruitless quest for an algebraic solution rather than simply solving the resulting equation with their GDC. Many of these candidates showed quite weak algebra manipulation skills, with errors involving the square root occurring in a myriad of ways.</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;">In part (c)(ii), a disappointing number of candidates embarked on a usually fruitless quest for an algebraic solution rather than simply solving the resulting equation with their GDC. Many of these candidates showed quite weak algebra manipulation skills, with errors involving the square root occurring in a myriad of ways.</span></p>
<div class="question_part_label">c.ii.</div>
</div>
<br><hr><br><div class="question">
<p><span style="font-family: times new roman,times; font-size: medium;">Let \({\boldsymbol{v}} = 3{\boldsymbol{i}} + 4{\boldsymbol{j}} + {\boldsymbol{k}}\) and \({\boldsymbol{w}} = {\boldsymbol{i}} + 2{\boldsymbol{j}} - 3{\boldsymbol{k}}\) . </span><span style="font-family: times new roman,times; font-size: medium;">The vector \({\boldsymbol{v}} + p{\boldsymbol{w}}\) is perpendicular to <strong><em>w</em></strong>. Find the value of <em>p</em>.</span></p>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\(p{\boldsymbol{w}} = p{\boldsymbol{i}} + 2p{\boldsymbol{j}} - 3p{\boldsymbol{k}}\) (seen anywhere) <strong><em>(A1)</em></strong></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">attempt to find \({\boldsymbol{v}} + p{\boldsymbol{w}}\) <strong><em>(M1)</em></strong></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">e.g. \(3{\boldsymbol{i}} + 4{\boldsymbol{j}} + {\boldsymbol{k}} + p({\boldsymbol{i}} + 2{\boldsymbol{j}} - 3{\boldsymbol{k}})\)</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">collecting terms \((3 + p){\boldsymbol{i}} + (4 + 2p){\boldsymbol{j}} + (1 - 3p){\boldsymbol{k}}\) <strong><em>A1</em></strong></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">attempt to find the dot product <em><strong>(M1)</strong></em></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">e.g. \(1(3 + p) + 2(4 + 2p) - 3(1 - 3p)\)</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">setting <strong>their</strong> dot product equal to 0 <em><strong>(M1)</strong></em></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">e.g. \(1(3 + p) + 2(4 + 2p) - 3(1 - 3p) = 0\)</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">simplifying <em><strong>A1</strong></em></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">e.g. \(3 + p + 8 + 4p - 3 + 9p = 0\) , \(14p + 8 = 0\)</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\(p = - 0.571\) \(\left( { - \frac{8}{{14}}} \right)\) <em><strong>A1 N3</strong></em></span></p>
<p><em><strong><span style="font-family: times new roman,times; font-size: medium;">[7 marks]</span></strong></em></p>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">This question was very poorly done with many leaving it blank. Of those that did attempt it, </span><span style="font-family: times new roman,times; font-size: medium;">most were able to find \({\boldsymbol{v}} + p{\boldsymbol{w}}\) but really did not know how to proceed from there. They tried </span><span style="font-family: times new roman,times; font-size: medium;">many approaches, such as, finding magnitudes, using negative reciprocals, or calculating the </span><span style="font-family: times new roman,times; font-size: medium;">angle between two vectors. A few had the idea that the scalar product should equal zero but </span><span style="font-family: times new roman,times; font-size: medium;">had trouble trying to set it up.</span></p>
</div>
<br><hr><br><div class="specification">
<p class="p1">The points A and B <span class="s1">lie on a line \(L\)</span>, and have position vectors \(\left( {\begin{array}{*{20}{c}} { - 3} \\ { - 2} \\ 2 \end{array}} \right)\) and \(\left( {\begin{array}{*{20}{c}} 6 \\ 4 \\ { - 1} \end{array}} \right)\) respectively. Let O <span class="s1">be the origin. This is shown on the following diagram.</span></p>
<p class="p1" style="text-align: center;"><span class="s1"><img src="images/Schermafbeelding_2017-02-01_om_15.56.14.png" alt="M16/5/MATME/SP2/ENG/TZ1/10"></span></p>
</div>
<div class="specification">
<p class="p1">The point C <span class="s1">also lies on \(L\)</span>, such that \(\overrightarrow {{\text{AC}}} = 2\overrightarrow {{\text{CB}}} \).</p>
</div>
<div class="specification">
<p class="p1"><span class="s1">Let \(\theta \) </span>be the angle between \(\overrightarrow {{\text{AB}}} \) and \(\overrightarrow {{\text{OC}}} \).</p>
</div>
<div class="specification">
<p class="p1"><span class="s1">Let D be a point such that \(\overrightarrow {{\text{OD}}} = k\overrightarrow {{\text{OC}}} \)</span>, where \(k > 1\)<span class="s1">. Let E </span>be a point on \(L\) <span class="s1">such that \({\rm{C\hat ED}}\) </span>is a right angle. This is shown on the following diagram.</p>
<p class="p1" style="text-align: center;"><img src="images/Schermafbeelding_2017-02-01_om_16.39.34.png" alt="M16/5/MATME/SP2/ENG/TZ1/10.d"></p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p class="p1">Find \(\overrightarrow {{\text{AB}}} \).</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">Show that \(\overrightarrow {{\text{OC}}} = \left( {\begin{array}{*{20}{c}} 3 \\ 2 \\ 0 \end{array}} \right)\).</p>
<div class="marks">[[N/A]]</div>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p class="p1">Find \(\theta \).</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">(i) <span class="Apple-converted-space"> </span>Show that \(\left| {\overrightarrow {{\text{DE}}} } \right| = (k - 1)\left| {\overrightarrow {{\text{OC}}} } \right|\sin \theta \).</p>
<p class="p2"><span class="s1">(ii) <span class="Apple-converted-space"> </span>The distance from D </span>to line \(L\) <span class="s1">is less than 3 </span>units. Find the possible values of \(k\).</p>
<div class="marks">[6]</div>
<div class="question_part_label">d.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p class="p1">valid approach (addition or subtraction) <span class="Apple-converted-space"> </span><strong><em>(M1)</em></strong></p>
<p class="p1"><em>eg</em>\(\,\,\,\,\,\)\({\text{AO}} + {\text{OB}},{\text{ B}} - {\text{A}}\)</p>
<p class="p1"><span class="Apple-converted-space">\(\overrightarrow {{\text{AB}}} = \left( {\begin{array}{*{20}{c}} 9 \\ 6 \\ { - 3} \end{array}} \right)\) </span><strong><em>A1 <span class="Apple-converted-space"> </span>N2</em></strong></p>
<p class="p1"><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">valid approach using \(\overrightarrow {{\text{OC}}} = \left( {\begin{array}{*{20}{c}} x \\ y \\ z \end{array}} \right)\) <span class="Apple-converted-space"> </span><span class="s1"><strong><em>(M1)</em></strong></span></p>
<p class="p1"><em>eg</em>\(\,\,\,\,\,\)\(\overrightarrow {{\text{AC}}} = \left( {\begin{array}{*{20}{c}} {x + 3} \\ {y + 2} \\ {z - 2} \end{array}} \right),{\text{ }}\overrightarrow {{\text{CB}}} = \left( {\begin{array}{*{20}{c}} {6 - x} \\ {4 - y} \\ { - 1 - z} \end{array}} \right)\)</p>
<p class="p1">correct working <span class="Apple-converted-space"> </span><strong><em>A1</em></strong></p>
<p class="p1"><em>eg</em>\(\,\,\,\,\,\)\(\left( {\begin{array}{*{20}{c}} {x + 3} \\ {y + 2} \\ {z - 2} \end{array}} \right) = \left( {\begin{array}{*{20}{c}} {12 - 2x} \\ {8 - 2y} \\ { - 2 - 2z} \end{array}} \right)\)</p>
<p class="p1">all three equations <span class="Apple-converted-space"> </span><strong><em>A1</em></strong></p>
<p class="p1"><em>eg</em>\(\,\,\,\,\,\)\(x + 3 = 12 - 2x,{\text{ }}y + 2 = 8 - 2y,{\text{ }}z - 2 = - 2 - 2z\)<span class="s2">,</span></p>
<p class="p1"><span class="s2">\(\overrightarrow {{\text{OC}}} = \left( {\begin{array}{*{20}{c}} 3 \\ 2 \\ 0 \end{array}} \right)\) <span class="Apple-converted-space"> </span></span><strong><em>AG <span class="Apple-converted-space"> </span>N0</em></strong></p>
<p class="p3"><strong>METHOD 2</strong></p>
<p class="p3">valid approach <span class="Apple-converted-space"> </span><strong><em>(M1)</em></strong></p>
<p class="p3"><em>eg</em>\(\,\,\,\,\,\)\(\overrightarrow {{\text{OC}}} - \overrightarrow {{\text{OA}}} = 2\left( {\overrightarrow {{\text{OB}}} - \overrightarrow {{\text{OC}}} } \right)\)</p>
<p class="p3">correct working <span class="Apple-converted-space"> </span><strong><em>A1</em></strong></p>
<p class="p3"><em>eg</em>\(\,\,\,\,\,\)\(3\overrightarrow {{\text{OC}}} = 2\overrightarrow {{\text{OB}}} + \overrightarrow {{\text{OA}}} \)</p>
<p class="p4">correct substitution of \(\overrightarrow {{\text{OB}}} \) <span class="s3">and \(\overrightarrow {{\text{OA}}} \) <span class="Apple-converted-space"> </span><strong><em>A1</em></strong></span></p>
<p class="p3"><em>eg</em>\(\,\,\,\,\,\)\(3\overrightarrow {{\text{OC}}} = 2\left( {\begin{array}{*{20}{c}} 6 \\ 4 \\ { - 1} \end{array}} \right) + \left( {\begin{array}{*{20}{c}} { - 3} \\ { - 2} \\ 2 \end{array}} \right),{\text{ }}3\overrightarrow {{\text{OC}}} = \left( {\begin{array}{*{20}{c}} 9 \\ 6 \\ 0 \end{array}} \right)\)</p>
<p class="p3"><span class="Apple-converted-space">\(\overrightarrow {{\text{OC}}} = \left( {\begin{array}{*{20}{c}} 3 \\ 2 \\ 0 \end{array}} \right)\) </span><strong><em>AG <span class="Apple-converted-space"> </span>N0</em></strong></p>
<p class="p3"><strong>METHOD 3</strong></p>
<p class="p3">valid approach <span class="Apple-converted-space"> </span><strong><em>(M1)</em></strong></p>
<p class="p3"><em>eg</em>\(\,\,\,\,\,\)\(\overrightarrow {{\text{AC}}} = \frac{2}{3}\overrightarrow {{\text{AB}}} \), diagram, \(\overrightarrow {{\text{CB}}} = \frac{1}{3}\overrightarrow {{\text{AB}}} \)</p>
<p class="p3"><img src="images/Schermafbeelding_2017-02-02_om_08.19.06.png" alt="M16/5/MATME/SP2/ENG/TZ1/10.b/M"></p>
<p class="p3">correct working <span class="Apple-converted-space"> </span><strong><em>A1</em></strong></p>
<p class="p3"><em>eg</em>\(\,\,\,\,\,\)\(\overrightarrow {{\text{AC}}} = \left( {\begin{array}{*{20}{c}} 6 \\ 4 \\ { - 2} \end{array}} \right),{\text{ }}\overrightarrow {{\text{CB}}} = \left( {\begin{array}{*{20}{c}} 3 \\ 2 \\ { - 1} \end{array}} \right)\)</p>
<p class="p3">correct working involving \(\overrightarrow {{\text{OC}}} \) <span class="Apple-converted-space"> </span><strong><em>A1</em></strong></p>
<p class="p3"><em>eg</em>\(\,\,\,\,\,\)\(\overrightarrow {{\text{OC}}} = \left( {\begin{array}{*{20}{c}} { - 3} \\ { - 2} \\ 2 \end{array}} \right) + \left( {\begin{array}{*{20}{c}} 6 \\ 4 \\ { - 2} \end{array}} \right),{\text{ }}\left( {\begin{array}{*{20}{c}} 6 \\ 4 \\ { - 1} \end{array}} \right) - \left( {\begin{array}{*{20}{c}} 3 \\ 2 \\ { - 1} \end{array}} \right)\)</p>
<p class="p3"><span class="Apple-converted-space">\(\overrightarrow {{\text{OC}}} = \left( {\begin{array}{*{20}{c}} 3 \\ 2 \\ 0 \end{array}} \right)\) </span><strong><em>AG <span class="Apple-converted-space"> </span>N0</em></strong></p>
<p class="p3"><strong><em>[3 marks]</em></strong></p>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">finding scalar product and magnitudes <span class="Apple-converted-space"> </span><strong><em>(A1)(A1)(A1)</em></strong></p>
<p class="p1">scalar product \( = (9 \times 3) + (6 \times 2) + ( - 3 \times 0){\text{ }}( = 39)\)</p>
<p class="p1">magnitudes \(\sqrt {81 + 36 + 9} {\text{ }}( = 11.22),{\text{ }}\sqrt {9 + 4} {\text{ }}( = 3.605)\)</p>
<p class="p1">substitution into formula <span class="Apple-converted-space"> </span><strong><em>M1</em></strong></p>
<p class="p1"><em>eg</em>\(\,\,\,\,\,\)\(\cos \theta = \frac{{(9 \times 3) + 12}}{{\sqrt {126} \times \sqrt {13} }}\)</p>
<p class="p1">\(\theta = 0.270549{\text{ }}({\text{accept }}15.50135^\circ )\)</p>
<p class="p1"><span class="Apple-converted-space">\(\theta = 0.271{\text{ }}({\text{accept }}15.5^\circ )\) </span><strong><em>A1 <span class="Apple-converted-space"> </span>N4</em></strong></p>
<p class="p1"><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">(i) <span class="Apple-converted-space"> </span>attempt to use a trig ratio <span class="Apple-converted-space"> </span><strong><em>M1</em></strong></p>
<p class="p1"><em>eg</em>\(\,\,\,\,\,\)\(\sin \theta = \frac{{{\text{DE}}}}{{{\text{CD}}}},{\text{ }}\left| {\overrightarrow {{\text{CE}}} } \right| = \left| {\overrightarrow {{\text{CD}}} } \right|\cos \theta \)</p>
<p class="p1">attempt to express \(\overrightarrow {{\text{CD}}} \) in terms of \(\overrightarrow {{\text{OC}}} \) <span class="Apple-converted-space"> </span><strong><em>M1</em></strong></p>
<p class="p1"><em>eg</em>\(\,\,\,\,\,\)\(\overrightarrow {{\text{OC}}} + \overrightarrow {{\text{CD}}} = \overrightarrow {{\text{OD}}} ,{\text{ OC}} + {\text{CD}} = {\text{OD}}\)</p>
<p class="p1">correct working <span class="Apple-converted-space"> </span><strong><em>A1</em></strong></p>
<p class="p2"><em>eg</em>\(\,\,\,\,\,\)\(\left| {k\overrightarrow {{\text{OC}}} - \overrightarrow {{\text{OC}}} } \right|\sin \theta \)</p>
<p class="p1"><span class="s1">\(\left| {\overrightarrow {{\text{DE}}} } \right| = (k - 1)\left| {\overrightarrow {{\text{OC}}} } \right|\sin \theta \) <span class="Apple-converted-space"> </span></span><strong><em>AG <span class="Apple-converted-space"> </span>N0</em></strong></p>
<p class="p2">(ii) <span class="Apple-converted-space"> </span>valid approach involving the segment DE <span class="Apple-converted-space"> </span><span class="s2"><strong><em>(M1)</em></strong></span></p>
<p class="p2"><span class="s2"><em>eg</em>\(\,\,\,\,\,\)</span>recognizing \(\left| {\overrightarrow {{\text{DE}}} } \right| < 3,{\text{ DE}} = 3\)</p>
<p class="p1">correct working (accept equation) <span class="Apple-converted-space"> </span><strong><em>(A1)</em></strong></p>
<p class="p1"><em>eg</em>\(\,\,\,\,\,\)\((k - 1)(\sqrt {13} )\sin 0.271 < 3,{\text{ }}k - 1 = 3.11324\)</p>
<p class="p1"><span class="Apple-converted-space">\(1 < k < 4.11{\text{ }}({\text{accept }}k < 4.11{\text{ but not }}k = 4.11)\) </span><strong><em>A1 <span class="Apple-converted-space"> </span>N2</em></strong></p>
<p class="p1"><strong><em>[6 marks]</em></strong></p>
<div class="question_part_label">d.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
<p class="p1">The majority of candidates had little difficulty with parts (a) and (c). The most common error in both these parts were unforced arithmetic errors and occasional misreads of the vectors.</p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">In part (b), candidates who were successful used a variety of different approaches, and it was pleasing to see the vast majority of these being well reasoned, however, there were numerous unsuccessful responses including those who attempted to use the given vector to work backwards. A lack of appropriate vector notation often meant that ideas were not always clearly communicated.</p>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">The majority of candidates had little difficulty with parts (a) and (c). The most common error in both these parts were unforced arithmetic errors and occasional misreads of the vectors.</p>
<div class="question_part_label">c.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">The majority of candidates struggled to make any progress in (d), with very few realizing that simple right-angled trigonometry could be used. Few were able to successfully express CD in terms of OC which was required to show the given result. Very few candidates attempted (d)(ii), with many unable to make the connection with results found in previous parts of the question.</p>
<div class="question_part_label">d.</div>
</div>
<br><hr><br><div class="specification">
<p><span style="font-family: times new roman,times; font-size: medium;">Line \({L_1}\) passes through points \({\text{A}}(1{\text{, }} - 1{\text{, }}4)\) and \({\text{B}}(2{\text{, }} - 2{\text{, }}5)\) .</span></p>
</div>
<div class="specification">
<p><span style="font-family: times new roman,times; font-size: medium;">Line \({L_2}\) has equation \({\boldsymbol{r}} = \left( \begin{array}{l}<br>2\\<br>4\\<br>7<br>\end{array} \right) + s\left( \begin{array}{l}<br>2\\<br>1\\<br>3<br>\end{array} \right)\) .</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 \(\overrightarrow {{\rm{AB}}} \) .</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 an equation for \({L_1}\) in the form \({\boldsymbol{r}} = {\boldsymbol{a}} + t{\boldsymbol{b}}\) .</span></p>
<div class="marks">[2]</div>
<div class="question_part_label">b.</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 angle between \({L_1}\) and \({L_2}\) .</span></p>
<div class="marks">[7]</div>
<div class="question_part_label">c.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">The lines \({L_1}\) and \({L_2}\) intersect at point C. Find the coordinates of C.</span></p>
<div class="marks">[6]</div>
<div class="question_part_label">d.</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;">appropriate approach <em><strong>(M1)</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g. \(\overrightarrow {{\rm{AO}}} + \overrightarrow {{\rm{OB}}} \) , \(B - A\)</span></p>
<p><span style="font-family: Times New Roman; font-size: medium;">\(\overrightarrow {{\rm{AB}}} = \left( {\begin{array}{*{20}{c}}<br>1\\<br>{ - 1}\\<br>1<br>\end{array}} \right)\) </span><em><strong><span style="font-family: times new roman,times; font-size: medium;">A1 N2</span></strong></em></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;">any correct equation in the form \({\boldsymbol{r}} = {\boldsymbol{a}} + t{\boldsymbol{b}}\) <em><strong>A2 N2</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">where \({\boldsymbol{b}}\) is a scalar multiple of \(\left( {\begin{array}{*{20}{c}}<br>1\\<br>{ - 1}\\<br>1<br>\end{array}} \right)\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g. \({\boldsymbol{r}} = \left( {\begin{array}{*{20}{c}}<br>1\\<br>{ - 1}\\<br>4<br>\end{array}} \right) + t\left( {\begin{array}{*{20}{c}}<br>1\\<br>{ - 1}\\<br>1<br>\end{array}} \right)\) , \({\boldsymbol{r}} = \left( {\begin{array}{*{20}{c}}<br>{2 + t}\\<br>{ - 2 - t}\\<br>{5 + t}<br>\end{array}} \right)\) , \({\boldsymbol{r}} = 2{\boldsymbol{i}} - 2{\boldsymbol{j}} + 5{\boldsymbol{k}} + t({\boldsymbol{i}} - {\boldsymbol{j}} + {\boldsymbol{k}})\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"> <em><strong>[2 marks]</strong></em></span></p>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">choosing correct direction vectors \(\left( \begin{array}{l}<br>2\\<br>1\\<br>3<br>\end{array} \right)\) , \(\left( {\begin{array}{*{20}{c}}<br>1\\<br>{ - 1}\\<br>1<br>\end{array}} \right)\) </span><em><strong><span style="font-family: times new roman,times; font-size: medium;"> (A1)(A1)</span></strong></em></p>
<p><span style="font-family: times new roman,times; font-size: medium;">finding scalar product and magnitudes <em><strong>(A1)(A1)(A1)</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">scalar product \( = 1 \times 2 + - 1 \times 1 + 1 \times 3\) \(\left( { = 4} \right)\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">magnitudes \(\sqrt {{1^2} + {{( - 1)}^2} + {1^2}}{\text{ }}( = 1.73 \ldots )\) , \(\sqrt {4 + 1 + 9}{\text{ }}( = 3.74 \ldots )\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">substitution into \(\frac{{{\boldsymbol{u}} \bullet {\boldsymbol{v}}}}{{|{\boldsymbol{u}}||{\boldsymbol{v}}|}}\) (</span><span style="font-family: times new roman,times; font-size: medium;">accept <span lang="EN-US">\(\theta = \frac{{{\boldsymbol{u}} \bullet {\boldsymbol{v}}}}{{|{\boldsymbol{u}}||{\boldsymbol{v}}|}}\) </span>, but not \(\sin \theta = \frac{{{\boldsymbol{u}} \bullet {\boldsymbol{v}}}}{{|{\boldsymbol{u}}||{\boldsymbol{v}}|}}\) ) </span><em><strong><span style="font-family: times new roman,times; font-size: medium;">M1</span></strong></em></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g. \(\cos \theta = \frac{{1 \times 2 + - 1 \times 1 + 1 \times 3}}{{\sqrt {{1^2} + {{( - 1)}^2} + {1^2}} \sqrt {{2^2} + {1^2} + {3^2}} }}\) , \(\cos \theta = \frac{4}{{\sqrt {42} }}\)</span></p>
<p><span style="font-family: Times New Roman; font-size: medium;">\(\theta = 0.906\) \(({51.9^ \circ })\) </span><strong><em><span style="font-family: times new roman,times; font-size: medium;">A1 N5</span></em></strong></p>
<p><strong><em><span style="font-family: times new roman,times; font-size: medium;"> [7 marks]</span></em></strong></p>
<div class="question_part_label">c.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><strong><span style="font-family: times new roman,times; font-size: medium;">METHOD 1</span></strong><span style="font-family: times new roman,times;"><span style="font-size: medium;"> (from \({\boldsymbol{r}} = \left( {\begin{array}{*{20}{c}}<br> 1 \\ <br> { - 1} \\ <br> 4 <br>\end{array}} \right) + t\left( {\begin{array}{*{20}{c}}<br> 1 \\ <br> { - 1} \\ <br> 1 <br>\end{array}} \right)\)</span><span style="font-size: medium;"> )</span></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">appropriate approach <strong><em>(M1)</em></strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g. \({\boldsymbol{p}} = {\boldsymbol{r}}\) , \(\left( {\begin{array}{*{20}{c}}<br>1\\<br>{ - 1}\\<br>4<br>\end{array}} \right) + \left( {\begin{array}{*{20}{c}}<br>1\\<br>{ - 1}\\<br>1<br>\end{array}} \right)t = \left( {\begin{array}{*{20}{c}}<br>2\\<br>4\\<br>7<br>\end{array}} \right) + \left( {\begin{array}{*{20}{c}}<br>2\\<br>1\\<br>3<br>\end{array}} \right)s\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">two <strong>correct</strong> equations <strong><em>A1A1</em></strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g. \(1 + t = 2 + 2s\) , <span lang="EN-US">\( - 1 - t = 4 + s\)</span> , \(4 + t = 7 + 3s\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">attempt to solve <strong><em>(M1)</em></strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">one correct parameter <em><strong>A1</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g. \(t = - 3\) , \(s = - 2\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">C is \(( - 2{\text{, }}2{\text{, }}1)\) <em><strong>A1 N3</strong></em></span></p>
<p><strong> <span style="font-family: times new roman,times; font-size: medium;">METHOD 2</span></strong><span style="font-family: times new roman,times;"><span style="font-size: medium;"> (from \({\boldsymbol{r}} = \left( {\begin{array}{*{20}{c}}<br> 2 \\ <br> { - 2} \\ <br> 5 <br>\end{array}} \right) + t\left( {\begin{array}{*{20}{c}}<br> 1 \\ <br> { - 1} \\ <br> 1 <br>\end{array}} \right)\)</span><span style="font-size: medium;"> )</span></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">appropriate approach <em><strong>(M1)</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g. \({\boldsymbol{p}} = {\boldsymbol{r}}\) , \(\left( {\begin{array}{*{20}{c}}<br>2\\<br>{ - 2}\\<br>5<br>\end{array}} \right) + t\left( {\begin{array}{*{20}{c}}<br>1\\<br>{ - 1}\\<br>1<br>\end{array}} \right) = \left( {\begin{array}{*{20}{c}}<br>2\\<br>4\\<br>7<br>\end{array}} \right) + \left( {\begin{array}{*{20}{c}}<br>2\\<br>1\\<br>3<br>\end{array}} \right)s\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">two <strong>correct</strong> equations <em><strong>A1A1</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g. \(2 + t = 2 + 2s\) , <span lang="EN-US">\( - 2 - t = 4 + s\)</span> , \(5 + t = 7 + 3s\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">attempt to solve <em><strong> (M1)</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">one correct parameter <em><strong>A1</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g. \(t = - 4\) , \(s = - 2\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">C is \(( - 2{\text{, }}2{\text{, }}1)\) <em><strong>A1 N3</strong></em></span></p>
<p><em><strong><span style="font-family: times new roman,times; font-size: medium;"> [6 marks]</span></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;">
<p><span style="font-family: times new roman,times; font-size: medium;">Finding \(\overrightarrow {{\rm{AB}}} \) was generally well done, although some candidates reversed the subtraction. </span></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">In part (b) not all the candidates recognized that \(\overrightarrow {{\rm{AB}}} \) </span><span style="font-family: times new roman,times; font-size: medium;">was the direction vector of the </span><span style="font-family: times new roman,times; font-size: medium;">line, as some used the position vector of point B as the direction vector.</span></p>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">Many candidates successfully used scalar product and magnitudes in part (c), although a large number did choose vectors other than the direction vectors and many did not state clearly which vectors they were using. </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;">Candidates who were comfortable on the first three parts often had little difficulty with the final part. While the resulting systems were easily solved algebraically, a surprising number of candidates did not check their solutions either manually or with technology. An occasionally seen error in the final part was using a midpoint to find C. Some candidates found the point of intersection in part (c) rather than in part (d), indicating a familiarity with the type of question but a lack of understanding of the concepts involved. </span></p>
<div class="question_part_label">d.</div>
</div>
<br><hr><br><div class="specification">
<p><span style="font-family: times new roman,times; font-size: medium;">The diagram shows a parallelogram ABCD.</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"><br><img src="images/tired.png" alt></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">The coordinates of A, B and D are A(1, 2, 3) , B(6, 4,4 ) and D(2, 5, 5) .</span></p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">(i) Show that \(\overrightarrow {{\rm{AB}}} = \left( {\begin{array}{*{20}{c}}<br>5\\<br>2\\<br>1<br>\end{array}} \right)\) .</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">(ii) Find \(\overrightarrow {{\rm{AD}}} \) .</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">(iii) <strong>Hence</strong> show that \(\overrightarrow {{\rm{AC}}} = \left( {\begin{array}{*{20}{c}}<br>6\\<br>5\\<br>3<br>\end{array}} \right)\) .</span></p>
<div class="marks">[5]</div>
<div class="question_part_label">a(i), (ii) and (iii).</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 point C.</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 align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">(i) Find \(\overrightarrow {{\rm{AB}}} \bullet \overrightarrow {{\rm{AD}}} \).</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">(ii) <strong>Hence</strong> find angle <em>A</em>.</span></p>
<div class="marks">[7]</div>
<div class="question_part_label">c(i) and (ii).</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, or otherwise, find the area of the parallelogram.</span></p>
<div class="marks">[3]</div>
<div class="question_part_label">d.</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;">(i) evidence of approach <em><strong>M1 </strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g. \({\text{B}} - {\text{A}}\) , \(\overrightarrow {{\rm{AO}}} + \overrightarrow {{\rm{OB}}} \) , \(\left( {\begin{array}{*{20}{c}}<br>6\\<br>4\\<br>4<br>\end{array}} \right) - \left( {\begin{array}{*{20}{c}}<br>1\\<br>2\\<br>3<br>\end{array}} \right)\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(\overrightarrow {{\rm{AB}}} = \left( {\begin{array}{*{20}{c}}<br>5\\<br>2\\<br>1<br>\end{array}} \right)\) <em><strong>AG N0</strong> </em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">(ii) evidence of approach <em><strong>(M1)</strong> </em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g. \({\text{D}} - {\text{A}}\) , \(\overrightarrow {{\rm{AO}}} + \overrightarrow {{\rm{OD}}} \) , \(\left( {\begin{array}{*{20}{c}}<br>2\\<br>5\\<br>5<br>\end{array}} \right) - \left( {\begin{array}{*{20}{c}}<br>1\\<br>2\\<br>3<br>\end{array}} \right)\) </span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(\overrightarrow {{\rm{AD}}}= \left( {\begin{array}{*{20}{c}}<br>1\\<br>3\\<br>2<br>\end{array}} \right)\) <em><strong> A1 N2</strong> </em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">(iii) evidence of approach <em><strong>(M1) </strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e</span><span style="font-family: times new roman,times; font-size: medium;">.g. \(\overrightarrow {{\rm{AC}}} = \overrightarrow {{\rm{AB}}} + \overrightarrow {{\rm{AD}}} \)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">correct substitution <em><strong>A1</strong> </em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g. \(\overrightarrow {{\rm{AC}}} = \left( {\begin{array}{*{20}{c}}<br>5\\<br>2\\<br>1<br>\end{array}} \right) + \left( {\begin{array}{*{20}{c}}<br>1\\<br>3\\<br>2<br>\end{array}} \right)\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(\overrightarrow {{\rm{AC}}} = \left( {\begin{array}{*{20}{c}}<br>6\\<br>5\\<br>3<br>\end{array}} \right)\) <em><strong>AG N0</strong> </em></span></p>
<p><em><span style="font-family: times new roman,times; font-size: medium;"><strong>[5 marks]</strong> </span></em></p>
<div class="question_part_label">a(i), (ii) and (iii).</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">evidence of combining vectors (there are at least 5 ways) <em><strong>(M1) </strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g. \(\overrightarrow {{\rm{OC}}} = \overrightarrow {{\rm{OA}}} + \overrightarrow {{\rm{AC}}} \) , \(\overrightarrow {{\rm{OC}}} = \overrightarrow {{\rm{OB}}} + \overrightarrow {{\rm{AD}}} \), \(\overrightarrow {{\rm{AB}}} = \overrightarrow {{\rm{OC}}} - \overrightarrow {{\rm{OD}}} \) </span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">correct substitution <em><strong>A1 </strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(\overrightarrow {{\rm{OC}}} = \left( {\begin{array}{*{20}{c}}<br>1\\<br>2\\<br>3<br>\end{array}} \right) + \left( {\begin{array}{*{20}{c}}<br>6\\<br>5\\<br>3<br>\end{array}} \right)\left( { = \left( {\begin{array}{*{20}{c}}<br>7\\<br>7\\<br>6<br>\end{array}} \right)} \right)\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g. coordinates of C are \((7{\text{, }}7{\text{, }}6)\) <em><strong>A1 N1</strong> </em></span></p>
<p><em><span style="font-family: times new roman,times; font-size: medium;"><strong>[3 marks]</strong> </span></em></p>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">(i) evidence of using scalar product on \(\overrightarrow {{\rm{AB}}} \) and \(\overrightarrow {{\rm{AD}}} \) <em><strong>(M1) </strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g. \(\overrightarrow {{\rm{AB}}} \bullet \overrightarrow {{\rm{AD}}} = 5(1) + 2(3) + 1(2)\) </span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(\overrightarrow {{\rm{AB}}} \bullet \overrightarrow {{\rm{AD}}} = 13\) <em><strong>A1 N2</strong> </em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">(ii) \(\left| {\overrightarrow {{\rm{AB}}} } \right| = 5.477 \ldots \) , \(\left| {\overrightarrow {{\rm{AD}}} } \right| = 3.741 \ldots \) <em><strong>(A1)(A1) </strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">evidence of using \(\cos A = \frac{{\overrightarrow {{\rm{AB}}} \bullet \overrightarrow {{\rm{AD}}} }}{{\left| {\overrightarrow {{\rm{AB}}} } \right|\left| {\overrightarrow {{\rm{AD}}} } \right|}}\) <em><strong>(M1) </strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">correct substitution <em><strong>A1 </strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g. \(\cos A = \frac{{13}}{{20.493}}\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(\widehat A = 0.884\) \((50.6^\circ )\) <em><strong>A1 N3</strong> </em></span></p>
<p><em><strong><span style="font-family: times new roman,times; font-size: medium;">[</span></strong><span style="font-family: times new roman,times; font-size: medium;"><strong>7 marks]</strong> </span></em></p>
<div class="question_part_label">c(i) and (ii).</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><strong><span style="font-family: times new roman,times; font-size: medium;">METHOD 1 </span></strong></p>
<p><span style="font-family: times new roman,times; font-size: medium;">evidence of using \({\rm{area}} = 2\left( {\frac{1}{2}\left| {\overrightarrow {{\rm{AD}}} } \right|\left| {\overrightarrow {{\rm{AB}}} } \right|\sin {\rm{D}}\widehat {\rm{A}}{\rm{B}}} \right)\) <strong><em> (M1) </em></strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">correct substitution <em><strong>A1</strong> </em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g. \({\rm{area}} = 2\left( {\frac{1}{2}(3,741 \ldots )(5.477 \ldots )\sin 0.883 \ldots } \right)\) </span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\({\rm{area}} = 15.8\) <em><strong>A1 N2</strong> </em></span></p>
<p><strong><span style="font-family: times new roman,times; font-size: medium;">METHOD 2 </span></strong></p>
<p><span style="font-family: times new roman,times; font-size: medium;">evidence of using \({\rm{area}} = b \times h\) <em><strong>(M1) </strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">finding height of parallelogram <em><strong>A1 </strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g. \(h = 3.741 \ldots \times \sin 0.883 \ldots ( = 2.892 \ldots )\) , \(h = 5.477 \ldots \times \sin 0.883 \ldots ( = 4.234 \ldots )\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\({\rm{area}} = 15.8\) <em><strong>A1 N2</strong> </em></span></p>
<p><em><span style="font-family: times new roman,times; font-size: medium;"><strong>[3 marks]</strong> </span></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;">
<p><span style="font-family: times new roman,times; font-size: medium;">Candidates performed very well in this question, showing a strong ability to work with the algebra and geometry of vectors. </span></p>
<div class="question_part_label">a(i), (ii) and (iii).</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">Candidates performed very well in this question, showing a strong ability to work with the algebra and geometry of vectors. </span></p>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">Some candidates were unable to find the scalar product in part (c), yet still managed to find the correct angle, able to use the formula in the information booklet without knowing that the scalar product is a part of that formula. </span></p>
<div class="question_part_label">c(i) and (ii).</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">Few candidates considered that the area of the parallelogram is twice the area of a triangle, which is conveniently found using \({\rm{B}}\widehat {\rm{A}}{\rm{D}}\) . In an effort to find base \(\times \) height , many candidates multiplied the magnitudes of \(\overrightarrow {{\rm{AB}}} \) and \(\overrightarrow {{\rm{AD}}} \) , missing that the height of a parallelogram is perpendicular to a base. </span></p>
<div class="question_part_label">d.</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;">The following diagram shows two perpendicular vectors <strong><em>u </em></strong>and <strong><em>v</em></strong>.</span></p>
<p style="font: normal normal normal 21px/normal 'Times New Roman'; text-align: center; margin: 0px;"><br><img src="images/maths_4.png" alt></p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p><span style="font-size: medium; font-family: 'times new roman', times;">Let \(w = u - v\). Represent \(w\) on the diagram above.</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><span><span style="font-family: 'times new roman', times; font-size: medium;"><span style="line-height: normal;">Given that \(u = \left( \begin{array}{c}3\\2\\1\end{array} \right)\) and </span></span></span></span><span style="font-family: 'times new roman', times; font-size: medium;"><span style="line-height: normal;">\(v = \left( \begin{array}{c}5\\n\\3\end{array} \right)\), where \(n \in \mathbb{Z}\), find \(</span></span><em style="font-family: 'times new roman', times; font-size: medium; font-style: normal; font-variant: normal; font-weight: normal; line-height: normal;">n\)</em><span style="font-family: 'times new roman', times; font-size: medium;"><span style="line-height: normal;">.</span></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><strong><span style="font-size: medium; font-family: 'times new roman', times;">METHOD 1</span></strong></p>
<p><strong><span style="font-size: medium; font-family: 'times new roman', times;"><br><img src="images/M1.jpg" alt> </span><em><span style="font-size: medium; font-family: 'times new roman', times;">A1A1 N2</span></em></strong></p>
<p><strong><span style="font-size: medium; font-family: 'times new roman', times;"> </span></strong></p>
<p><span style="font-size: medium; font-family: 'times new roman', times;"><strong>Note</strong>: Award <em><strong>A1</strong> </em>for segment connecting endpoints and <em><strong>A1</strong> </em>for direction (must see arrow).</span></p>
<p><strong><span style="font-size: medium; font-family: 'times new roman', times;">METHOD 2</span></strong></p>
<p><strong style="background-color: #f7f7f7;"><span style="font-size: medium; font-family: 'times new roman', times;"><br><img src="images/M2.jpg" alt><em> </em></span></strong><em><strong style="background-color: #f7f7f7;"><span style="font-size: medium; font-family: 'times new roman', times;">A1A1 N2</span></strong></em></p>
<p> </p>
<p><span style="font-size: medium; font-family: 'times new roman', times;"><strong>Notes</strong>: Award <em><strong>A1</strong> </em>for segment connecting endpoints and <em><strong>A1</strong> </em>for direction (must see arrow).</span></p>
<p><span style="font-size: medium; font-family: 'times new roman', times;">Additional lines not required.</span></p>
<p><strong><em><span style="font-size: medium; font-family: 'times new roman', times;">[2 marks]</span></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: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">evidence of setting scalar product equal to zero (seen anywhere) <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;"><em>eg </em><strong><em>u </em></strong>\( \cdot \)<strong><em> v </em></strong>\( = 0,{\text{ }}15 + 2n + 3 = 0\)</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;">correct expression for scalar product <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;"><em>eg </em>\(3 \times 5 + 2 \times n + 1 \times 3,{\text{ }}2n + 18 = 0\)</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 solve equation <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> \(2n = - 18\)</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;">\(n = - 9\) <strong><em>A1 N3</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">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 lines \({L_1}\) , \({L_2}\) , \({L_2}\) , and \({L_4}\) , with respective equations.</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\({L_1}\) : \(\left( {\begin{array}{*{20}{c}}<br>x\\<br>y\\<br>z<br>\end{array}} \right) = \left( {\begin{array}{*{20}{c}}<br>1\\<br>2\\<br>3<br>\end{array}} \right) + t\left( {\begin{array}{*{20}{c}}<br>3\\<br>{ - 2}\\<br>1<br>\end{array}} \right)\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\({L_2}\) : \(\left( \begin{array}{l}<br>x\\<br>y\\<br>z<br>\end{array} \right) = \left( \begin{array}{l}<br>1\\<br>2\\<br>3<br>\end{array} \right) + p\left( \begin{array}{l}<br>3\\<br>2\\<br>1<br>\end{array} \right)\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\({L_3}\) : \(\left( {\begin{array}{*{20}{c}}<br>x\\<br>y\\<br>z<br>\end{array}} \right) = \left( {\begin{array}{*{20}{c}}<br>0\\<br>1\\<br>0<br>\end{array}} \right) + s\left( {\begin{array}{*{20}{c}}<br>{ - 1}\\<br>2\\<br>{ - a}<br>\end{array}} \right)\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\({L_4}\) : \(\left( {\begin{array}{*{20}{c}}<br>x\\<br>y\\<br>z<br>\end{array}} \right) = q\left( {\begin{array}{*{20}{c}}<br>{ - 6}\\<br>4\\<br>{ - 2}<br>\end{array}} \right)\)</span></p>
<p> </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 the line that is parallel to \({L_4}\) .</span></p>
<div class="marks">[1]</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;">Write down the position vector of the point of intersection of \({L_1}\) and \({L_2}\) .</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><span style="font-family: times new roman,times; font-size: medium;">Given that \({L_1}\) is perpendicular to \({L_3}\) , find the value of <em>a</em> .</span></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 align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\({L_1}\) <em><strong>A1 N1</strong></em></span></p>
<p><em><strong><span style="font-family: times new roman,times; font-size: medium;">[1 mark]</span></strong></em></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\(\left( \begin{array}{l}<br>1\\<br>2\\<br>3<br>\end{array} \right)\) <em><strong>A1 N1</strong></em></span></p>
<p><em><strong><span style="font-family: times new roman,times; font-size: medium;">[1 mark]</span></strong></em></p>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">choosing correct direction vectors <em><strong>A1A1</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g. \(\left( {\begin{array}{*{20}{c}}<br>3\\<br>{ - 2}\\<br>1<br>\end{array}} \right)\) , \(\left( {\begin{array}{*{20}{c}}<br>{ - 1}\\<br>2\\<br>{ - a}<br>\end{array}} \right)\) </span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">recognizing that \({\boldsymbol{a}} \bullet {\boldsymbol{b}} = 0\) <em><strong>M1</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">correct substitution <em><strong>A1</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g. \( - 3 - 4 - a = 0\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(a = - 7\) <em><strong>A1 N3</strong></em></span></p>
<p><em><strong><span style="font-family: times new roman,times; font-size: medium;"> [5 marks]</span></strong></em></p>
<div class="question_part_label">c.</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>
<br><hr><br><div class="specification">
<p>Two points P and Q have coordinates (3, 2, 5) and (7, 4, 9) respectively.</p>
</div>
<div class="specification">
<p>Let \({\mathop {{\text{PR}}}\limits^ \to }\) = 6<em><strong>i</strong></em> − <em><strong>j</strong></em> + 3<em><strong>k</strong></em>.</p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Find \(\mathop {{\text{PQ}}}\limits^ \to \).</p>
<div class="marks">[2]</div>
<div class="question_part_label">a.i.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Find \(\left| {\mathop {{\text{PQ}}}\limits^ \to } \right|\).</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 the angle between PQ and PR.</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 area of triangle PQR.</p>
<div class="marks">[2]</div>
<div class="question_part_label">c.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Hence or otherwise find the shortest distance from R to the line through P and Q.</p>
<div class="marks">[3]</div>
<div class="question_part_label">d.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p>valid approach <em><strong>(M1)</strong></em></p>
<p><em>eg </em>(7, 4, 9) − (3, 2, 5) <em>A − B</em></p>
<p>\(\mathop {{\text{PQ}}}\limits^ \to = \) 4<em><strong>i</strong></em> + 2<em><strong>j</strong></em> + 4<em><strong>k</strong></em> \(\left( { = \left( \begin{gathered}<br> 4 \hfill \\<br> 2 \hfill \\<br> 4 \hfill \\ <br>\end{gathered} \right)} \right)\) <em><strong>A1 N2</strong></em></p>
<p><em><strong>[2 marks]</strong></em></p>
<div class="question_part_label">a.i.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p>correct substitution into magnitude formula <em><strong>(A1)</strong></em><br><em>eg</em> \(\sqrt {{4^2} + {2^2} + {4^2}} \)</p>
<p>\(\left| {\mathop {{\text{PQ}}}\limits^ \to } \right| = 6\) <em><strong>A1 N2</strong></em></p>
<p><em><strong>[2 marks]</strong></em></p>
<div class="question_part_label">a.ii.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p>finding scalar product and magnitudes <em><strong>(A1)(A1)</strong></em></p>
<p>scalar product = (4 × 6) + (2 × (−1) + (4 × 3) (= 34)</p>
<p>magnitude of PR = \(\sqrt {36 + 1 + 9} = \left( {6.782} \right)\)</p>
<p>correct substitution of <strong>their</strong> values to find cos \({\text{Q}}\mathop {\text{P}}\limits^ \wedge {\text{R}}\) <em> <strong>M1</strong></em></p>
<p><em>eg </em>cos \({\text{Q}}\mathop {\text{P}}\limits^ \wedge {\text{R}}\,\,{\text{ = }}\frac{{24 - 2 + 12}}{{\left( 6 \right) \times \left( {\sqrt {46} } \right)}},\,\,0.8355\)</p>
<p>0.581746</p>
<p>\({\text{Q}}\mathop {\text{P}}\limits^ \wedge {\text{R}}\) = 0.582 radians or \({\text{Q}}\mathop {\text{P}}\limits^ \wedge {\text{R}}\) = 33.3° <em><strong>A1 N3</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>correct substitution <em><strong>(A1)</strong></em><br><em>eg</em> \(\frac{1}{2} \times \left| {\mathop {{\text{PQ}}}\limits^ \to } \right| \times \left| {\mathop {{\text{PR}}}\limits^ \to } \right| \times \,\,{\text{sin}}\,P,\,\,\frac{1}{2} \times 6 \times \sqrt {46} \times \,\,{\text{sin}}\,0.582\)</p>
<p>area is 11.2 (sq. units) <em><strong> A1 N2</strong></em></p>
<p><em><strong>[2 marks]</strong></em></p>
<div class="question_part_label">c.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p>recognizing shortest distance is perpendicular distance from R to line through P and Q <em><strong>(M1)</strong></em></p>
<p><em>eg</em> sketch, height of triangle with base \(\left[ {{\text{PQ}}} \right],\,\,\frac{1}{2} \times 6 \times h,\,\,{\text{sin}}\,33.3^\circ = \frac{h}{{\sqrt {46} }}\)</p>
<p>correct working <em><strong> (A1)</strong></em></p>
<p><em>eg </em>\(\frac{1}{2} \times 6 \times d = 11.2,\,\,\left| {\mathop {{\text{PR}}}\limits^ \to } \right| \times \,\,{\text{sin}}\,P,\,\,\sqrt {46} \times \,\,{\text{sin}}\,33.3^\circ \)</p>
<p>3.72677</p>
<p>distance = 3.73 (units) <em><strong>A1 N2</strong></em></p>
<p><em><strong>[3 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.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>
<br><hr><br><div class="question">
<p><span style="font-family: times new roman,times; font-size: medium;">The line <em>L<sub>1</sub></em> is represented by \({{\boldsymbol{r}}_1} = \left( {\begin{array}{*{20}{c}}<br>2\\<br>5\\<br>3<br>\end{array}} \right) + s\left( {\begin{array}{*{20}{c}}<br>1\\<br>2\\<br>3<br>\end{array}} \right)\) and the line <em>L<sub>2</sub></em> by \({{\boldsymbol{r}}_2} = \left( {\begin{array}{*{20}{c}}<br>3\\<br>{ - 3}\\<br>8<br>\end{array}} \right) + t\left( {\begin{array}{*{20}{c}}<br>{ - 1}\\<br>3\\<br>{ - 4}<br>\end{array}} \right)\) .</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"> The lines <em>L<sub>1</sub></em> and <em>L<sub>2</sub></em> intersect at point T. Find the coordinates of T. </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;">evidence of equating vectors <em><strong>(M1)</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g. \({L_1} = {L_2}\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">for any <strong>two</strong> correct equations <em><strong>A1A1</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g. \(2 + s = 3 - t\) , \(5 + 2s = - 3 + 3t\) , \(3 + 3s = 8 - 4t\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">attempting to solve the equations <em><strong>(M1)</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">finding <strong>one</strong> correct parameter \((2 = - 1{\text{, }}t = 2)\) <em><strong>A1</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">the coordinates of T are \((1{\text{, }}3{\text{, }}0)\) <em><strong>A1 N3</strong></em></span></p>
<p><em><strong><span style="font-family: times new roman,times; font-size: medium;"> [6 marks] </span></strong></em></p>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">Those candidates prepared in this topic area answered the question particularly well, often </span><span style="font-family: times new roman,times; font-size: medium;">only making some calculation error when solving the resulting system of equations. </span><span style="font-family: times new roman,times; font-size: medium;">Curiously, a few candidates found correct values for <em>s</em> and <em>t</em>, but when substituting back into </span><span style="font-family: times new roman,times; font-size: medium;">one of the vector equations, neglected to find the <em>z</em>-coordinate of T.</span></p>
</div>
<br><hr><br><div class="question">
<p><span style="font-family: times new roman,times; font-size: medium;">Line \({L_1}\) </span><span style="font-family: times new roman,times; font-size: medium;">has equation \({\boldsymbol{r}_1} = \left( {\begin{array}{*{20}{c}}<br>{10}\\<br>6\\<br>{ - 1}<br>\end{array}} \right) + s\left( {\begin{array}{*{20}{c}}<br>2\\<br>{ - 5}\\<br>{ - 2}<br>\end{array}} \right)\) </span><span style="font-family: times new roman,times; font-size: medium;">and line \({L_2}\) has equation \({\boldsymbol{r}_2} = \left( {\begin{array}{*{20}{c}}<br>2\\<br>1\\<br>{ - 3}<br>\end{array}} \right) + t\left( {\begin{array}{*{20}{c}}<br>3\\<br>5\\<br>2<br>\end{array}} \right)\) .</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">Lines \({L_1}\) and \({L_2}\) intersect at point A. Find the coordinates of A.</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;">appropriate approach <em><strong> (M1)</strong> </em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"><em>eg </em> \(\left( {\begin{array}{*{20}{c}}<br>{10}\\<br>6\\<br>{ - 1}<br>\end{array}} \right) + s\left( {\begin{array}{*{20}{c}}<br>2\\<br>{ - 5}\\<br>{ - 2}<br>\end{array}} \right) = \left( {\begin{array}{*{20}{c}}<br>2\\<br>1\\<br>{ - 3}<br>\end{array}} \right) + t\left( {\begin{array}{*{20}{c}}<br>3\\<br>5\\<br>2<br>\end{array}} \right)\) , \({L_1} = {L_2}\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">any two correct equations <em><strong> A1A1</strong> </em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"><em>eg </em> \(10 + 2s = 2 + 3t\) , \(6 - 5s = 1 + 5t\) , \( - 1 - 2s = - 3 + 2t\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">attempt to solve <em><strong>(M1)</strong> </em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"><em>eg </em>substituting one equation into another </span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">one correct parameter <em><strong>A1</strong> </em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"><em>eg </em> \(s = - 1\) , \(t = 2\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">correct substitution <em><strong> (A1)</strong> </em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"><em>eg </em> \(2 + 3(2)\) , \(1 + 5(2)\) , \( - 3 + 2(2)\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">A \( = \) (\(8\), \(11\), \(1\)) (accept column vector) <em><strong>A1 N4 </strong></em></span></p>
<p><em><span style="font-family: times new roman,times; font-size: medium;"><strong>[7 marks]</strong> </span></em></p>
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<h2 style="margin-top: 1em">Examiners report</h2>
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<p><span style="font-family: times new roman,times; font-size: medium;">Most students were able to set up one or more equations, but few chose to use their GDCs to solve the resulting system. Algebraic errors prevented many of these candidates from obtaining the final three marks. Some candidates stopped after finding the value of s and/or \(t\).</span></p>
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