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</div><h2>SL Paper 1</h2><div class="specification">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">In the following diagram, \(\boldsymbol{u} = \overrightarrow {{\rm{AB}}} \) and \(\boldsymbol{v} = \overrightarrow {{\rm{BD}}} \) .</span></p>
<p> </p>
<p><span style="font-family: times new roman,times; font-size: medium;"><br><img src="images/spec.png" alt></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">The midpoint of \(\overrightarrow {{\rm{AD}}} \) is E and \(\frac{{{\rm{BD}}}}{{{\rm{DC}}}} = \frac{1}{3}\)</span><span style="font-family: times new roman,times; font-size: medium;"> .</span></p>
<p><span style="font-size: medium;"><span style="font-family: times new roman,times;">Express each of the following vectors in terms of</span></span><span style="font-family: times new roman,times; font-size: medium;"> \(\boldsymbol{u}\)</span><span style="font-family: times new roman,times; font-size: medium;"> and \(\boldsymbol{v}\) .</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;">\(\overrightarrow {{\rm{AE}}} \)</span></p>
<div class="marks">[3]</div>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">\(\overrightarrow {{\rm{EC}}} \)</span></p>
<div class="marks">[4]</div>
<div class="question_part_label">b.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">\(\overrightarrow {{\rm{AE}}} = \frac{1}{2}\overrightarrow {{\rm{AD}}} \) </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;">attempt to find \(\overrightarrow {{\rm{AD}}} \) </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. \(\overrightarrow {{\rm{AB}}} + \overrightarrow {{\rm{BD}}} \) </span><span style="font-family: times new roman,times; font-size: medium;">, \({\boldsymbol{u}} + {\boldsymbol{v}}\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(\overrightarrow {{\rm{AE}}} = \frac{1}{2}(u + v)\) \(\left( { = \frac{1}{2}{\boldsymbol{u}} + \frac{1}{2}{\boldsymbol{v}}} \right)\) <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">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;"><span style="color: #000000;">\(\overrightarrow {{\rm{EC}}} = \overrightarrow {{\rm{AE}}} = \frac{1}{2}({\boldsymbol{u}} + {\boldsymbol{v}})\) </span></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;"> \(\overrightarrow {{\rm{DC}}} = 3{\boldsymbol{v}}\) <em><strong>A1</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">attempt to find \(\overrightarrow {{\rm{EC}}} \) </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. \(\overrightarrow {{\rm{ED}}} + \overrightarrow {{\rm{DC}}} \) , \(\frac{1}{2}({\boldsymbol{u}} + {\boldsymbol{v}}) + 3{\boldsymbol{v}}\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(\overrightarrow {{\rm{EC}}} = \frac{1}{2}{\boldsymbol{u}} + \frac{7}{2}{\boldsymbol{v}}\) \(\left( { = \frac{1}{2}({\boldsymbol{u}} + 7{\boldsymbol{v}})} \right)\) <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">b.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">b.</div>
</div>
<br><hr><br><div class="question">
<p class="p1">Let <span class="s1"><strong><em>u </em></strong>\( = - 3\)<em><strong>i </strong></em>\( + \) <strong><em>j </em></strong>\( + \) <strong><em>k </em></strong></span>and <span class="s1"><strong><em>v </em></strong>\( = m\)<strong><em>j </em></strong>\( + {\text{ }}n\)<strong><em>k </em></strong></span>, where \(m,{\text{ }}n \in \mathbb{R}\). Given that <span class="s1"><strong><em>v </em></strong></span>is a unit vector perpendicular to <span class="s1"><strong><em>u</em></strong></span>, find the possible values of \(m\) and of \(n\).</p>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question">
<p>correct scalar product <strong><em>(A1)</em></strong></p>
<p><em>eg</em>\(\,\,\,\,\,\)\(m + n\)</p>
<p>setting up their scalar product equal to 0 (seen anywhere) <strong><em>(M1)</em></strong></p>
<p><em>eg</em>\(\,\,\,\,\,\)<strong><em>u</em></strong> \( \bullet \) <strong><em>v</em></strong> \( = 0,{\text{ }} - 3(0) + 1(m) + 1(n) = 0,{\text{ }}m = - n\)</p>
<p>correct interpretation of unit vector <strong><em>(A1)</em></strong></p>
<p><em>eg</em>\(\,\,\,\,\,\)\(\sqrt {{0^2} + {m^2} + {n^2}} = 1,{\text{ }}{m^2} + {n^2} = 1\)</p>
<p>valid attempt to solve their equations (must be in one variable) <strong><em>M1</em></strong></p>
<p><em>eg</em>\(\,\,\,\,\,\)\({( - n)^2} + {n^2} = 1,{\text{ }}\sqrt {1 - {n^2}} + n = 0,{\text{ }}{m^2} + {( - m)^2} = 1,{\text{ }}m - \sqrt {1 - {m^2}} = 0\)</p>
<p>correct working <strong><em>A1</em></strong></p>
<p><em>eg</em>\(\,\,\,\,\,\)\(2{n^2} = 1,{\text{ }}2{m^2} = 1,{\text{ }}\sqrt 2 = \frac{1}{n},{\text{ }}m = \pm \frac{1}{{\sqrt 2 }}\)</p>
<p><strong>both </strong>correct pairs <strong><em>A2 N3</em></strong></p>
<p><em>eg</em>\(\,\,\,\,\,\)\(m = \frac{1}{{\sqrt 2 }}\) and \(n = - \frac{1}{{\sqrt 2 }},{\text{ }}m = - \frac{1}{{\sqrt 2 }}\) and \(n = \frac{1}{{\sqrt 2 }}\),</p>
<p>\(m = {(0.5)^{\frac{1}{2}}}\) and \(n = - {(0.5)^{\frac{1}{2}}},{\text{ }}m = - \sqrt {\frac{1}{2}} \) and \(n = \sqrt {\frac{1}{2}} \)</p>
<p><strong>Note: </strong>Award <strong><em>A0 </em></strong>for \(m = \pm \frac{1}{{\sqrt 2 }},{\text{ }}n = \pm \frac{1}{{\sqrt 2 }}\), or any other answer that does not clearly indicate the correct pairs.</p>
<p><strong><em>[7 marks]</em></strong></p>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question">
<p class="p1">Most of the candidates recognized that the scalar product of the vectors must be zero. However, some did not find the correct scalar product because they did not multiply the correct corresponding vector components of <strong><em>u</em></strong> and <strong><em>v</em></strong>. In addition, the majority of candidates did not attempt to use the fact that the unit vector <strong><em>v</em></strong> has a magnitude of 1. For the small number of candidates who were successful in solving for \(m\) and/or \(n\), some did not clearly present the correct pairs of answers.</p>
</div>
<br><hr><br><div class="specification">
<p>The vectors <strong><em>a</em></strong> = \(\left( {\begin{array}{*{20}{c}} 4 \\ 2 \end{array}} \right)\) and <strong><em>b</em></strong> = \(\left( {\begin{array}{*{20}{c}} {k + 3} \\ k \end{array}} \right)\) are perpendicular to each other.</p>
<p> </p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Find the value of \(k\).</p>
<div class="marks">[4]</div>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Given that <strong><em>c</em></strong> = <strong><em>a</em></strong> + 2<strong><em>b</em></strong>, find <strong><em>c</em></strong>.</p>
<div class="marks">[3]</div>
<div class="question_part_label">b.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p>evidence of scalar product <strong><em>M1</em></strong></p>
<p><em>eg</em>\(\,\,\,\,\,\)<strong><em>a</em></strong> \( \bullet \) <strong><em>b</em></strong>, \(4(k + 3) + 2k\)</p>
<p>recognizing scalar product must be zero <strong><em>(M1)</em></strong></p>
<p><em>eg</em>\(\,\,\,\,\,\)<strong><em>a</em></strong> \( \bullet \) <strong><em>b</em></strong> \( = 0,{\text{ }}4k + 12 + 2k = 0\)</p>
<p>correct working (must involve combining terms) <strong><em>(A1)</em></strong></p>
<p><em>eg </em>\(6k + 12,\,\,\,6k = - 12\)</p>
<p>\(\,\,\,\,\,\)\(k = - 2\) <strong><em>A1</em></strong> <strong><em>N2</em></strong></p>
<p><strong><em>[4 marks]</em></strong></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p>attempt to substitute <strong>their </strong>value of \(k\) (seen anywhere) <strong><em>(M1)</em></strong></p>
<p><em>eg</em>\(\,\,\,\,\,\)<strong><em>b</em></strong> = \(\left( {\begin{array}{*{20}{c}} { - 2 + 3} \\ { - 2} \end{array}} \right)\), 2<strong><em>b</em></strong> = \(\left( {\begin{array}{*{20}{c}} 2 \\ { - 4} \end{array}} \right)\)</p>
<p>correct working <strong><em>(A1)</em></strong></p>
<p><em>eg</em>\(\,\,\,\,\,\)\(\left( {\begin{array}{*{20}{c}} 4 \\ 2 \end{array}} \right) + \left( {\begin{array}{*{20}{c}} 2 \\ { - 4} \end{array}} \right),{\text{ }}\left( {\begin{array}{*{20}{c}} {4 + 2k + 6} \\ {2 + 2k} \end{array}} \right)\)</p>
<p><strong><em>c</em></strong> = \(\left( {\begin{array}{*{20}{c}} 6 \\ { - 2} \end{array}} \right)\) <strong><em>A1</em></strong> <strong><em>N2</em></strong></p>
<p><strong><em>[3 marks]</em></strong></p>
<div class="question_part_label">b.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">b.</div>
</div>
<br><hr><br><div class="specification">
<p><strong>Note: In this question, distance is in metres and time is in seconds.</strong></p>
<p>Two particles \({P_1}\) and \({P_2}\) start moving from a point A at the same time, along different straight lines.</p>
<p>After \(t\) seconds, the position of \({P_1}\) is given by <strong><em>r</em></strong> = \(\left( {\begin{array}{*{20}{c}} 4 \\ { - 1} \\ 3 \end{array}} \right) + t\left( {\begin{array}{*{20}{c}} 1 \\ 2 \\ { - 1} \end{array}} \right)\).</p>
</div>
<div class="specification">
<p>Two seconds after leaving A, \({P_1}\) is at point B.</p>
</div>
<div class="specification">
<p>Two seconds after leaving A, \({P_2}\) is at point C, where \(\overrightarrow {{\text{AC}}} = \left( {\begin{array}{*{20}{c}} 3 \\ 0 \\ 4 \end{array}} \right)\).</p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Find the coordinates of A.</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>Find \(\overrightarrow {{\text{AB}}} \);</p>
<div class="marks">[3]</div>
<div class="question_part_label">b.i.</div>
</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">b.ii.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Find \(\cos {\rm{B\hat AC}}\).</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>Hence or otherwise, find the distance between \({P_1}\) and \({P_2}\) two seconds after they leave A.</p>
<div class="marks">[4]</div>
<div class="question_part_label">d.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p>recognizing \(t = 0\) at A <strong><em>(M1)</em></strong></p>
<p>A is \((4,{\text{ }} - 1,{\text{ }}3)\) <strong><em>A1</em></strong> <strong><em>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><strong>METHOD 1</strong></p>
<p>valid approach <strong><em>(M1)</em></strong></p>
<p><em>eg</em>\(\,\,\,\,\,\)\(\left( {\begin{array}{*{20}{c}} 4 \\ { - 1} \\ 3 \end{array}} \right) + 2\left( {\begin{array}{*{20}{c}} 1 \\ 2 \\ { - 2} \end{array}} \right),{\text{ }}(6,{\text{ }}3,{\text{ }} - 1)\)</p>
<p>correct approach to find \(\overrightarrow {{\text{AB}}} \) <strong><em>(A1)</em></strong></p>
<p><em>eg</em>\(\,\,\,\,\,\)\({\text{AO}} + {\text{OB}},{\text{ B}} - {\text{A, }}\left( {\begin{array}{*{20}{c}} 6 \\ 3 \\ { - 1} \end{array}} \right) - \left( {\begin{array}{*{20}{c}} 4 \\ { - 1} \\ 3 \end{array}} \right)\)</p>
<p>\(\overrightarrow {{\text{AB}}} = \left( {\begin{array}{*{20}{c}} 2 \\ 4 \\ { - 4} \end{array}} \right)\) <strong><em>A1</em></strong> <strong><em>N2</em></strong></p>
<p><strong>METHOD 2</strong></p>
<p>recognizing \(\overrightarrow {{\text{AB}}} \) is two times the direction vector <strong><em>(M1)</em></strong></p>
<p>correct working <strong><em>(A1)</em></strong></p>
<p><em>eg</em>\(\,\,\,\,\,\)\(\overrightarrow {{\text{AB}}} = 2\left( {\begin{array}{*{20}{c}} 1 \\ 2 \\ { - 2} \end{array}} \right)\)</p>
<p>\(\overrightarrow {{\text{AB}}} = \left( {\begin{array}{*{20}{c}} 2 \\ 4 \\ { - 4} \end{array}} \right)\) <strong><em>A1</em></strong> <strong><em>N2</em></strong></p>
<p><strong><em>[3 marks]</em></strong></p>
<div class="question_part_label">b.i.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p>correct substitution <strong><em>(A1)</em></strong></p>
<p><em>eg</em>\(\,\,\,\,\,\)\(\left| {\overrightarrow {{\text{AB}}} } \right| = \sqrt {{2^2} + {4^2} + {4^2}} ,{\text{ }}\sqrt {4 + 16 + 16} ,{\text{ }}\sqrt {36} \)</p>
<p>\(\left| {\overrightarrow {{\text{AB}}} } \right| = 6\) <strong><em>A1</em></strong> <strong><em>N2</em></strong></p>
<p><strong><em>[2 marks]</em></strong></p>
<div class="question_part_label">b.ii.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><strong>METHOD 1 (vector approach)</strong></p>
<p>valid approach involving \(\overrightarrow {{\text{AB}}} \) and \(\overrightarrow {{\text{AC}}} \) <strong><em>(M1)</em></strong></p>
<p><em>eg</em>\(\,\,\,\,\,\)\(\overrightarrow {{\text{AB}}} \bullet \overrightarrow {{\text{AC}}} ,{\text{ }}\frac{{\overrightarrow {{\text{BA}}} \bullet \overrightarrow {{\text{AC}}} }}{{{\text{AB}} \times {\text{AC}}}}\)</p>
<p>finding scalar product and \(\left| {\overrightarrow {{\text{AC}}} } \right|\) <strong><em>(A1)(A1)</em></strong></p>
<p>scalar product \(2(3) + 4(0) - 4(4){\text{ }}( = - 10)\)</p>
<p>\(\left| {\overrightarrow {{\text{AC}}} } \right| = \sqrt {{3^2} + {0^2} + {4^2}} {\text{ }}( = 5)\)</p>
<p>substitution of <strong>their </strong>scalar product and magnitudes into cosine formula <strong><em>(M1)</em></strong></p>
<p><em>eg</em>\(\,\,\,\,\,\)\(\cos {\rm{B\hat AC = }}\frac{{6 + 0 - 16}}{{6\sqrt {{3^2} + {4^2}} }}\)</p>
<p>\({\text{cos}}\,B\hat AC = - \frac{{10}}{{30}}\left( { = - \frac{1}{3}} \right)\) <strong><em>A1</em></strong> <strong><em>N2</em></strong></p>
<p> </p>
<p><strong>METHOD 2 (triangle approach)</strong></p>
<p>valid approach involving cosine rule <strong><em>(M1)</em></strong></p>
<p><em>eg</em>\(\,\,\,\,\,\)\(\cos {\rm{B\hat AC = }}\frac{{{\text{A}}{{\text{B}}^2} + {\text{A}}{{\text{C}}^2} - {\text{B}}{{\text{C}}^2}}}{{2 \times {\text{AB}} \times {\text{AC}}}}\)</p>
<p>finding lengths AC and BC <strong><em>(A1)(A1)</em></strong></p>
<p>\({\text{AC}} = 5,{\text{ BC}} = 9\)</p>
<p>substitution of <strong>their </strong>lengths into cosine formula <strong><em>(M1)</em></strong></p>
<p><em>eg</em>\(\,\,\,\,\,\)\(\cos {\rm{B\hat AC}} = \frac{{{5^2} + {6^2} - {9^2}}}{{2 \times 5 \times 6}}\)</p>
<p>\(\cos {\rm{B\hat AC}} = - \frac{{20}}{{60}}{\text{ }}\left( { = - \frac{1}{3}} \right)\) <strong><em>A1</em></strong> <strong><em>N2</em></strong></p>
<p><strong><em>[5 marks]</em></strong></p>
<div class="question_part_label">c.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><strong>Note:</strong> Award relevant marks for working seen to find BC in part (c) (if cosine rule used in part (c)).</p>
<p> </p>
<p><strong>METHOD 1 (using cosine rule)</strong></p>
<p>recognizing need to find BC <strong><em>(M1)</em></strong></p>
<p>choosing cosine rule <strong><em>(M1)</em></strong></p>
<p><em>eg</em>\(\,\,\,\,\,\)\({c^2} = {a^2} + {b^2} - 2ab\cos {\text{C}}\)</p>
<p>correct substitution into RHS <strong><em>A1</em></strong></p>
<p><em>eg</em>\(\,\,\,\,\,\)\({\text{B}}{{\text{C}}^2} = {(6)^2} + {(5)^2} - 2(6)(5)\left( { - \frac{1}{3}} \right),{\text{ }}36 + 25 + 20\)</p>
<p>distance is 9 <strong><em>A1</em></strong> <strong><em>N2</em></strong></p>
<p> </p>
<p><strong>METHOD 2 (finding magnitude of </strong>\(\overrightarrow {BC} \)<strong>) </strong></p>
<p>recognizing need to find BC <strong><em>(M1)</em></strong></p>
<p>valid approach <strong><em>(M1)</em></strong></p>
<p><em>eg</em>\(\,\,\,\,\,\)attempt to find \(\overrightarrow {{\rm{OB}}} \) or \(\overrightarrow {{\rm{OC}}} \), \(\overrightarrow {{\rm{OB}}} = \left( {\begin{array}{*{20}{c}} 6 \\ 3 \\ { - 1} \end{array}} \right)\) or \(\overrightarrow {{\rm{OC}}} = \left( {\begin{array}{*{20}{c}} 7 \\ { - 1} \\ 7 \end{array}} \right),{\text{ }}\overrightarrow {{\rm{BA}}} + \overrightarrow {{\rm{AC}}} \)</p>
<p>correct working <strong><em>A1</em></strong></p>
<p><em>eg</em>\(\,\,\,\,\,\)\(\overrightarrow {{\rm{BC}}} = \left( {\begin{array}{*{20}{c}} 1 \\ { - 4} \\ 8 \end{array}} \right),{\text{ }}\overrightarrow {{\rm{CB}}} = \left( {\begin{array}{*{20}{c}} { - 1} \\ 4 \\ { - 8} \end{array}} \right),{\text{ }}\sqrt {{1^2} + {4^2} + {8^2}} = \sqrt {81} \)</p>
<p>distance is 9 <strong><em>A1</em></strong> <strong><em>N2</em></strong></p>
<p> </p>
<p><strong>METHOD 3 (finding coordinates and using distance formula)</strong></p>
<p>recognizing need to find BC <strong><em>(M1)</em></strong></p>
<p>valid approach <strong><em>(M1)</em></strong></p>
<p><em>eg</em>\(\,\,\,\,\,\)attempt to find coordinates of B or C, \({\text{B}}(6,{\text{ }}3,{\text{ }} - 1)\) or \({\text{C}}(7,{\text{ }} - 1,{\text{ }}7)\)</p>
<p>correct substitution into distance formula <strong><em>A1</em></strong></p>
<p><em>eg</em>\(\,\,\,\,\,\)\({\text{BC}} = \sqrt {{{(6 - 7)}^2} + {{\left( {3 - ( - 1)} \right)}^2} + {{( - 1 - 7)}^2}} ,{\text{ }}\sqrt {{1^2} + {4^2} + {8^2}} = \sqrt {81} \)</p>
<p>distance is 9 <strong><em>A1</em></strong> <strong><em>N2</em></strong></p>
<p><strong><em>[4 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;">
[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.i.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">b.ii.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">c.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">d.</div>
</div>
<br><hr><br><div class="specification">
<p><span style="font-family: times new roman,times; font-size: medium;">The following diagram shows quadrilateral ABCD, with \(\overrightarrow {{\rm{AD}}} = \overrightarrow {{\rm{BC}}} \) , \(\overrightarrow {{\rm{AB}}} = \left( \begin{array}{l}<br>3\\<br>1<br>\end{array} \right)\) , and \(\overrightarrow {{\rm{AC}}} = \left( \begin{array}{l}<br>4\\<br>4<br>\end{array} \right)\) .</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"><br><img src="images/trig.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;">Find \(\overrightarrow {{\rm{BC}}} \) .</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;">Show that \(\overrightarrow {{\rm{BD}}} = \left( {\begin{array}{*{20}{c}}<br>{ - 2}\\<br>2<br>\end{array}} \right)\) .</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 align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">Show that vectors \(\overrightarrow {{\rm{BD}}} \) </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;">are perpendicular.</span></p>
<div class="marks">[3]</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;">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. \(\overrightarrow {{\rm{AC}}} - \overrightarrow {{\rm{AB}}} \) , \(\left( \begin{array}{l}<br>4 - 3\\<br>4 - 1<br>\end{array} \right)\)</span></p>
<p><span style="font-family: Times New Roman; font-size: medium;">\(\overrightarrow {{\rm{BC}}} = \left( \begin{array}{l}<br>1\\<br>3<br>\end{array} \right)\) </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><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; font-size: medium;">\(\overrightarrow {{\rm{AD}}} = \left( {\begin{array}{*{20}{c}}<br>1\\<br>3<br>\end{array}} \right)\) </span><span style="font-family: times new roman,times; font-size: medium;"> <em><strong>(A1)</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">correct approach <strong><em>A1</em></strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g. \(\overrightarrow {{\rm{AD}}} - \overrightarrow {{\rm{AB}}} \) , \(\left( \begin{array}{l}<br>1 - 3\\<br>3 - 1<br>\end{array} \right)\)</span></p>
<p><span style="font-family: Times New Roman; font-size: medium;">\(\overrightarrow {{\rm{BD}}} = \left( {\begin{array}{*{20}{c}}<br>{ - 2}\\<br>2<br>\end{array}} \right)\) </span><em><strong><span style="font-family: times new roman,times; font-size: medium;">AG N0</span></strong></em></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;">recognizing \(\overrightarrow {{\rm{CD}}} = \overrightarrow {{\rm{BA}}} \) </span><span style="font-family: times new roman,times; font-size: medium;"> <strong><em>(A1)</em></strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">correct approach <em><strong>A1</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g. \(\overrightarrow {{\rm{BC}}} + \overrightarrow {{\rm{CD}}} \) , \(\left( \begin{array}{l}<br>1 - 3\\<br>3 - 1<br>\end{array} \right)\)</span></p>
<p><span style="font-family: Times New Roman; font-size: medium;">\(\overrightarrow {{\rm{BD}}} = \left( {\begin{array}{*{20}{c}}<br>{ - 2}\\<br>2<br>\end{array}} \right)\) </span><em><strong><span style="font-family: times new roman,times; font-size: medium;">AG N0</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">b.</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 scalar product <em><strong>(M1)</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g. \(\overrightarrow {{\rm{BD}}} \bullet \overrightarrow {{\rm{AC}}} \) , \(\left( {\begin{array}{*{20}{c}}<br>{ - 2}\\<br>2<br>\end{array}} \right) \bullet \left( \begin{array}{l}<br>4\\<br>4<br>\end{array} \right)\)</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. \(( - 2)(4) + (2)(4)\) , \( - 8 + 8\)</span></p>
<p><span style="font-family: Times New Roman; font-size: medium;">\(\overrightarrow {{\rm{BD}}} \bullet \overrightarrow {{\rm{AC}}} = 0\) </span><strong><em><span style="font-family: times new roman,times; font-size: medium;">A1</span></em></strong></p>
<p><span style="font-family: times new roman,times; font-size: medium;">therefore vectors \(\overrightarrow {{\rm{BD}}} \) and \(\overrightarrow {{\rm{AC}}} \) are perpendicular <strong><em>AG N0</em></strong></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;">attempt to find angle between two vectors <strong><em>(M1)</em></strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g. \(\frac{{{\boldsymbol{a}} \bullet {\boldsymbol{b}}}}{{{\boldsymbol{ab}}}}\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">correct substitution <em><strong>A1</strong></em><br></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g. \(\frac{{( - 2)(4) + (2)(4)}}{{\sqrt 8 \sqrt {32} }}\) , \(\cos \theta = 0\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(\theta = {90^ \circ }\) <em><strong>A1</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"><span style="font-family: times new roman,times; font-size: medium;">therefore vectors \(\overrightarrow {{\rm{BD}}} \) and \(\overrightarrow {{\rm{AC}}} \) are perpendicular <strong><em>AG N0</em></strong></span></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>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">This question on two-dimensional vectors was generally very well done. </span></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">This question on two-dimensional vectors was generally very well done. A very small number of candidates had trouble with the "show that" in part (b) of the question. </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;">Nearly all candidates knew to use the scalar product in part (c) to show that the vectors are perpendicular. </span></p>
<div class="question_part_label">c.</div>
</div>
<br><hr><br><div class="specification">
<p>Let \(\mathop {{\text{OA}}}\limits^ \to = \left( \begin{gathered}<br> 2 \hfill \\<br> 1 \hfill \\<br> 3 \hfill \\ <br>\end{gathered} \right)\) and \(\mathop {{\text{AB}}}\limits^ \to = \left( \begin{gathered}<br> 1 \hfill \\<br> 3 \hfill \\<br> 1 \hfill \\ <br>\end{gathered} \right)\), where O is the origin. <em>L</em><sub>1</sub> is the line that passes through A and B.</p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Find a vector equation for <em>L</em><sub>1</sub>.</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>The vector \(\left( \begin{gathered}<br> 2 \hfill \\<br> p \hfill \\<br> 0 \hfill \\ <br>\end{gathered} \right)\) is perpendicular to \(\mathop {{\text{AB}}}\limits^ \to \). Find the value of <em>p</em>.</p>
<div class="marks">[3]</div>
<div class="question_part_label">b.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p>any correct equation in the form <em><strong>r</strong> = <strong>a</strong> + t<strong>b</strong></em> (accept any parameter for <em>t</em>)</p>
<p>where <strong><em>a</em></strong> is \(\left( \begin{gathered}<br> 2 \hfill \\<br> 1 \hfill \\<br> 3 \hfill \\ <br>\end{gathered} \right)\), and <strong><em>b</em></strong> is a scalar multiple of \(\left( \begin{gathered}<br> 1 \hfill \\<br> 3 \hfill \\<br> 1 \hfill \\ <br>\end{gathered} \right)\) <em><strong>A2 N2</strong></em></p>
<p>eg <em><strong>r</strong> = </em>\(\left( \begin{gathered}<br> 2 \hfill \\<br> 1 \hfill \\<br> 3 \hfill \\ <br>\end{gathered} \right) = t\left( \begin{gathered}<br> 1 \hfill \\<br> 3 \hfill \\<br> 1 \hfill \\ <br>\end{gathered} \right)\), <em><strong>r</strong> = 2<strong>i</strong> + <strong>j</strong> + 3<strong>k</strong> + s</em>(<em><strong>i</strong> + 3<strong>j</strong> + <strong>k</strong></em>)</p>
<p><strong>Note:</strong> Award<em><strong> A1</strong></em> for the form<em> <strong>a</strong> + t<strong>b</strong>, <strong>A1 </strong></em>for the form<em> L = <strong>a</strong> + t<strong>b</strong></em>, A0 for the form <em><strong>r</strong> = <strong>b</strong> + t<strong>a</strong></em>.</p>
<p><em><strong>[2 marks]</strong></em></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><strong>METHOD 1</strong></p>
<p>correct scalar product <em><strong>(A1)</strong></em></p>
<p><em>eg </em> (1 × 2) + (3 × <em>p</em>) + (1 × 0), 2 + 3<em>p</em></p>
<p>evidence of equating <strong>their</strong> scalar product to zero <em><strong>(M1)</strong></em></p>
<p><em>eg</em> <em><strong>a•b</strong></em> = 0, 2 + 3p = 0, 3<em>p</em> = −2</p>
<p>\(p = - \frac{2}{3}\) <em><strong>A1 N3</strong></em></p>
<p> </p>
<p><strong>METHOD 2</strong></p>
<p>valid attempt to find angle between vectors <em><strong>(M1)</strong></em></p>
<p>correct substitution into numerator and/or angle <em><strong>(A1)</strong></em></p>
<p><em>eg </em>\({\text{cos}}\,\theta = \frac{{\left( {1 \times 2} \right) + \left( {3 \times p} \right) + \left( {1 \times 0} \right)}}{{\left| a \right|\left| b \right|}},\,\,{\text{cos}}\,\theta = 0\)</p>
<p>\(p = - \frac{2}{3}\) <em><strong>A1 N3</strong></em></p>
<p><em><strong>[3 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;">
[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 class="p1"><span class="s1">A line \(L\) </span>passes through points \({\text{A}}( - 2,{\text{ }}4,{\text{ }}3)\) and \({\text{B}}( - 1,{\text{ }}3,{\text{ }}1)\)<span class="s1">.</span></p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>(i) Show that \(\overrightarrow {{\text{AB}}} = \left( {\begin{array}{*{20}{c}} 1 \\ { - 1} \\ { - 2} \end{array}} \right)\).</p>
<p>(ii) Find \(\left| {\overrightarrow {{\text{AB}}} } \right|\).</p>
<div class="marks">[3]</div>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p class="p1">Find a vector equation for \(L\).</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">The following diagram shows the line \(L\) and the origin \(O\). The point \(C\) also lies on \(L\).</p>
<p class="p1" style="text-align: center;"><img src="images/Schermafbeelding_2016-01-13_om_08.10.35.png" alt></p>
<p class="p1">Point \(C\) has position vector \(\left( {\begin{array}{*{20}{c}} 0 \\ y \\ { - 1} \end{array}} \right)\).</p>
<p class="p1">Show that \(y = 2\).</p>
<div class="marks">[4]</div>
<div class="question_part_label">c.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p class="p1">(i) Find \(\overrightarrow {{\text{OC}}} \bullet \overrightarrow {{\text{AB}}} \).</p>
<p class="p1">(ii) Hence, write down the size of the angle between \(C\) and \(L\).</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">Hence or otherwise, find the area of triangle \(OAB\).</p>
<div class="marks">[4]</div>
<div class="question_part_label">e.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p>(i) correct approach <strong><em>A1</em></strong></p>
<p><em>eg</em>\(\;\;\;{\text{B}} - {\text{A, AO}} + {\text{OB}}\)</p>
<p>\(\overrightarrow {{\text{AB}}} = \left( {\begin{array}{*{20}{c}} 1 \\ { - 1} \\ { - 2} \end{array}} \right)\) <strong><em>AG N0</em></strong></p>
<p>(ii) correct substitution <strong><em>(A1)</em></strong></p>
<p><em>eg</em>\(\;\;\;\sqrt {{{(1)}^2} + {{( - 1)}^2} + {{( - 2)}^2}} ,{\text{ }}\sqrt {1 + 1 + 4} \)</p>
<p>\(\left| {\overrightarrow {{\text{AB}}} } \right| = \sqrt 6 \) <strong><em>A1 N2</em></strong></p>
<p><strong><em>[3 marks]</em></strong></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">any correct equation in the form \(r = a + tb\) (any parameter for \(t\))</p>
<p class="p1">where \(a\) is \(\left( {\begin{array}{*{20}{c}} { - 2} \\ 4 \\ 3 \end{array}} \right)\) or \(\left( {\begin{array}{*{20}{c}} { - 1} \\ 3 \\ 1 \end{array}} \right)\) and \(b\) is a scalar multiple of \(\left( {\begin{array}{*{20}{c}} 1 \\ { - 1} \\ { - 2} \end{array}} \right)\) <span class="Apple-converted-space"> </span><strong><em>A2 <span class="Apple-converted-space"> </span>N2</em></strong></p>
<p class="p1"><em>eg</em>\(\;\;\;\(r\) = \left( {\begin{array}{*{20}{c}} { - 2} \\ 4 \\ 3 \end{array}} \right) + t\left( {\begin{array}{*{20}{c}} 1 \\ { - 1} \\ { - 2} \end{array}} \right),{\text{ }}(x,{\text{ }}y,{\text{ }}z) = ( - 1,{\text{ }}3,{\text{ }}1) + t(1,{\text{ }} - 1,{\text{ }} - 2),{\text{ }}{\mathbf{r}} = \left( {\begin{array}{*{20}{c}} { - 1 + t} \\ {3 - t} \\ {1 - 2t} \end{array}} \right)\)</p>
<p class="p1"> </p>
<p class="p1"><strong>Note:</strong> <span class="Apple-converted-space"> </span>Award <strong><em>A1 </em></strong>for the form \({\mathbf{a}} + t{\mathbf{b}}\), <strong><em>A1 </em></strong>for the form \(L = {\mathbf{a}} + t{\mathbf{b}}\), <strong><em>A0 </em></strong>for the form \({\mathbf{r}} = {\mathbf{b}} + t{\mathbf{a}}\).</p>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1"><strong>METHOD 1</strong></p>
<p class="p1">valid approach <strong><em>(M1)</em></strong></p>
<p class="p1"><em>eg</em>\(\;\;\;\left( {\begin{array}{*{20}{c}} { - 1} \\ 3 \\ 1 \end{array}} \right) + t\left( {\begin{array}{*{20}{c}} 1 \\ { - 1} \\ { - 2} \end{array}} \right) = \left( {\begin{array}{*{20}{c}} 0 \\ y \\ { - 1} \end{array}} \right),{\text{ }}\left( {\begin{array}{*{20}{c}} 0 \\ y \\ { - 1} \end{array}} \right) = \left( {\begin{array}{*{20}{c}} { - 2} \\ 4 \\ 3 \end{array}} \right) + s\left( {\begin{array}{*{20}{c}} 1 \\ { - 1} \\ { - 2} \end{array}} \right)\)</p>
<p class="p1">one correct equation from <strong>their </strong>approach <strong><em>A1</em></strong></p>
<p class="p1"><em>eg</em>\(\;\;\; - 1 + t = 0,{\text{ }}1 - 2t = - 1,{\text{ }} - 2 + s = 0,{\text{ }}3 - 2s = - 1\)</p>
<p class="p1">one correct value for <strong>their </strong>parameter and equation <strong><em>A1</em></strong></p>
<p class="p1"><em>eg</em>\(\;\;\;t = 1,{\text{ }}s = 2\)</p>
<p class="p1">correct substitution <strong><em>A1</em></strong></p>
<p class="p1"><em>eg</em>\(\;\;\;3 + 1( - 1),{\text{ }}4 + 2( - 1)\)</p>
<p class="p1">\(y = 2\) <strong><em>AG N0</em></strong></p>
<p class="p1"><strong>METHOD 2</strong></p>
<p class="p1">valid approach <strong><em>(M1)</em></strong></p>
<p class="p1"><em>eg</em>\(\;\;\;\overrightarrow {{\text{AC}}} = k\overrightarrow {{\text{AB}}} \)</p>
<p class="p1">correct working <strong><em>A1</em></strong></p>
<p class="p1"><em>eg</em>\(\;\;\;\overrightarrow {{\text{AC}}} = \left( {\begin{array}{*{20}{c}} 2 \\ {y - 4} \\ { - 4} \end{array}} \right),{\text{ }}\left( {\begin{array}{*{20}{c}} 2 \\ {y - 4} \\ { - 4} \end{array}} \right) = k\left( {\begin{array}{*{20}{c}} 1 \\ { - 1} \\ { - 2} \end{array}} \right)\)</p>
<p class="p1">\(k = 2\) <strong><em>A1</em></strong></p>
<p class="p1">correct substitution <strong><em>A1</em></strong></p>
<p class="p1"><em>eg</em>\(\;\;\;y - 4 = - 2\)</p>
<p class="p1">\(y = 2\) <strong><em>AG N0</em></strong></p>
<p class="p1"><strong><em>[4 marks]</em></strong></p>
<div class="question_part_label">c.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p>(i) correct substitution <strong><em>A1</em></strong></p>
<p><em>eg</em>\(\;\;\;0(1) + 2( - 1) - 1( - 2),{\text{ }}0 - 2 + 2\)</p>
<p>\(\overrightarrow {{\text{OC}}} \bullet \overrightarrow {{\text{AB}}} = 0\) <strong><em>A1 N1</em></strong></p>
<p>(ii) \(9{0^ \circ }\) or \(\frac{\pi }{2}\) <strong><em>A1 N1</em></strong></p>
<p><strong><em>[3 marks]</em></strong></p>
<div class="question_part_label">d.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><strong>METHOD 1</strong> \({\text{(area}} = 0.5 \times {\text{height}} \times {\text{base)}}\)</p>
<p>\(\left| {\overrightarrow {{\text{OC}}} } \right| = \sqrt {0 + {2^2} + {{( - 1)}^2}} \;\;\;\left( { = \sqrt 5 } \right)\;\;\;\)(seen anywhere) <strong><em>A1</em></strong></p>
<p>valid approach <strong><em>(M1)</em></strong></p>
<p><em>eg</em>\(\;\;\;\frac{1}{2} \times \left| {\overrightarrow {{\text{AB}}} } \right| \times \left| {\overrightarrow {{\text{OC}}} } \right|,{\text{ }}\left| {\overrightarrow {{\text{OC}}} } \right|\) is height of triangle</p>
<p>correct substitution <strong><em>A1</em></strong></p>
<p><em>eg</em>\(\;\;\;\frac{1}{2} \times \sqrt 6 \times \sqrt {0 + {{(2)}^2} + {{( - 1)}^2}} ,{\text{ }}\frac{1}{2} \times \sqrt 6 \times \sqrt 5 \)</p>
<p>area is \(\frac{{\sqrt {30} }}{2}\) <strong><em>A1 N2</em></strong></p>
<p><strong>METHOD 2 </strong>(difference of two areas)</p>
<p>one correct magnitude (seen anywhere) <strong><em>A1</em></strong></p>
<p><em>eg</em>\(\;\;\;\left| {\overrightarrow {{\text{OC}}} } \right| = \sqrt {{2^2} + {{( - 1)}^2}} \;\;\;\left( { = \sqrt 5 } \right),\;\;\;\left| {\overrightarrow {{\text{AC}}} } \right| = \sqrt {4 + 4 + 16} \;\;\;\left( { = \sqrt {24} } \right),\;\;\;\left| {\overrightarrow {{\text{BC}}} } \right| = \sqrt 6 \)</p>
<p>valid approach <strong><em>(M1)</em></strong></p>
<p><em>eg</em>\(\;\;\;\Delta {\text{OAC}} - \Delta {\text{OBC}}\)</p>
<p>correct substitution <strong><em>A1</em></strong></p>
<p><em>eg</em>\(\;\;\;\frac{1}{2} \times \sqrt {24} \times \sqrt 5 - \frac{1}{2} \times \sqrt 5 \times \sqrt 6 \)</p>
<p>area is \(\frac{{\sqrt {30} }}{2}\) <strong><em>A1 N2</em></strong></p>
<p><strong>METHOD 3 </strong>\({\text{(area}} = \frac{1}{2}ab\sin C{\text{ for }}\Delta {\text{OAB)}}\)</p>
<p>one correct magnitude of \(\overrightarrow {{\text{OA}}} \) or \(\overrightarrow {{\text{OB}}} \) (seen anywhere) <strong><em>A1</em></strong></p>
<p><em>eg</em>\(\;\;\;\left| {\overrightarrow {{\text{OA}}} } \right| = \sqrt {{{( - 2)}^2} + {4^2} + {3^2}} \;\;\;\left( { = \sqrt {29} } \right),\;\;\;\left| {\overrightarrow {{\text{OB}}} } \right| = \sqrt {1 + 9 + 1} \;\;\;\left( { = \sqrt {11} } \right)\)</p>
<p>valid attempt to find \(\cos \theta \) <strong><em>or</em></strong> \(\sin \theta \) <strong><em>(M1)</em></strong></p>
<p><em>eg</em>\(\;\;\;\cos {\text{C}} = \frac{{ - 1 - 3 - 2}}{{\sqrt 6 \times \sqrt {11} }}\;\;\;\left( { = \frac{{ - 6}}{{\sqrt {66} }}} \right),\;\;\;29 = 6 + 11 - 2\sqrt 6 \sqrt {11} \cos \theta ,{\text{ }}\frac{{\sin \theta }}{{\sqrt 5 }} = \frac{{\sin 90}}{{\sqrt {29} }}\)</p>
<p>correct substitution into \(\frac{1}{2}ab\sin {\text{C}}\) <strong><em>A1</em></strong></p>
<p><em>eg</em>\(\;\;\;\frac{1}{2} \times \sqrt 6 \times \sqrt {11} \times \sqrt {1 - \frac{{36}}{{66}}} ,{\text{ }}0.5 \times \sqrt 6 \times \sqrt {29} \times \frac{{\sqrt 5 }}{{\sqrt {29} }}\)</p>
<p>area is \(\frac{{\sqrt {30} }}{2}\) <strong><em>A1 N2</em></strong></p>
<p><strong><em>[4 marks]</em></strong></p>
<p><strong><em>Total [16 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">Finding \(\overrightarrow {{\text{AB}}} \) and its magnitude were mostly well done.</p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">Mostly correct answers with common errors being using both position vectors or writing it as “\(L = \)” instead of “\({\mathbf{r}} = \)”.</p>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p>Many candidates assumed that \(\overrightarrow {{\text{AB}}} = \overrightarrow {{\text{BC}}} \), although this was not indicated on the diagram nor given in the question.</p>
<div class="question_part_label">c.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">Mostly this was well answered. A surprising number of candidates wrote the scalar product as a vector \((0,{\text{ }} - 2,{\text{ }}2)\). In part b) many missed the clue given by the phrase “hence, write down” and carried out a calculation for cosine theta using the scalar product again.</p>
<div class="question_part_label">d.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">This part was poorly done. Few candidates realised how to directly calculate the area based on their previous work and could not see the “height” of the obtuse triangle as \(\left| {\overrightarrow {{\text{OC}}} } \right|\). Those who tried to use \(A = \frac{1}{2}ab\sin C\) had trouble generating the angle. Those who subtracted areas \((\Delta {\text{OAC}} - \Delta {\text{OBC)}}\) were usually successful.</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;">Let \(\overrightarrow {{\rm{AB}}} = \left( {\begin{array}{*{20}{c}}<br>6\\<br>{ - 2}\\<br>3<br>\end{array}} \right)\) and \(\overrightarrow {{\rm{AC}}} = \left( {\begin{array}{*{20}{c}}<br>{ - 2}\\<br>{ - 3}\\<br>2<br>\end{array}} \right)\) .</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 \(\overrightarrow {{\rm{BC}}} \) .</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 a unit vector in the direction of \(\overrightarrow {{\rm{AB}}} \) .</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;">Show that \(\overrightarrow {{\rm{AB}}} \) </span><span style="font-family: times new roman,times; font-size: medium;">is perpendicular to \(\overrightarrow {{\rm{AC}}} \) </span><span style="font-family: times new roman,times; font-size: medium;">.</span></p>
<div class="marks">[3]</div>
<div class="question_part_label">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;">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. \(\overrightarrow {{\rm{BC}}} {\rm{ = }}\overrightarrow {{\rm{BA}}} + \overrightarrow {{\rm{AC}}} \) , \(\left( {\begin{array}{*{20}{c}}<br>{ - 2}\\<br>{ - 3}\\<br>2<br>\end{array}} \right) - \left( {\begin{array}{*{20}{c}}<br>6\\<br>{ - 2}\\<br>3<br>\end{array}} \right)\)</span></p>
<p><span style="font-family: Times New Roman; font-size: medium;">\(\overrightarrow {{\rm{BC}}} = \left( {\begin{array}{*{20}{c}}<br>{ - 8}\\<br>{ - 1}\\<br>{ - 1}<br>\end{array}} \right)\) </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;">attempt to find the length of \(\overrightarrow {{\rm{AB}}} \) <em><strong>(M1)</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(\overrightarrow {|{\rm{AB}}} | = \sqrt {{6^2} + {{( - 2)}^2} + {3^2}} \) \(( = \sqrt {36 + 4 + 9} = \sqrt {49} = 7)\) <em><strong>(A1)</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">unit vector is \(\frac{1}{7}\left( {\begin{array}{*{20}{c}}<br>6\\<br>{ - 2}\\<br>3<br>\end{array}} \right)\) \(\left( { = \left( {\begin{array}{*{20}{c}}<br>{\frac{6}{7}}\\<br>{ - \frac{2}{7}}\\<br>{\frac{3}{7}}<br>\end{array}} \right)} \right)\) </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;"> [3 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;">recognizing that the dot product or \(\cos \theta \) being 0 implies perpendicular <em><strong>(M1)</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">correct substitution in a scalar product formula <em><strong>A1</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g. \((6) \times ( - 2) + ( - 2) \times ( - 3) + (3) \times (2)\) , \(\cos \theta = \frac{{ - 12 + 6 + 6}}{{7 \times \sqrt {17} }}\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">correct calculation <em><strong>A1</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g. \(\overrightarrow {{\rm{AB}}} \bullet \overrightarrow {{\rm{AC}}} = 0\) , \(\cos \theta = 0\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">therefore, they are perpendicular <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">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;">Part (a) was generally done well with candidates employing different correct methods to find </span><span style="font-family: times new roman,times; font-size: medium;">the vector \(\overrightarrow {{\rm{BC}}} \) . Some candidates subtracted the given vectors in the wrong order and others </span><span style="font-family: times new roman,times; font-size: medium;">simply added them. Calculation errors were seen with some frequency.</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;">Many candidates did not appear to know how to find a unit vector in part (b). Some tried to write down the vector equation of a line, indicating no familiarity with the concept of unit vectors while others gave the vector (1, 1, 1) or wrote the same vector \(\overrightarrow {{\rm{AB}}} \) as a linear combination of <strong><em>i</em></strong>, <strong><em>j</em></strong> and <strong><em>k</em></strong>. A number of candidates correctly found the magnitude but did not continue on to write the unit vector.</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;">Candidates were generally successful in showing that the vectors in part (c) were </span><span style="font-family: times new roman,times; font-size: medium;">perpendicular. Many used the efficient approach of showing that the scalar product equalled </span><span style="font-family: times new roman,times; font-size: medium;">zero, while others worked a little harder than necessary and used the cosine rule to find the </span><span style="font-family: times new roman,times; font-size: medium;">angle between the two vectors.</span></p>
<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;">Consider points A(\(1\), \( - 2\), \( -1\)) , B(\(7\), \( - 4\), \(3\)) and C(\(1\), \( -2\), \(3\)) . The line \({L_1}\) passes </span><span style="font-family: times new roman,times; font-size: medium;">through C and is parallel to \(\overrightarrow {{\rm{AB}}} \) </span><span style="font-family: times new roman,times; font-size: medium;">.</span></p>
</div>
<div class="specification">
<p><span style="font-family: times new roman,times; font-size: medium;">A second line, \({L_2}\) , is given by \(\boldsymbol{r} = \left( {\begin{array}{*{20}{c}}<br>{ - 1}\\<br>2\\<br>{15}<br>\end{array}} \right) + s\left( {\begin{array}{*{20}{c}}<br>3\\<br>{ - 3}\\<br>p<br>\end{array}} \right)\) .</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;">Find \(\overrightarrow {{\rm{AB}}} \) </span><span style="font-family: times new roman,times; font-size: medium;">.</span></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><span style="font-family: times new roman,times; font-size: medium;">Hence, write down a vector equation for \({L_1}\) .</span></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><span style="font-family: times new roman,times; font-size: medium;">Given that \({L_1}\) is perpendicular to \({L_2}\) , show that \(p = - 6\) .</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 line \({L_1}\) intersects the line \({L_2}\) at point Q. Find the \(x\)-coordinate of Q.</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;">valid approach <em><strong>(M1)</strong> </em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"><em>eg</em> \(\left( {\begin{array}{*{20}{c}}<br>7\\<br>{ - 4}\\<br>3<br>\end{array}} \right) - \left( {\begin{array}{*{20}{c}}<br>1\\<br>{ - 2}\\<br>{ - 1}<br>\end{array}} \right)\) , \({\rm{A}} - {\rm{B}}\) , \(\overrightarrow {{\rm{AB}}} = \overrightarrow {{\rm{AO}}} + \overrightarrow {{\rm{OB}}} \)</span></p>
<p><span style="font-family: Times New Roman; font-size: medium;">\(\overrightarrow {{\rm{AB}}} = \left( {\begin{array}{*{20}{c}}<br>6\\<br>{ - 2}\\<br>4<br>\end{array}} \right)\) </span><em><span style="font-family: times new roman,times; font-size: medium;"><strong>A1 N2</strong> </span></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.i.</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 \(\boldsymbol{r} = \boldsymbol{a} + t\boldsymbol{b}\) (accept any parameter for \(t\)) </span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">where \(\boldsymbol{a} = \left( {\begin{array}{*{20}{c}}<br>1\\<br>{ - 2}\\<br>3<br>\end{array}} \right)\) and \(\boldsymbol{b}\) is a scalar multiple of \(\overrightarrow {{\rm{AB}}} \) <strong><em>A2 N2</em> </strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"><em>eg</em> \(\boldsymbol{r} = \left( {\begin{array}{*{20}{c}}<br>1\\<br>{ - 2}\\<br>3<br>\end{array}} \right) + t\left( {\begin{array}{*{20}{c}}<br>6\\<br>{ - 2}\\<br>4<br>\end{array}} \right)\) , \((x,y,z) = (1, - 2,3) + t(3, - 1,2)\) , \(\boldsymbol{r} = \left( {\begin{array}{*{20}{c}}<br>{1 + 6t}\\<br>{ - 2 - 2t}\\<br>{3 + 4t}<br>\end{array}} \right)\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"><strong>Note</strong>: Award <em><strong>A1</strong></em> for \(\boldsymbol{a} + t\boldsymbol{b}\) , <strong><em>A1</em></strong> for \({L_1} = \boldsymbol{a} + t\boldsymbol{b}\) , <strong><em>A0</em></strong> for \(\boldsymbol{r} = \boldsymbol{b} + t\boldsymbol{a}\) . </span></p>
<p><em><strong><span style="font-family: times new roman,times; font-size: medium;"> [2 marks]</span></strong></em></p>
<div class="question_part_label">a.ii.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">recognizing that scalar product \( = 0\) (seen anywhere) <strong><em>R1</em> </strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">correct calculation of scalar product <strong><em> (A1)</em> </strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"><em>eg</em> \(6(3) - 2( - 3) + 4p\) , \(18 + 6 + 4p\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">correct working <em><strong>A1 </strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"><em>eg</em> \(24 + 4p = 0\) , \(4p = - 24\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(p = - 6\) <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">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">setting lines equal <em><strong>(M1)</strong> </em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"><em>eg</em> \({L_1} = {L_2}\) , \(\left( {\begin{array}{*{20}{c}}<br>1\\<br>{ - 2}\\<br>3<br>\end{array}} \right) + t\left( {\begin{array}{*{20}{c}}<br>6\\<br>{ - 2}\\<br>4<br>\end{array}} \right) = \left( {\begin{array}{*{20}{c}}<br>{ - 1}\\<br>2\\<br>{15}<br>\end{array}} \right) + s\left( {\begin{array}{*{20}{c}}<br>3\\<br>{ - 3}\\<br>{ - 6}<br>\end{array}} \right)\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">any two correct equations with <strong>different</strong> parameters <strong><em>A1A1</em> </strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"><em>eg</em> \(1 + 6t = 1 + 3s\) , \( - 2 - 2t = 2 - 3s\) , \(3 + 4t = 15 - 6s\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">attempt to solve <strong>their</strong> simultaneous equations <em><strong>(M1)</strong> </em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"><strong>one</strong> correct parameter <em><strong>A1</strong> </em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"><em>eg</em> \(t = \frac{1}{2}\) , \(s = \frac{5}{3}\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">attempt to substitute parameter into vector equation <strong><em>(M1)</em> </strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"><em>eg</em> \(\left( {\begin{array}{*{20}{c}}<br>1\\<br>{ - 2}\\<br>3<br>\end{array}} \right) + \frac{1}{2}\left( {\begin{array}{*{20}{c}}<br>6\\<br>{ - 2}\\<br>4<br>\end{array}} \right)\) , \(1 + \frac{1}{2} \times 6\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(x = 4\) (accept (4, -3, 5), ignore incorrect values for \(y\) and \(z\)) <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 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;">While many candidates can find a vector given two points, few could write down a fully correct vector equation of a line.</span></p>
<div class="question_part_label">a.i.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">While many candidates can find a vector given two points, few could write down a fully correct vector equation of a line. Most candidates wrote their equation as “ \({L_1} = \) ”, which misrepresents that the resulting equation must still be a vector.</span></p>
<div class="question_part_label">a.ii.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">Those who recognized that vector perpendicularity means the scalar product is zero found little difficulty answering part (b). Occasionally a candidate would use the given \(p = 6\) to show the scalar product is zero. However, working backward from the given answer earns no marks in a question that requires candidates to show that this value is achieved.</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;">While many candidates knew to set the lines equal to find an intersection point, a surprising number could not carry the process to correct completion. Some could not solve a simultaneous pair of equations, and for those who did, some did not know what to do with the parameter value. Another common error was to set the vector equations equal using the same parameter, from which the candidates did not recognize a system to solve. Furthermore, it is interesting to note that while only one parameter value is needed to answer the question, most candidates find or attempt to find both, presumably out of habit in the algorithm.</span></p>
<div class="question_part_label">c.</div>
</div>
<br><hr><br><div class="specification">
<p>A line \({L_1}\) passes through the points \({\text{A}}(0,{\text{ }}1,{\text{ }}8)\) and \({\text{B}}(3,{\text{ }}5,{\text{ }}2)\).</p>
</div>
<div class="specification">
<p>Given that \({L_1}\) and \({L_2}\) are perpendicular, show that \(p = 2\).</p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Find \(\overrightarrow {AB} \).</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>Hence, write down a vector equation for \({L_1}\).</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>A second line \({L_2}\), has equation <strong><em>r</em></strong> = \(\left( {\begin{array}{*{20}{c}} 1 \\ {13} \\ { - 14} \end{array}} \right) + s\left( {\begin{array}{*{20}{c}} p \\ 0 \\ 1 \end{array}} \right)\).</p>
<p>Given that \({L_1}\) and \({L_2}\) are perpendicular, show that \(p = 2\).</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>The lines \({L_1}\) and \({L_1}\) intersect at \(C(9,{\text{ }}13,{\text{ }}z)\). Find \(z\).</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>Find a unit vector in the direction of \({L_2}\).</p>
<div class="marks">[2]</div>
<div class="question_part_label">d.i.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Hence or otherwise, find one point on \({L_2}\) which is \(\sqrt 5 \) units from C.</p>
<div class="marks">[3]</div>
<div class="question_part_label">d.ii.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p>valid approach <strong><em>(M1)</em></strong></p>
<p><em>eg \(A - B,\,\, - \left( \begin{gathered}<br> 0 \hfill \\<br> 1 \hfill \\<br> 8 \hfill \\ <br>\end{gathered} \right) + \left( \begin{gathered}<br> 3 \hfill \\<br> 5 \hfill \\<br> 2 \hfill \\ <br>\end{gathered} \right)\)</em></p>
<p>\(\overrightarrow {AB} = \left( \begin{gathered}<br> 3 \hfill \\<br> 4 \hfill \\<br> - 6 \hfill \\ <br>\end{gathered} \right)\) <strong><em>A1</em></strong> <strong><em>N2</em></strong></p>
<p><strong><em>[2 marks]</em></strong></p>
<div class="question_part_label">a.i.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><strong>any </strong>correct equation in the form <strong><em>r</em></strong> = <strong><em>a</em></strong> + <em>t</em><strong><em>b</em></strong> (any parameter for \(t\)) <strong><em>A2</em></strong> <strong><em>N2</em></strong></p>
<p>where <strong><em>a</em></strong> is \(\left( \begin{gathered}<br> 0 \hfill \\<br> 1 \hfill \\<br> 8 \hfill \\ <br>\end{gathered} \right)\) or \(\left( \begin{gathered}<br> 3 \hfill \\<br> 5 \hfill \\<br> 2 \hfill \\ <br>\end{gathered} \right)\), and <strong><em>b </em></strong>is a scalar multiple of \(\left( \begin{gathered}<br> 3 \hfill \\<br> 4 \hfill \\<br> - 6 \hfill \\ <br>\end{gathered} \right)\)</p>
<p> </p>
<p><em>eg</em>\(\,\,\,\,\,\)<strong><em>r</em></strong> = \(\left( {\begin{array}{*{20}{c}} 0 \\ 1 \\ 8 \end{array}} \right) + t\left( {\begin{array}{*{20}{c}} 3 \\ 4 \\ { - 6} \end{array}} \right)\), <strong><em>r</em></strong> = \(\left( {\begin{array}{*{20}{c}} {3 + 3t} \\ {5 + 4t} \\ {2 - 6t} \end{array}} \right)\), <strong><em>r</em></strong> = <strong><em>j</em></strong> + 8<strong><em>k</em></strong> + <em>t</em>(3<strong><em>i</em></strong> + 4<strong><em>j</em></strong> – 6<strong><em>k</em></strong>)</p>
<p> </p>
<p><strong>Note:</strong> Award <strong><em>A1</em> </strong>for the form <strong><em>a</em></strong> + <em>t</em><strong><em>b</em></strong>, <strong><em>A1</em></strong> for the form <strong><em>L</em></strong> = <strong><em>a</em></strong> + <em>t</em><strong><em>b</em></strong>, <strong><em>A0</em></strong> for the form <strong><em>r</em></strong> = <strong><em>b</em></strong> + <em>t</em><strong><em>a</em></strong>.</p>
<p> </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>valid approach <strong><em>(M1)</em></strong></p>
<p><em>eg</em>\(\,\,\,\,\,\)\(a \bullet b = 0\)</p>
<p>choosing correct direction vectors (may be seen in scalar product) <strong><em>A1</em></strong></p>
<p><em>eg</em>\(\,\,\,\,\,\)\(\left( {\begin{array}{*{20}{c}} 3 \\ 4 \\ { - 6} \end{array}} \right)\) and \(\left( {\begin{array}{*{20}{c}} p \\ 0 \\ 1 \end{array}} \right),{\text{ }}\left( {\begin{array}{*{20}{c}} 3 \\ 4 \\ { - 6} \end{array}} \right) \bullet \left( {\begin{array}{*{20}{c}} p \\ 0 \\ 1 \end{array}} \right) = 0\)</p>
<p>correct working/equation <strong><em>A1</em></strong></p>
<p><em>eg</em>\(\,\,\,\,\,\)\(3p - 6 = 0\)</p>
<p>\(p = 2\) <strong><em>AG</em></strong> <strong><em>N0</em></strong></p>
<p><strong><em>[3 marks]</em></strong></p>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p>valid approach <strong><em>(M1)</em></strong></p>
<p><em>eg</em>\(\,\,\,\,\,\)\({L_1} = \left( {\begin{array}{*{20}{c}} 9 \\ {13} \\ z \end{array}} \right),{\text{ }}{L_1} = {L_2}\)</p>
<p>one correct equation (must be different parameters if both lines used) <strong><em>(A1)</em></strong></p>
<p><em>eg</em>\(\,\,\,\,\,\)\(3t = 9,{\text{ }}1 + 2s = 9,{\text{ }}5 + 4t = 13,{\text{ }}3t = 1 + 2s\)</p>
<p>one correct value <strong><em>A1</em></strong></p>
<p><em>eg</em>\(\,\,\,\,\,\)\(t = 3,{\text{ }}s = 4,{\text{ }}t = 2\)</p>
<p>valid approach to substitute their \(t\) or \(s\) value <strong><em>(M1)</em></strong></p>
<p><em>eg</em>\(\,\,\,\,\,\)\(8 + 3( - 6),{\text{ }} - 14 + 4(1)\)</p>
<p>\(z = - 10\) <strong><em>A1</em></strong> <strong><em>N3</em></strong></p>
<p><strong><em>[5 marks]</em></strong></p>
<div class="question_part_label">c.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p>\(\left| {\vec d} \right| = \sqrt {{2^2} + 1} \,\,\left( { = \sqrt 5 } \right)\) <strong><em>(A1)</em></strong></p>
<p>\(\frac{1}{{\sqrt 5 }}\left( {\begin{array}{*{20}{c}} 2 \\ 0 \\ 1 \end{array}} \right)\,\,\,\,\,\left( {{\text{accept}}\left( {\begin{array}{*{20}{c}} {\frac{2}{{\sqrt 5 }}} \\ {\frac{0}{{\sqrt 5 }}} \\ {\frac{1}{{\sqrt 5 }}} \end{array}} \right)} \right)\) <strong><em>A1</em></strong> <strong><em>N2</em></strong></p>
<p><strong><em>[2 marks]</em></strong></p>
<div class="question_part_label">d.i.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><strong>METHOD 1 (using unit vector) </strong></p>
<p>valid approach <strong><em>(M1)</em></strong></p>
<p><em>eg</em>\(\,\,\,\,\,\)\(\left( {\begin{array}{*{20}{c}} 9 \\ {13} \\ { - 10} \end{array}} \right) \pm \sqrt 5 \hat d\)</p>
<p>correct working <strong><em>(A1)</em></strong></p>
<p><em>eg</em>\(\,\,\,\,\,\)\(\left( {\begin{array}{*{20}{c}} 9 \\ {13} \\ { - 10} \end{array}} \right) + \left( {\begin{array}{*{20}{c}} 2 \\ 0 \\ 1 \end{array}} \right),{\text{ }}\left( {\begin{array}{*{20}{c}} 9 \\ {13} \\ { - 10} \end{array}} \right) - \left( {\begin{array}{*{20}{c}} 2 \\ 0 \\ 1 \end{array}} \right)\)</p>
<p>one correct point <strong><em>A1</em></strong> <strong><em>N2</em></strong></p>
<p><em>eg</em>\(\,\,\,\,\,\)\((11,{\text{ }}13,{\text{ }} - 9),{\text{ }}(7,{\text{ }}13,{\text{ }} - 11)\)</p>
<p><strong>METHOD 2 (distance between points) </strong></p>
<p>attempt to use distance between \((1 + 2s,{\text{ }}13,{\text{ }} - 14 + s)\) and \((9,{\text{ }}13,{\text{ }} - 10)\) <strong><em>(M1)</em></strong></p>
<p><em>eg</em>\(\,\,\,\,\,\)\({(2s - 8)^2} + {0^2} + {(s - 4)^2} = 5\)</p>
<p>solving \(5{s^2} - 40s + 75 = 0\) leading to \(s = 5\) or \(s = 3\) <strong><em>(A1)</em></strong></p>
<p>one correct point <strong><em>A1</em></strong> <strong><em>N2</em></strong></p>
<p><em>eg</em>\(\,\,\,\,\,\)\((11,{\text{ }}13,{\text{ }} - 9),{\text{ }}(7,{\text{ }}13,{\text{ }} - 11)\)</p>
<p><strong><em>[3 marks]</em></strong></p>
<div class="question_part_label">d.ii.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">a.i.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">a.ii.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">c.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">d.i.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">d.ii.</div>
</div>
<br><hr><br><div class="specification">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">The diagram shows quadrilateral ABCD with vertices A(1, 0), B(1, 5), C(5, 2) </span><span style="font-family: times new roman,times; font-size: medium;">and D(4, −1) .</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;"><br><img src="images/111.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) Show that \(\overrightarrow {{\rm{AC}}} = \left( {\begin{array}{*{20}{c}}<br>4\\<br>2<br>\end{array}} \right)\) .</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">(ii) Find \(\overrightarrow {{\rm{BD}}} \) .</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">(iii) Show that \(\overrightarrow {{\rm{AC}}} \) is perpendicular to \(\overrightarrow {{\rm{BD}}} \) .</span></p>
<p> </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;">The line (AC) has equation \({\boldsymbol{r}} = {\boldsymbol{u}} + s{\boldsymbol{v}}\) .</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">(i) Write down vector <strong><em>u</em></strong> and vector <strong><em>v</em></strong> .</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">(ii) Find a vector equation for the line (BD).</span></p>
<p> </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;">The lines (AC) and (BD) intersect at the point \({\text{P}}(3{\text{, }}k)\) .</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">Show that \(k = 1\) .</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 lines (AC) and (BD) intersect at the point \({\text{P}}(3{\text{, }}k)\) .</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"><strong>Hence</strong> find the area of triangle ACD.</span></p>
<div class="marks">[5]</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) correct approach <em><strong>A1</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g. \(\overrightarrow {{\rm{OC}}} - \overrightarrow {{\rm{OA}}} \) , \(\left( {\begin{array}{*{20}{c}}<br>5\\<br>2<br>\end{array}} \right) - \left( {\begin{array}{*{20}{c}}<br>1\\<br>0<br>\end{array}} \right)\)</span></p>
<p><span style="font-family: Times New Roman; font-size: medium;">\(\overrightarrow {{\rm{AC}}} = \left( {\begin{array}{*{20}{c}}<br>4\\<br>2<br>\end{array}} \right)\) </span><em><strong><span style="font-family: times new roman,times; font-size: medium;">AG N0</span></strong></em></p>
<p><span style="font-family: times new roman,times; font-size: medium;">(ii) appropriate approach <em><strong>(M1)</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g. \({\mathop{\rm D}\nolimits} - {\rm{B}}\) , \(\left( {\begin{array}{*{20}{c}}<br>4\\<br>{ - 1}<br>\end{array}} \right) - \left( {\begin{array}{*{20}{c}}<br>1\\<br>5<br>\end{array}} \right)\) , move 3 to the right and 6 down</span></p>
<p><span style="font-family: Times New Roman; font-size: medium;">\(\overrightarrow {{\rm{BD}}} = \left( {\begin{array}{*{20}{c}}<br>3\\<br>{ - 6}<br>\end{array}} \right)\) </span><em><strong><span style="font-family: times new roman,times; font-size: medium;">A1 N2</span></strong></em></p>
<p><span style="font-family: times new roman,times; font-size: medium;">(iii) finding the scalar product <em><strong>A1</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g. \(4(3) + 2( - 6)\) , \(12 - 12\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">valid reasoning <em><strong>R1</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g. \(4(3) + 2( - 6) = 0\) , scalar product is zero</span></p>
<p><span style="font-family: Times New Roman; font-size: medium;">\(\overrightarrow {{\rm{AC}}} \) is perpendicular to \(\overrightarrow {{\rm{BD}}} \) </span><em><strong><span style="font-family: times new roman,times; font-size: medium;">AG N0</span></strong></em></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">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) correct “position” vector for <strong><em>u</em></strong>; “direction” vector for <strong><em>v</em></strong> <em><strong>A1A1 N2</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g. \({\boldsymbol{u}} = \left( {\begin{array}{*{20}{c}}<br>5\\<br>2<br>\end{array}} \right)\) , \({\boldsymbol{u}} = \left( {\begin{array}{*{20}{c}}<br>1\\<br>0<br>\end{array}} \right)\) ; \({\boldsymbol{v}} = \left( {\begin{array}{*{20}{c}}<br>4\\<br>2<br>\end{array}} \right)\) , \({\boldsymbol{v}} = \left( {\begin{array}{*{20}{c}}<br>{ - 2}\\<br>{ - 1}<br>\end{array}} \right)\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">accept in equation e.g. \(\left( {\begin{array}{*{20}{c}}<br>5\\<br>2<br>\end{array}} \right) + t\left( {\begin{array}{*{20}{c}}<br>{ - 4}\\<br>{ - 2}<br>\end{array}} \right)\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">(ii) any correct equation in the form \({\boldsymbol{r}} = {\boldsymbol{a}} + t{\boldsymbol{b}}\) , where \({\boldsymbol{b}} = \overrightarrow {{\rm{BD}}} \)</span></p>
<p><span style="font-family: Times New Roman; font-size: medium;">\({\boldsymbol{r}} = \left( {\begin{array}{*{20}{c}}<br>1\\<br>5<br>\end{array}} \right) + t\left( {\begin{array}{*{20}{c}}<br>3\\<br>{ - 6}<br>\end{array}} \right)\) , \(\left( {\begin{array}{*{20}{c}}<br>x\\<br>y<br>\end{array}} \right) = \left( {\begin{array}{*{20}{c}}<br>4\\<br>{ - 1}<br>\end{array}} \right) + t\left( {\begin{array}{*{20}{c}}<br>{ - 1}\\<br>2<br>\end{array}} \right)\) </span><em><strong><span style="font-family: times new roman,times; font-size: medium;">A2 N2</span></strong></em></p>
<p><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><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;">substitute (3, <em>k</em>) into equation for (AC) or (BD) <em><strong>(M1)</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g. \(3 = 1 + 4s\) , \(3 = 1 + 3t\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">value of <em>t</em> or <em>s</em> <em><strong>A1</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g. \(s = \frac{1}{2}\) , \( - \frac{1}{2}\) , \(t = \frac{2}{3}\) , \( - \frac{1}{3}\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">substituting <em><strong>A1</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g. \(k = 0 + \frac{1}{2}(2)\)</span></p>
<p><span style="font-family: Times New Roman; font-size: medium;">\(k = 1\) </span><em><strong><span style="font-family: times new roman,times; font-size: medium;">AG N0</span></strong></em></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;">setting up two equations <em><strong>(M1)</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g. \(1 + 4s = 4 + 3t\) , \(2s = - 1 - 6t\) ; setting vector equations of lines equal</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">value of <em>t</em> or <em>s</em> <em><strong>A1</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g. \(s = \frac{1}{2}\) , \( - \frac{1}{2}\) , \(t = \frac{2}{3}\) , \( - \frac{1}{3}\) </span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">substituting <em><strong>A1</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g. \({\boldsymbol{r}} = \left( {\begin{array}{*{20}{c}}<br>4\\<br>{ - 1}<br>\end{array}} \right) - \frac{1}{3}\left( {\begin{array}{*{20}{c}}<br>3\\<br>{ - 6}<br>\end{array}} \right)\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(k = 1\) <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">c.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;"><span style="color: #000000;">\(\overrightarrow {{\rm{PD}}} = \left( {\begin{array}{*{20}{c}}<br>1\\<br>{ - 2}<br>\end{array}} \right)\) </span></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;"> \(|\overrightarrow {{\rm{PD}}} | = \sqrt {{2^2} + {1^2}} \) \(( = \sqrt 5 )\) </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; font-size: medium;">\(|\overrightarrow {{\rm{AC}}} | = \sqrt {{4^2} + {2^2}} \) \(( = \sqrt {20} )\) </span><span style="font-family: times new roman,times; font-size: medium;"><em><strong>(A1)</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">area \( = \frac{1}{2} \times |\overrightarrow {{\rm{AC}}} | \times |\overrightarrow {{\rm{PD}}} |\) (\( = \frac{1}{2} \times \sqrt {20} \times \sqrt 5 \)) </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;">\( = 5\) <em><strong>A1 N4</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">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;">The majority of candidates were successful on part (a), finding vectors between two points and using the scalar product to show two vectors to be perpendicular. </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;">Although a large number of candidates answered part (b) correctly, there were many who had trouble with the vector equation of a line. Most notably, there were those who confused the position vector with the direction vector, and those who wrote their equation in an incorrect form. </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;">In part (c), most candidates seemed to know what was required, though there were many who made algebraic errors when solving for the parameters. A few candidates worked backward, using \(k = 1\) , which is not allowed on a "show that" question. </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 (d), candidates attempted many different geometric and vector methods to find the area of the triangle. As the question said "hence", it was required that candidates should use answers from their previous working - i.e. \({\rm{AC}} \bot {\rm{BD}}\) and \({\text{P}}(3{\text{, }}1)\) . Some geometric approaches, while leading to the correct answer, did not use "hence" or lacked the required justification. </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: 12.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">The line \({L_1}\) passes through the points \(\rm{A}(2, 1, 4)\) and \(\rm{B}(1, 1, 5)\).</span></p>
</div>
<div class="specification">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 12.0px 'Times New Roman';"><span><span style="font-family: 'times new roman', times; font-size: medium;"><span style="line-height: normal;">Another line \({L_2}\) has equation </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;"><strong>r</strong></em><span style="font-family: 'times new roman', times; font-size: medium;"><span style="line-height: normal;"> = \(\left( \begin{array}{c}4\\7\\ - 4\end{array} \right) + s\left( \begin{array}{c}0\\ - 1\\1\end{array} \right)\) . The lines \({L_1}\) and \({L_2}\) intersect at the point P.</span></span></span></p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 12.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">Show that \(\overrightarrow {{\text{AB}}} = \) </span><span style="font-family: 'times new roman', times; font-size: medium; background-color: #f7f7f7;"> \(\left( \begin{array}{c} - 1\\0\\1\end{array} \right)\)</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 style="margin: 0.0px 0.0px 0.0px 0.0px; font: 12.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">Hence, write down </span><span style="font-family: 'times new roman', times; font-size: medium;">a direction vector for \({L_1}\);</span></p>
<div class="marks">[1]</div>
<div class="question_part_label">b(i).</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 12.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">Hence, write down a vector equation for \({L_1}\).</span></p>
<div class="marks">[2]</div>
<div class="question_part_label">b(ii).</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 12.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">Find the coordinates of P.</span></p>
<div class="marks">[6]</div>
<div class="question_part_label">c.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 12.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">Write down a direction vector for \({L_2}\).</span></p>
<div class="marks">[1]</div>
<div class="question_part_label">d(i).</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 12.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">Hence, find the angle between \({L_1}\) and \({L_2}\).</span></p>
<div class="marks">[6]</div>
<div class="question_part_label">d(ii).</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman'; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;">correct approach <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman'; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;"><em>eg</em> \(\left( \begin{array}{c}1\\1\\5\end{array} \right) - \left( \begin{array}{l}2\\1\\4\end{array} \right)\), \({\text{AO}} + {\text{OB}}\), \(b - a\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman'; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;">\(\overrightarrow {{\text{AB}}} = \) <span style="background-color: #f7f7f7;"> \(\left( \begin{array}{c} - 1\\0\\1\end{array} \right)\)</span> <strong><em>AG N0</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman'; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[1 mark]</em></strong></span></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman'; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;">correct vector (or any multiple) <strong><em>A1 N1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman'; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;"><em>eg </em> <strong><em>d</em></strong> = </span><span style="background-color: #f7f7f7; font-family: 'times new roman', times; font-size: medium;"> \(\left( \begin{array}{c} - 1\\0\\1\end{array} \right)\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman'; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[1 mark]</em></strong></span></p>
<div class="question_part_label">b(i).</div>
</div>
<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 10.0px 0.0px; font: 21.0px 'Times New Roman'; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>any</strong> correct equation in the form <strong><em>r</em></strong> = <strong><em>a</em></strong> + <em>t</em><strong><em>b</em></strong> (accept any parameter for <em>t</em>)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman'; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;">where <strong><em>a</em></strong> is \(\left( \begin{array}{c}2\\1\\4\end{array} \right)\) or \(\left( \begin{array}{c}1\\1\\5\end{array} \right)\) , and <strong><em>b</em></strong> is a scalar multiple of <span style="background-color: #f7f7f7;"> \(\left( \begin{array}{c} - 1\\0\\1\end{array} \right)\)</span> <strong><em>A2 N2</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman'; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;"><em>eg</em> <strong><em>r</em></strong> = \(\left( \begin{array}{c}1\\1\\5\end{array} \right) + t\left( \begin{array}{c} - 1\\0\\1\end{array} \right),\left( \begin{array}{c}x\\y\\z\end{array} \right) = \left( \begin{array}{c}2 - s\\1\\4 + s\end{array} \right)\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman'; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;"> </span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman'; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>Note: </strong>Award <strong><em>A1</em></strong> for <strong><em>a</em></strong> + <em>t</em><strong><em>b</em></strong>, <strong><em>A1</em></strong> for \({L_1}\) = <strong><em>a</em></strong> + <em>t</em><strong><em>b</em></strong>, <strong><em>A0</em></strong> for <strong><em>r</em></strong> = <strong><em>b</em></strong> + <em>t</em><strong><em>a</em></strong>.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman'; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;"> </span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman'; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[2 marks]</em></strong></span></p>
<div class="question_part_label">b(ii).</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'; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;">valid approach <strong><em>(M1)</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman'; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;"><em>eg</em> \({r_1} = {r_2}\), \(\left( \begin{array}{c}2\\1\\4\end{array} \right) + t\left( \begin{array}{c} - 1\\0\\1\end{array} \right) = \left( \begin{array}{c}4\\7\\ - 4\end{array} \right) + s\left( \begin{array}{c}0\\ - 1\\1\end{array} \right)\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman'; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;">one correct equation in one parameter <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman'; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;"><em>eg</em> \(2 - t = 4, 1 = 7 - s, 1 - t = 4\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman'; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;">attempt to solve <strong><em>(M1)</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman'; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;"><em>eg</em> \(2 - 4 = t, s = 7 - 1, t = 1 - 4\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman'; color: #3f3f3f;"><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: 21.0px 'Times New Roman'; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;"><em>eg</em> \(t = -2, s = 6, t = -3\),</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman'; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;">attempt to substitute <strong>their </strong>parameter into vector 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'; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;"><em>eg</em> \(\left( \begin{array}{c}4\\7\\ - 4\end{array} \right) + 6\left( \begin{array}{c}0\\ - 1\\1\end{array} \right)\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman'; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;">P(4, 1, 2) (accept position vector) <strong><em>A1 N2</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman'; color: #3f3f3f;"><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>
<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman'; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;">correct direction vector for \({L_2}\) <strong><em>A1 N1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman'; color: #3f3f3f;"><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; background-color: #f7f7f7;">\(\left( \begin{array}{c}0\\-1\\ 1\end{array} \right)\), </span><span style="font-family: 'times new roman', times; font-size: medium; background-color: #f7f7f7;">\(\left( \begin{array}{c}0\\2\\ - 2\end{array} \right)\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman'; color: #3f3f3f;"> </p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman'; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[1 mark]</em></strong></span></p>
<div class="question_part_label">d(i).</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'; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;">correct scalar product and magnitudes for <strong>their </strong>direction vectors <strong><em>(A1)(A1)(A1)</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman'; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;">scalar product \( = 0 \times - 1 + - 1 \times 0 + 1 \times 1{\text{ }}( = 1)\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman'; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;">magnitudes \( = \sqrt {{0^2} + {{( - 1)}^2} + {1^2}} ,{\text{ }}\sqrt { - {1^2} + {0^2} + {1^2}} \left( {\sqrt 2 ,{\text{ }}\sqrt 2 } \right)\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman'; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;">attempt to substitute <strong>their </strong>values into formula <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman'; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;"><em>eg</em> \(\frac{{0 + 0 + 1}}{{\left( {\sqrt {{0^2} + {{( - 1)}^2} + {1^2}} } \right) \times \left( {\sqrt { - {1^2} + {0^2} + {1^2}} } \right)}},{\text{ }}\frac{1}{{\sqrt 2 \times \sqrt 2 }}\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman'; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;">correct value for cosine, \(\frac{1}{2}\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman'; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;">angle is \(\frac{\pi }{3}{\text{ }}( = {60^ \circ })\) <strong><em>A1 N1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman'; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[6 marks]</em></strong></span></p>
<div class="question_part_label">d(ii).</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">b(i).</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">b(ii).</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">c.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">d(i).</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">d(ii).</div>
</div>
<br><hr><br><div class="specification">
<p><span style="font-family: times new roman,times; font-size: medium;">Consider the points A (1 , 5 , 4) , B (3 , 1 , 2) and D (3 , <em>k</em> , 2) , with (AD) perpendicular to (AB) .</span></p>
</div>
<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 align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">Find</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">(i) \(\overrightarrow {{\rm{AB}}} \) ;</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">(ii) \(\overrightarrow {{\rm{AD}}} \) giving your answer in terms of <em>k</em> .</span></p>
<p align="LEFT"><em><strong><span><span style="font-family: times new roman,times; font-size: medium;">[3 marks]</span></span></strong></em></p>
<div class="marks">[3]</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;">Show that \(k = 7\) .</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 C is such that \(\overrightarrow {{\rm{BC}}} = \frac{1}{2}\overrightarrow {{\rm{AD}}} \) .</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"> Find the position vector of C.</span></p>
<div class="marks">[4]</div>
<div class="question_part_label">c.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">Find \(\cos {\rm{A}}\widehat {\rm{B}}{\rm{C}}\) .</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 combining vectors <em><strong>(M1)</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g. \(\overrightarrow {{\rm{AB}}} = \overrightarrow {{\rm{OB}}} - \overrightarrow {{\rm{OA}}} \) (or \(\overrightarrow {{\rm{AD}}} = \overrightarrow {{\rm{AO}}} + \overrightarrow {{\rm{OD}}} \) in part (ii)) </span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"> \(\overrightarrow {{\rm{AB}}} = \left( {\begin{array}{*{20}{c}}<br>2\\<br>{ - 4}\\<br>{ - 2}<br>\end{array}} \right)\)</span><span style="font-family: times new roman,times; font-size: medium;"> </span> <em><strong><span style="font-family: times new roman,times; font-size: medium;">A1 N2</span></strong></em></p>
<p><span style="font-family: times new roman,times; font-size: medium;">(ii) \(\overrightarrow {{\rm{AD}}} = \left( {\begin{array}{*{20}{c}}<br>2\\<br>{k - 5}\\<br>{ - 2}<br>\end{array}} \right)\) </span><em><strong><span style="font-family: times new roman,times; font-size: medium;">A1 N1</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(i) and (ii).</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">evidence of using perpendicularity \( \Rightarrow \) scalar product = 0 <em><strong>(M1)</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g. \(\left( {\begin{array}{*{20}{c}}<br>2\\<br>{ - 4}\\<br>{ - 2}<br>\end{array}} \right) \bullet \left( {\begin{array}{*{20}{c}}<br>2\\<br>{k - 5}\\<br>{ - 2}<br>\end{array}} \right) = 0\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(4 - 4(k - 5) + 4 = 0\) <em><strong>A1</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\( - 4k + 28 = 0\) (accept any correct equation clearly leading to \(k = 7\) ) <em><strong>A1</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(k = 7\) <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">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;"><span style="color: #000000;">\(\overrightarrow {{\rm{AD}}} = \left( {\begin{array}{*{20}{c}}<br>2\\<br>2\\<br>{ - 2}<br>\end{array}} \right)\) <em><strong>(A1)</strong></em></span></span></p>
<p> <span style="font-family: times new roman,times; font-size: medium;">\(\overrightarrow {{\rm{BC}}} = \left( {\begin{array}{*{20}{c}}<br>1\\<br>1\\<br>{ - 1}<br>\end{array}} \right)\)</span><span style="font-family: times new roman,times; font-size: medium;"> </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;">evidence of correct approach <em><strong>(M1)</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g. \(\overrightarrow {{\rm{OC}}} = \overrightarrow {{\rm{OB}}} + \overrightarrow {{\rm{BC}}} \) , \(\left( {\begin{array}{*{20}{c}}<br>3\\<br>1\\<br>2<br>\end{array}} \right) + \left( {\begin{array}{*{20}{c}}<br>1\\<br>1\\<br>{ - 1}<br>\end{array}} \right)\) , \(\left( {\begin{array}{*{20}{c}}<br>{x - 3}\\<br>{y - 1}\\<br>{z - 2}<br>\end{array}} \right) = \left( {\begin{array}{*{20}{c}}<br>1\\<br>1\\<br>{ - 1}<br>\end{array}} \right)\)</span></p>
<p><span style="font-family: Times New Roman; font-size: medium;">\(\overrightarrow {{\rm{OC}}} = \left( {\begin{array}{*{20}{c}}<br>4\\<br>2\\<br>1<br>\end{array}} \right)\) </span><em><strong><span style="font-family: times new roman,times; font-size: medium;">A1 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>
<p> </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></p>
<p><span style="font-family: times new roman,times; font-size: medium;">choosing appropriate vectors, \(\overrightarrow {{\rm{BA}}} \) , \(\overrightarrow {{\rm{BC}}} \) </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;">finding the scalar product <em><strong>M1</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g. \( - 2(1) + 4(1) + 2( - 1)\) , \(2(1) + ( - 4)(1) + ( - 2)( - 1)\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(\cos {\rm{A}}\widehat {\rm{B}}{\rm{C = 0}}\) <em><strong> A1 N1</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;">\(\overrightarrow {{\rm{BC}}} \) parallel to \(\overrightarrow {{\rm{AD}}} \) </span><span style="font-family: times new roman,times; font-size: medium;">(may show this on a diagram with points labelled) <em><strong>R1</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"> \(\overrightarrow {{\rm{BC}}} \bot \overrightarrow {{\rm{AB}}} \) </span><span style="font-family: times new roman,times; font-size: medium;">(may show this on a diagram with points labelled) <em><strong>R1</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"> \({\rm{A}}\widehat {\rm{B}}{\rm{C}} = 90^\circ \)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(\cos {\rm{A}}\widehat {\rm{B}}{\rm{C = 0}}\) <em><strong>A1 N1</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>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">This question was well done by many candidates. Most found \(\overrightarrow {{\rm{AB}}} \) and \(\overrightarrow {{\rm{AD}}} \) correctly.</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;">The majority of candidates correctly used the scalar product to show \(k = 7\) .</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 confusion arose in substituting \(k = 7\) into \(\overrightarrow {{\rm{AD}}} \) , but otherwise part (c) was well done, though finding the position vector of C presented greater difficulty.</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;">Owing to \(\overrightarrow {{\rm{AB}}} \) and \(\overrightarrow {{\rm{BC}}} \) being perpendicular, no problems were created by using these two vectors to find \(\cos {\rm{A}}\widehat {\rm{B}}{\rm{C}} = 0\) , and the majority of candidates answering part (d) did exactly that.</span></p>
<div class="question_part_label">d.</div>
</div>
<br><hr><br><div class="specification">
<p class="p1"><span class="s1">A line \({L_1}\) </span>passes through the points \({\text{A}}(0,{\text{ }} - 3,{\text{ }}1)\) and \({\text{B}}( - 2,{\text{ }}5,{\text{ }}3)\)<span class="s1">.</span></p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>(i) Show that \(\overrightarrow {{\text{AB}}} = \left( {\begin{array}{*{20}{c}} { - 2} \\ 8 \\ 2 \end{array}} \right)\).</p>
<p>(ii) Write down a vector equation for \({L_1}\).</p>
<div class="marks">[3]</div>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p class="p1">A line \({L_2}\) has equation \({\mathbf{r}} = \left( {\begin{array}{*{20}{c}} { - 1} \\ 7 \\ { - 4} \end{array}} \right) + s\left( {\begin{array}{*{20}{c}} 0 \\ 1 \\ { - 1} \end{array}} \right)\). The lines \({L_1}\) and \({L_2}\) <span class="s1">intersect at a point \(C\).</span></p>
<p class="p2">Show that the coordinates of \(C\) are \(( - 1,{\text{ }}1,{\text{ }}2)\).</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>A point \(D\) lies on line \({L_2}\) so that \(\left| {\overrightarrow {{\text{CD}}} } \right| = \sqrt {18} \) and \(\overrightarrow {{\text{CA}}} \bullet \overrightarrow {{\text{CD}}} = - 9\). Find \({\rm{A\hat CD}}\).</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>(i) correct approach <strong><em>A1</em></strong></p>
<p><em>eg</em>\(\;\;\;{\text{OB}} - {\text{OA, }}\left( {\begin{array}{*{20}{c}} { - 2} \\ 5 \\ 3 \end{array}} \right) - \left( {\begin{array}{*{20}{c}} 0 \\ { - 3} \\ 1 \end{array}} \right),{\text{ B}} - {\text{A}}\)</p>
<p>\(\overrightarrow {{\text{AB}}} = \left( {\begin{array}{*{20}{c}} { - 2} \\ 8 \\ 2 \end{array}} \right)\) <strong><em>AG N0</em></strong></p>
<p>(ii) <strong>any </strong>correct equation in the form \(r = a + \) <em>t</em>\(b\) (accept any parameter for \(t\))</p>
<p>where \(a\) is \(\left( {\begin{array}{*{20}{c}} 0 \\ { - 3} \\ 1 \end{array}} \right)\) or \(\left( {\begin{array}{*{20}{c}} { - 2} \\ 5 \\ 3 \end{array}} \right)\), and \(b\) is a scalar multiple of \(\left( {\begin{array}{*{20}{c}} { - 2} \\ 8 \\ 2 \end{array}} \right)\) <strong><em>A2 N2</em></strong></p>
<p><em>eg</em>\(r\) = \left( {\begin{array}{*{20}{c}} 0 \\ { - 3} \\ 1 \end{array}} \right) + <em>t</em>\left( {\begin{array}{*{20}{c}} { - 2} \\ 8 \\ 2 \end{array}} \right),\(r\) = \left( {\begin{array}{*{20}{c}} { - 2 - 2s} \\ {5 + 8s} \\ {3 + 2s} \end{array}} \right),\(r = 2i + 5j + 3k + \) <em>t</em>\(( - 2i + 8j + 2k)\)</p>
<p> </p>
<p><strong>Note: </strong>Award <strong><em>A1 </em></strong>for the form \(a\) + <em>t</em>\(b\), <strong><em>A1</em></strong> for the form \(L = \(a\) + <em>t</em>\(b\),</p>
<p><strong><em>A0</em></strong> for the form \(r\) = \(b\) + <em>t</em>\(a\).</p>
<p><em><strong>[3 marks]</strong></em></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p>valid approach <strong><em>(M1)</em></strong></p>
<p><em>eg</em>\(\;\;\;\)equating lines, \({L_1} = {L_2}\)</p>
<p><strong>one </strong>correct equation in one variable <strong><em>A1</em></strong></p>
<p><em>eg</em>\(\;\;\; - 2t = - 1,{\text{ }} - 2 - 2t = - 1\)</p>
<p>valid attempt to solve <strong><em>(M1)</em></strong></p>
<p><em>eg</em>\(\;\;\;2t = 1,{\text{ }} - 2t = 1\)</p>
<p><strong>one </strong>correct parameter <strong><em>A1</em></strong></p>
<p><em>eg</em>\(\;\;\;t = \frac{1}{2},{\text{ }}t = - \frac{1}{2},{\text{ }}s = - 6\)</p>
<p>correct substitution of either parameter <strong><em>A1</em></strong></p>
<p><em>eg</em>\(\;\;\;r = \left( {\begin{array}{*{20}{c}} 0 \\ { - 3} \\ 1 \end{array}} \right) + \frac{1}{2}\left( {\begin{array}{*{20}{c}} { - 2} \\ 8 \\ 2 \end{array}} \right),{\text{ }}r = \left( {\begin{array}{*{20}{c}} { - 2} \\ 5 \\ 3 \end{array}} \right) - \frac{1}{2}\left( {\begin{array}{*{20}{c}} { - 2} \\ 8 \\ 2 \end{array}} \right),{\text{ }}r = \left( {\begin{array}{*{20}{c}} { - 1} \\ 7 \\ { - 4} \end{array}} \right) - 6\left( {\begin{array}{*{20}{c}} 0 \\ 1 \\ { - 1} \end{array}} \right)\)</p>
<p>the coordinates of \(C\) are \(( - 1,{\text{ }}1,{\text{ }}2)\), or position vector of \(C\) is \(\left( {\begin{array}{*{20}{c}} { - 1} \\ 1 \\ 2 \end{array}} \right)\) <strong><em>AG N0</em></strong></p>
<p> </p>
<p><strong>Note: </strong>If candidate uses the same parameter in both vector equations and working shown, award <strong><em>M1A1M1A0A0</em></strong>.</p>
<p><em><strong>[5 marks]</strong></em></p>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p>valid approach <strong><em>(M1)</em></strong></p>
<p><em>eg</em>\(\;\;\;\)attempt to find \(\overrightarrow {{\text{CA}}} ,{\text{ }}\cos {\rm{A\hat CD}} = \frac{{\overrightarrow {{\text{CA}}} \bullet \overrightarrow {{\text{CD}}} }}{{\left| {\overrightarrow {{\text{CA}}} } \right|\left| {\overrightarrow {{\text{CD}}} } \right|}},{\rm{ A\hat CD}}\) formed by \(\overrightarrow {{\text{CA}}} \) and \(\overrightarrow {{\text{CD}}} \)</p>
<p>\(\overrightarrow {{\text{CA}}} = \left( {\begin{array}{*{20}{c}} 1 \\ { - 4} \\ { - 1} \end{array}} \right)\) <strong><em>(A1)</em></strong></p>
<p> </p>
<p><strong>Notes: </strong>Exceptions to <strong><em>FT:</em></strong></p>
<p>1 if candidate indicates that they are finding \(\overrightarrow {{\text{CA}}} \), but makes an error, award <strong><em>M1A0</em></strong>;</p>
<p>2 if candidate finds an incorrect vector (including \(\overrightarrow {{\text{AC}}} \)), award <strong><em>M0A0</em></strong>.</p>
<p>In both cases, if working shown, full <strong><em>FT </em></strong>may be awarded for subsequent correct <strong><em>FT </em></strong>work.</p>
<p>Award the final (<strong><em>A1</em></strong>) for simplification of <strong>their </strong>value for \({\rm{A\hat CD}}\).</p>
<p>Award the final <strong><em>A2 </em></strong>for finding <strong>their </strong>arc cos. If their value of cos does not allow them to find an angle, they cannot be awarded this <strong><em>A2</em></strong>.</p>
<p> </p>
<p>finding \(\left| {\overrightarrow {{\text{CA}}} } \right|\) (may be seen in cosine formula) <strong><em>A1</em></strong></p>
<p><em>eg</em>\(\;\;\;\sqrt {{1^2} + {{( - 4)}^2} + {{( - 1)}^2}} ,{\text{ }}\sqrt {18} \)</p>
<p>correct substitution into cosine formula <strong><em>(A1)</em></strong></p>
<p><em>eg</em>\(\;\;\;\frac{{ - 9}}{{\sqrt {18} \sqrt {18} }}\)</p>
<p>finding \(\cos {\rm{A\hat CD}} - \frac{1}{2}\) <strong><em>(A1)</em></strong></p>
<p>\({\rm{A\hat CD}} = \frac{{2\pi }}{3}\;\;\;(120^\circ )\) <strong><em>A2 N2</em></strong></p>
<p> </p>
<p><strong>Notes: </strong>Award <strong><em>A1 </em></strong>if additional answers are given.</p>
<p>Award <strong><em>A1 </em></strong>for answer \(\frac{\pi }{3}{\rm{ (60^\circ )}}\).</p>
<p><em><strong>[7 marks]</strong></em></p>
<p><em><strong>Total [15 marks]</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 class="p1">Let \(\overrightarrow {{\text{OA}}} = \left( {\begin{array}{*{20}{c}} { - 1} \\ 0 \\ 4 \end{array}} \right)\) and \(\overrightarrow {{\text{OB}}} = \left( {\begin{array}{*{20}{c}} 4 \\ 1 \\ 3 \end{array}} \right)\).</p>
</div>
<div class="specification">
<p class="p1">The point <span class="s1">C </span>is such that \(\overrightarrow {{\text{AC}}} = \left( {\begin{array}{*{20}{c}} { - 1} \\ 1 \\ { - 1} \end{array}} \right)\).</p>
</div>
<div class="specification">
<p class="p1">The following diagram shows triangle <span class="s1">ABC</span>. Let <span class="s1">D </span>be a point on <span class="s1">[BC]</span>, with acute angle \({\text{ADC}} = \theta \).</p>
<p class="p1" style="text-align: center;"><img src="images/Schermafbeelding_2017-03-03_om_05.46.25.png" alt="N16/5/MATME/SP1/ENG/TZ0/08.c.d.e"></p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>(i) Find \(\overrightarrow {{\text{AB}}} \).</p>
<p>(ii) Find \(\left| {\overrightarrow {{\text{AB}}} } \right|\).</p>
<div class="marks">[4]</div>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p class="p1">Show that the coordinates of <span class="s1">C </span><span class="s2">are \(( - 2,{\text{ }}1,{\text{ }}3)\).</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 class="p1">Write down an expression in terms of \(\theta \) <span class="s1">for</span></p>
<p class="p2">(i) <span class="Apple-converted-space"> </span>angle <span class="s2">ADB</span><span class="s1">;</span></p>
<p class="p2">(ii) <span class="Apple-converted-space"> </span>area of triangle <span class="s2">ABD</span>.</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"><span class="s1">Given that \(\frac{{{\text{area }}\Delta {\text{ABD}}}}{{{\text{area }}\Delta {\text{ACD}}}} = 3\)</span>, show that \(\frac{{{\text{BD}}}}{{{\text{BC}}}} = \frac{3}{4}\).</p>
<div class="marks">[5]</div>
<div class="question_part_label">d.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p class="p1">Hence or otherwise, find the coordinates of point <span class="s1">D</span>.</p>
<div class="marks">[4]</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">(i) <span class="Apple-converted-space"> </span>valid approach to find \(\overrightarrow {{\text{AB}}} \)</p>
<p class="p2"><em>eg</em>\(\,\,\,\,\,\)\(\overrightarrow {{\text{OB}}} - \overrightarrow {{\text{OA}}} ,{\text{ }}\left( {\begin{array}{*{20}{c}} {4 - ( - 1)} \\ {1 - 0} \\ {3 - 4} \end{array}} \right)\)</p>
<p class="p2"><span class="Apple-converted-space">\(\overrightarrow {{\text{AB}}} = \left( {\begin{array}{*{20}{c}} 5 \\ 1 \\ { - 1} \end{array}} \right)\) </span><strong><em>A1 <span class="Apple-converted-space"> </span>N2</em></strong></p>
<p class="p1">(ii) <span class="Apple-converted-space"> </span>valid approach to find \(\left| {\overrightarrow {{\text{AB}}} } \right|\) <span class="Apple-converted-space"> </span><span class="s1"><strong><em>(M1)</em></strong></span></p>
<p class="p2"><em>eg</em>\(\,\,\,\,\,\)\(\sqrt {{{(5)}^2} + {{(1)}^2} + {{( - 1)}^2}} \)</p>
<p class="p2"><span class="s2">\(\left| {\overrightarrow {{\text{AB}}} } \right| = \sqrt {27} \) <span class="Apple-converted-space"> </span></span><strong><em>A1 <span class="Apple-converted-space"> </span>N2</em></strong></p>
<p class="p2"><strong><em>[4 marks]</em></strong></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">correct approach <span class="Apple-converted-space"> </span><span class="s1"><strong><em>A1</em></strong></span></p>
<p class="p2"><em>eg</em>\(\,\,\,\,\,\)\(\overrightarrow {{\text{OC}}} = \left( {\begin{array}{*{20}{c}} { - 1} \\ 1 \\ { - 1} \end{array}} \right) + \left( {\begin{array}{*{20}{c}} { - 1} \\ 0 \\ 4 \end{array}} \right)\)</p>
<p class="p3"><span class="s1">\(C\) </span>has coordinates \(( - 2,{\text{ }}1,{\text{ }}3)\) <span class="Apple-converted-space"> </span><span class="s1"><strong><em>AG <span class="Apple-converted-space"> </span>N0</em></strong></span></p>
<p class="p2"><strong><em>[1 mark]</em></strong></p>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">(i) <span class="Apple-converted-space"> \({\rm{A\hat DB}} = \pi - \theta ,{\rm{ \hat D}} = 180 - \theta \)</span> <span class="Apple-converted-space"> </span><span class="s1"><strong><em>A1 <span class="Apple-converted-space"> </span>N1</em></strong></span></p>
<p class="p2">(ii) <span class="Apple-converted-space"> </span>any correct expression for the area involving \(\theta \) <span class="Apple-converted-space"> </span><span class="s1"><strong><em>A1 <span class="Apple-converted-space"> </span>N1</em></strong></span></p>
<p class="p3"><em>eg</em>\(\,\,\,\,\,\)\({\text{area}} = \frac{1}{2} \times {\text{AD}} \times {\text{BD}} \times \sin (180 - \theta ),{\text{ }}\frac{1}{2}ab\sin \theta ,{\text{ }}\frac{1}{2}\left| {\overrightarrow {{\text{DA}}} } \right|\left| {\overrightarrow {{\text{DB}}} } \right|\sin (\pi - \theta )\)</p>
<p class="p3"><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"><span class="s1"><strong>METHOD 1 </strong></span>(using sine formula for area)</p>
<p class="p1">correct expression for the area of triangle ACD <span class="s1">(seen anywhere) <span class="Apple-converted-space"> </span></span><span class="s2"><strong><em>(A1)</em></strong></span></p>
<p class="p2"><em>eg</em>\(\,\,\,\,\,\)\(\frac{1}{2}{\text{AD}} \times {\text{DC}} \times \sin \theta \)</p>
<p class="p3">correct equation involving areas <span class="Apple-converted-space"> </span><span class="s2"><strong><em>A1</em></strong></span></p>
<p class="p2"><em>eg</em>\(\,\,\,\,\,\)\(\frac{{\frac{1}{2}{\text{AD}} \times {\text{BD}} \times \sin (\pi - \theta )}}{{\frac{1}{2}{\text{AD}} \times {\text{DC}} \times \sin \theta }} = 3\)</p>
<p class="p1">recognizing that \(\sin (\pi - \theta ) = \sin \theta \) <span class="s1">(seen anywhere) <span class="Apple-converted-space"> </span></span><span class="s2"><strong><em>(A1)</em></strong></span></p>
<p class="p3"><span class="s3">\(\frac{{{\text{BD}}}}{{{\text{DC}}}} = 3\) </span>(seen anywhere) <span class="Apple-converted-space"> </span><span class="s2"><strong><em>(A1)</em></strong></span></p>
<p class="p3">correct approach using ratio <span class="Apple-converted-space"> </span><span class="s2"><strong><em>A1</em></strong></span></p>
<p class="p2"><em>eg</em>\(\,\,\,\,\,\)\(3\overrightarrow {{\text{DC}}} + \overrightarrow {{\text{DC}}} = \overrightarrow {{\text{BC}}} ,{\text{ }}\overrightarrow {{\text{BC}}} = 4\overrightarrow {{\text{DC}}} \)</p>
<p class="p1">correct ratio \(\frac{{{\text{BD}}}}{{{\text{BC}}}} = \frac{3}{4}\) <span class="Apple-converted-space"> </span><span class="s2"><strong><em>AG <span class="Apple-converted-space"> </span>N0</em></strong></span></p>
<p class="p3"><strong>METHOD 2 </strong>(Geometric approach)</p>
<p class="p3">recognising \(\Delta {\text{ABD}}\) and \(\Delta {\text{ACD}}\) have same height <span class="Apple-converted-space"> </span><span class="s2"><strong><em>(A1)</em></strong></span></p>
<p class="p1"><span class="s1"><em>eg</em>\(\,\,\,\,\,\)use of \(h\) </span>for both triangles, \(\frac{{\frac{1}{2}{\text{BD}} \times h}}{{\frac{1}{2}{\text{CD}} \times h}} = 3\)</p>
<p class="p3">correct approach <span class="Apple-converted-space"> </span><span class="s2"><strong><em>A2</em></strong></span></p>
<p class="p1"><span class="s2"><em>eg</em>\(\,\,\,\,\,\)\({\text{BD}} = 3x\) </span>and \({\text{DC}} = x,{\text{ }}\frac{{{\text{BD}}}}{{{\text{DC}}}} = 3\)</p>
<p class="p3">correct working <span class="Apple-converted-space"> </span><span class="s2"><strong><em>A2</em></strong></span></p>
<p class="p2"><em>eg</em>\(\,\,\,\,\,\)\({\text{BC}} = 4x,{\text{ BD}} + {\text{DC}} = 4{\text{DC, }}\frac{{{\text{BD}}}}{{{\text{BC}}}} = \frac{{3x}}{{4x}},{\text{ }}\frac{{{\text{BD}}}}{{{\text{BC}}}} = \frac{{3{\text{DC}}}}{{4{\text{DC}}}}\)</p>
<p class="p2"><span class="Apple-converted-space">\(\frac{{{\text{BD}}}}{{{\text{BC}}}} = \frac{3}{4}\) </span><strong><em>AG <span class="Apple-converted-space"> </span>N0</em></strong></p>
<p class="p2"><strong><em>[5 marks]</em></strong></p>
<div class="question_part_label">d.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">correct working (seen anywhere) <span class="Apple-converted-space"> </span><span class="s1"><strong><em>(A1)</em></strong></span></p>
<p class="p2"><em>eg</em>\(\,\,\,\,\,\)\(\overrightarrow {{\text{BD}}} = \frac{3}{4}\overrightarrow {{\text{BC}}} ,{\text{ }}\overrightarrow {{\text{OD}}} = \overrightarrow {{\text{OB}}} + \frac{3}{4}\left( {\begin{array}{*{20}{c}} { - 6} \\ 0 \\ 0 \end{array}} \right),{\text{ }}\overrightarrow {{\text{CD}}} = \frac{1}{4}\overrightarrow {{\text{CB}}} \)</p>
<p class="p1">valid approach (seen anywhere) <span class="Apple-converted-space"> </span><span class="s1"><strong><em>(M1)</em></strong></span></p>
<p class="p2"><em>eg</em>\(\,\,\,\,\,\)\(\overrightarrow {{\text{OD}}} = \overrightarrow {{\text{OB}}} + \overrightarrow {{\text{BD}}} ,{\text{ }}\overrightarrow {{\text{BC}}} = \left( {\begin{array}{*{20}{c}} { - 6} \\ 0 \\ 0 \end{array}} \right)\)</p>
<p class="p1">correct working to find \(x\)-coordinate <span class="Apple-converted-space"> </span><span class="s1"><strong><em>(A1)</em></strong></span></p>
<p class="p2"><em>eg</em>\(\,\,\,\,\,\)\(\left( {\begin{array}{*{20}{c}} 4 \\ 1 \\ 3 \end{array}} \right) + \frac{3}{4}\left( {\begin{array}{*{20}{c}} { - 6} \\ 0 \\ 0 \end{array}} \right),{\text{ }}x = 4 + \frac{3}{4}( - 6),{\text{ }} - 2 + \frac{1}{4}(6)\)</p>
<p class="p3">D is \(\left( { - \frac{1}{2},{\text{ }}1,{\text{ }}3} \right)\) <span class="Apple-converted-space"> </span><span class="s1"><strong><em>A1 <span class="Apple-converted-space"> </span>N3</em></strong></span></p>
<p class="p2"><strong><em>[4 marks]</em></strong></p>
<div class="question_part_label">e.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">c.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">d.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">e.</div>
</div>
<br><hr><br><div class="specification">
<p><span style="font-family: times new roman,times; font-size: medium;">A line \({L_1}\) passes though points P(−1, 6, −1) and Q(0, 4, 1) .</span></p>
</div>
<div class="specification">
<p><span style="font-family: times new roman,times; font-size: medium;">A second line \({L_2}\) has equation \(r = \left( {\begin{array}{*{20}{c}}<br>4\\<br>2\\<br>{ - 1}<br>\end{array}} \right) + s\left( {\begin{array}{*{20}{c}}<br>3\\<br>0\\<br>{ - 4}<br>\end{array}} \right)\) .</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) Show that \(\overrightarrow {{\rm{PQ}}} = \left( {\begin{array}{*{20}{c}}<br>1\\<br>{ - 2}\\<br>2<br>\end{array}} \right)\) .</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">(ii) Hence, write down an equation for \({L_1}\) in the form \({\boldsymbol{r}} = {\boldsymbol{a}} + t{\boldsymbol{b}}\) .</span></p>
<div class="marks">[3]</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;">Find the cosine of the angle between \(\overrightarrow {{\rm{PQ}}} \) </span><span style="font-family: times new roman,times; font-size: medium;">and \({L_2}\) .</span></p>
<div class="marks">[7]</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 lines \({L_1}\) and \({L_2}\) intersect at the point R. Find the coordinates of R.</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) 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. \(\overrightarrow {{\rm{PQ}}} = \overrightarrow {{\rm{OQ}}} - \overrightarrow {{\rm{OP}}} \) , \(Q - P\)</span></p>
<p><span style="font-family: Times New Roman; font-size: medium;">\(\overrightarrow {{\rm{PQ}}} = \left( {\begin{array}{*{20}{c}}<br>1\\<br>{ - 2}\\<br>2<br>\end{array}} \right)\) </span><em><span style="font-family: times new roman,times; font-size: medium;"><strong>AG N0</strong> </span></em></p>
<p><span style="font-family: times new roman,times; font-size: medium;">(ii) 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 <em><strong>a</strong></em> is either \(\overrightarrow {{\rm{OP}}} \) or \(\overrightarrow {{\rm{OQ}}} \) and <strong><em>b</em></strong> is a scalar multiple of \(\overrightarrow {{\rm{PQ}}} \) </span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g. \({\boldsymbol{r}} = \left( {\begin{array}{*{20}{c}}<br>{ - 1}\\<br>6\\<br>{ - 1}<br>\end{array}} \right) + t\left( {\begin{array}{*{20}{c}}<br>1\\<br>{ - 2}\\<br>2<br>\end{array}} \right)\) , \({\boldsymbol{r}} = \left( {\begin{array}{*{20}{c}}<br> t \\ <br> {4 - 2t} \\ <br> {1 + 2t} <br>\end{array}} \right)\), \({\boldsymbol{r}} = 4{\boldsymbol{j}} + {\boldsymbol{k}} + t({\boldsymbol{i}} - 2{\boldsymbol{j}} + 2{\boldsymbol{k}})\)</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">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;">choosing a correct direction vector for \({L_2}\) <em><strong>(A1)</strong> </em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g. \(\left( {\begin{array}{*{20}{c}}<br>3\\<br>0\\<br>{ - 4}<br>\end{array}} \right)\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">finding scalar products 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 \( = 1(3) - 2(0) + 2( - 4)\) \(( = - 5)\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">magnitudes \( = \sqrt {{1^2} + {{( - 2)}^2} + {2^2}} \) \(( = 3)\) , \(\sqrt {{3^2} + {0^2} + {{( - 4)}^2}} \) \(( = 5)\)</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. \(\cos \theta = \frac{{ - 5}}{{\sqrt 9 \times \sqrt {25} }}\)</span></p>
<p><span style="font-family: Times New Roman; font-size: medium;">\(\cos \theta = - \frac{1}{3}\) </span><em><span style="font-family: times new roman,times; font-size: medium;"><strong>A2 N5</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">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">evidence of valid approach <em><strong>(M1)</strong> </em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g. equating lines, \({L_1} = {L_2}\) </span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"><strong>EITHER</strong> </span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"><strong>one</strong> correct equation in one variable <em><strong>A2 </strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g. \(6 - 2t = 2\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"><strong>OR</strong> </span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"><strong>two</strong> correct equations in two variables <em><strong>A1A1 </strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g. \(2t + 4s = 0\) , \(t - 3s = 5\)</span></p>
<p><strong><span style="font-family: times new roman,times; font-size: medium;">THEN </span></strong></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;"><strong>one</strong> correct parameter <em><strong>A1</strong> </em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g. \(t = 2\) , \(s = - 1\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">correct substitution of either parameter <em><strong>(A1)</strong> </em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g. \({\boldsymbol{r}} = \left( {\begin{array}{*{20}{c}}<br>4\\<br>2\\<br>{ - 1}<br>\end{array}} \right) + ( - 1)\left( {\begin{array}{*{20}{c}}<br>3\\<br>0\\<br>{ - 4}<br>\end{array}} \right)\) , \({\boldsymbol{r}} = \left( {\begin{array}{*{20}{c}}<br>{ - 1}\\<br>6\\<br>{ - 1}<br>\end{array}} \right) + ( + 2)\left( {\begin{array}{*{20}{c}}<br>1\\<br>{ - 2}\\<br>2<br>\end{array}} \right)\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">coordinates \({\text{R}}(1{\text{, }}2{\text{, }}3)\) <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 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;">A pleasing number of candidates were successful on this straightforward vector and line question. Part (a) was generally well answered, although a few candidates still labelled their line \(L =\) or used a position vector for the direction vector. Follow-through marking allowed full recovery from the latter error. </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;">Few candidates wrote down their direction vector in part (b) which led to lost follow-through marks, and a common error was finding an incorrect scalar product due to difficulty multiplying by zero. </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;">Part (c) was generally well understood with some candidates realizing that the equation in just one variable led to the correct parameter more quickly than solving a system of two equations to find both parameters. Some candidates gave the answer as (<em>s</em>, <em>t</em>) instead of substituting those parameters, indicating a more rote understanding of the problem. Another common error was using the same parameter for both lines. </span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">There were an alarming number of misreads of negative signs from the question or from the candidate 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;">The line \({L_1}\) is parallel to the <em>z</em>-axis. The point P has position vector \(\left( {\begin{array}{*{20}{c}}<br>8\\<br>1\\<br>0<br>\end{array}} \right)\) and lies on \({L_1}\).</span></p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">Write down the equation of \({L_1}\) in the form \({\boldsymbol{r}} = {\boldsymbol{a}} + t{\boldsymbol{b}}\)</span><span style="font-family: times new roman,times; font-size: medium;">.</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;">The line \({L_2}\) has equation \({\boldsymbol{r}} = \left( {\begin{array}{*{20}{c}}<br>2\\<br>4\\<br>{ - 1}<br>\end{array}} \right) + s\left( {\begin{array}{*{20}{c}}<br>2\\<br>{ - 1}\\<br>5<br>\end{array}} \right)\) . The point A has position vector \(\left( {\begin{array}{*{20}{c}}<br>6\\<br>2\\<br>9<br>\end{array}} \right)\) .</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">Show that A lies on \({L_2}\) .<br></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><span style="font-family: times new roman,times; font-size: medium;">Let B be the point of intersection of lines \({L_1}\) and \({L_2}\) .</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">(i) Show that \(\overrightarrow {{\rm{OB}}} = \left( {\begin{array}{*{20}{c}}<br>8\\<br>1\\<br>{14}<br>\end{array}} \right)\) .</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">(ii) Find \(\overrightarrow {{\rm{AB}}} \) .</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 align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">The point C is at (2, 1, − 4). Let D be the point such that ABCD is a parallelogram.</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">Find \(\overrightarrow {{\rm{OD}}} \) .</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;">\({L_1}:{\boldsymbol{r}} = \left( {\begin{array}{*{20}{c}}<br>8\\<br>1\\<br>0<br>\end{array}} \right) + t\left( {\begin{array}{*{20}{c}}<br>0\\<br>0\\<br>1<br>\end{array}} \right)\) <em><strong>A2 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">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">evidence of equating \({\boldsymbol{r}}\) and \(\overrightarrow {{\rm{OA}}} \) <em><strong>(M1)</strong> </em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g. \(\left( {\begin{array}{*{20}{c}}<br>6\\<br>2\\<br>9<br>\end{array}} \right) = \left( {\begin{array}{*{20}{c}}<br>2\\<br>4\\<br>{ - 1}<br>\end{array}} \right) + s\left( {\begin{array}{*{20}{c}}<br>2\\<br>{ - 1}\\<br>5<br>\end{array}} \right)\) , \(A = r\)</span></p>
<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. \(6 = 2 + 2s\) , \(2 = 4 - s\) , \(9 = - 1 + 5s\)</span></p>
<p><span style="font-family: Times New Roman; font-size: medium;">\(s = 2\) </span><em><span style="font-family: times new roman,times; font-size: medium;"><strong>A1</strong> </span></em></p>
<p><span style="font-family: times new roman,times; font-size: medium;">evidence of confirming for other <strong>two</strong> equations <em><strong>A1</strong> </em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g. \(6 = 2 + 4\) , \(2 = 4 - 2\) , \(9 = - 1 + 10\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">so A lies on \({L_2}\) <em><strong>AG N0</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;">(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. \(\left( {\begin{array}{*{20}{c}}<br>2\\<br>4\\<br>{ - 1}<br>\end{array}} \right) + s\left( {\begin{array}{*{20}{c}}<br>2\\<br>{ - 1}\\<br>5<br>\end{array}} \right) = \left( {\begin{array}{*{20}{c}}<br>8\\<br>1\\<br>0<br>\end{array}} \right) + t\left( {\begin{array}{*{20}{c}}<br>0\\<br>0\\<br>1<br>\end{array}} \right)\) \({L_1} = {L_2}\) </span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">one correct equation <em><strong>A1</strong> </em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g. \(2 + 2s = 8\) , \(4 - s = 1\) , \( - 1 + 5s = t\) </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;">finding \(s = 3\) <em><strong>A1</strong> </em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">substituting <em><strong>M1</strong> </em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g. \(\overrightarrow {{\rm{OB}}} = \left( {\begin{array}{*{20}{c}}<br>2\\<br>4\\<br>{ - 1}<br>\end{array}} \right) + 3\left( {\begin{array}{*{20}{c}}<br>2\\<br>{ - 1}\\<br>5<br>\end{array}} \right)\)</span></p>
<p><span style="font-family: Times New Roman; font-size: medium;">\(\overrightarrow {{\rm{OB}}} = \left( {\begin{array}{*{20}{c}}<br>8\\<br>1\\<br>{14}<br>\end{array}} \right)\) </span><em><span style="font-family: times new roman,times; font-size: medium;"><strong>AG N0</strong> </span></em></p>
<p><span style="font-family: times new roman,times; font-size: medium;">(ii) 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. \(\overrightarrow {{\rm{AB}}} = \overrightarrow {{\rm{AO}}} + \overrightarrow {{\rm{OB}}} \) , \(\overrightarrow {{\rm{AB}}} = \overrightarrow {{\rm{OB}}} - \overrightarrow {{\rm{OA}}} \)</span></p>
<p><span style="font-family: Times New Roman; font-size: medium;">\(\overrightarrow {{\rm{AB}}} = \left( {\begin{array}{*{20}{c}}<br>2\\<br>{ - 1}\\<br>5<br>\end{array}} \right)\) </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;"> [7 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;">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. \(\overrightarrow {{\rm{AB}}} = \overrightarrow {{\rm{DC}}} \)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">correct values <em><strong>A1 </strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g. \(\overrightarrow {{\rm{OD}}} + \left( {\begin{array}{*{20}{c}}<br>2\\<br>{ - 1}\\<br>5<br>\end{array}} \right) = \left( {\begin{array}{*{20}{c}}<br>2\\<br>1\\<br>{ - 4}<br>\end{array}} \right)\) , \(\left( {\begin{array}{*{20}{c}}<br>x\\<br>y\\<br>z<br>\end{array}} \right) + \left( {\begin{array}{*{20}{c}}<br>2\\<br>{ - 1}\\<br>5<br>\end{array}} \right) = \left( {\begin{array}{*{20}{c}}<br>2\\<br>1\\<br>{ - 4}<br>\end{array}} \right)\) , \(\left( {\begin{array}{*{20}{c}}<br>2\\<br>{ - 1}\\<br>5<br>\end{array}} \right) = \left( {\begin{array}{*{20}{c}}<br>{2 - x}\\<br>{1 - y}\\<br>{ - 4 - z}<br>\end{array}} \right)\)</span></p>
<p><span style="font-family: Times New Roman; font-size: medium;">\(\overrightarrow {{\rm{OD}}} = \left( {\begin{array}{*{20}{c}}<br>0\\<br>2\\<br>{ - 9}<br>\end{array}} \right)\) </span><em><span style="font-family: times new roman,times; font-size: medium;"><strong>A1 N2</strong> </span></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">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;">Very few candidates gave a correct direction vector parallel to the <em>z</em>-axis. Provided they wrote down an equation here they were able to earn most of subsequent marks on follow through. </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 (b), many found the correct parameter but neglected to confirm it in the other two equations. </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;">In (c) some performed a trial and error approach to obtaining an integer parameter and thus did not "show" the mathematical origin of the result. Finding vector \(\overrightarrow {{\rm{AB}}} \) proved accessible. </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;">A good number of candidates had an appropriate approach to (d), although surprisingly many subtracted \(\overrightarrow {{\rm{OC}}} \) from \(\overrightarrow {{\rm{AB}}} \) in finding \(\overrightarrow {{\rm{OD}}} \) . </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;">Consider the vectors \(\boldsymbol{a} = \left( {\begin{array}{*{20}{c}}<br>2\\<br>{ - 3}<br>\end{array}} \right)\) and \(\boldsymbol{b} = \left( {\begin{array}{*{20}{c}}<br>1\\<br>4<br>\end{array}} \right)\) .</span></p>
</div>
<div class="specification">
<p><span style="font-family: times new roman,times; font-size: medium;">Let \(2\boldsymbol{a} + \boldsymbol{b} + \boldsymbol{c} = 0\) , where \(0\) is the zero vector.</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;">(a) Find</span></p>
<p style="margin-left: 30px;"><span style="font-family: times new roman,times; font-size: medium;"> (i) \(2\boldsymbol{a} + \boldsymbol{b}\) ;</span></p>
<p style="margin-left: 30px;"><span style="font-family: times new roman,times; font-size: medium;"> (ii) \(\left| {2\boldsymbol{a} + \boldsymbol{b}} \right|\) .</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;">Let \(2\boldsymbol{a} + \boldsymbol{b} + \boldsymbol{c} = 0\) , where \(0\) is the zero vector.</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">(b) Find \(\boldsymbol{c}\) .</span></p>
<div class="marks">[6]</div>
<div class="question_part_label">.</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</span></p>
<p style="margin-left: 30px;"><span style="font-family: times new roman,times; font-size: medium;"> (i) \(2\boldsymbol{a} + \boldsymbol{b}\) ;</span></p>
<p style="margin-left: 30px;"><span style="font-family: times new roman,times; font-size: medium;"> (ii) \(\left| {2\boldsymbol{a} + \boldsymbol{b}} \right|\) .</span></p>
<div class="marks">[4]</div>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">Find \(\boldsymbol{c}\) .</span></p>
<div class="marks">[2]</div>
<div class="question_part_label">b.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">(a) (i) \(2\boldsymbol{a} = \left( {\begin{array}{*{20}{c}}<br>4\\<br>{ - 6}<br>\end{array}} \right)\) <em><strong>(A1)</strong> </em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">correct expression for \(2\boldsymbol{a} + \boldsymbol{b}\) <em><strong>A1 N2</strong> </em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"><em>eg</em> \(\left( {\begin{array}{*{20}{c}}<br>5\\<br>{ - 2}<br>\end{array}} \right)\) , \((5, - 2)\) , \(5\boldsymbol{i} - 2\boldsymbol{j}\)</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) correct substitution into length formula <em><strong>(A1)</strong> </em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"><em>eg</em> \(\sqrt {{5^2} + {2^2}} \) , \(\sqrt {{5^2} + - {2^2}} \)</span></p>
<p><span style="font-family: Times New Roman; font-size: medium;">\(\left| {2\boldsymbol{a} + \boldsymbol{b}} \right| = \sqrt {29} \) </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;"> [4 marks]</span></strong></em></p>
<p><em><strong><span style="font-family: times new roman,times; font-size: medium;"> </span></strong></em></p>
<p><span style="font-family: times new roman,times; font-size: medium;">(b) valid approach <em><strong>(M1)</strong> </em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"><em>eg</em> \(\boldsymbol{c} = - (2\boldsymbol{a} + \boldsymbol{b})\) , \(5 + x = 0\) , \( - 2 + y = 0\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(\boldsymbol{c} = \left( {\begin{array}{*{20}{c}}<br>{ - 5}\\<br>2<br>\end{array}} \right)\) <strong><em>A1 N2 </em></strong></span></p>
<p><strong><em><span style="font-family: times new roman,times; font-size: medium;">[2 marks] </span></em></strong></p>
<div class="question_part_label">.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">(i) \(2\boldsymbol{a} = \left( {\begin{array}{*{20}{c}}<br>4\\<br>{ - 6}<br>\end{array}} \right)\) <em><strong>(A1)</strong> </em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">correct expression for \(2\boldsymbol{a} + \boldsymbol{b}\) <em><strong>A1 N2</strong> </em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"><em>eg</em> \(\left( {\begin{array}{*{20}{c}}<br>5\\<br>{ - 2}<br>\end{array}} \right)\) , \((5, - 2)\) , \(5\boldsymbol{i} - 2\boldsymbol{j}\)</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) correct substitution into length formula <em><strong>(A1)</strong> </em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"><em>eg</em> \(\sqrt {{5^2} + {2^2}} \) , \(\sqrt {{5^2} + - {2^2}} \)</span></p>
<p><span style="font-family: Times New Roman; font-size: medium;">\(\left| {2\boldsymbol{a} + \boldsymbol{b}} \right| = \sqrt {29} \) </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;"> [4 marks]</span></strong></em></p>
<p><em><strong><span style="font-family: times new roman,times; font-size: medium;"> </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 approach <em><strong>(M1)</strong> </em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"><em>eg</em> \(\boldsymbol{c} = - (2\boldsymbol{a} + \boldsymbol{b})\) , \(5 + x = 0\) , \( - 2 + y = 0\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(\boldsymbol{c} = \left( {\begin{array}{*{20}{c}}<br>{ - 5}\\<br>2<br>\end{array}} \right)\) <strong><em>A1 N2 </em></strong></span></p>
<p><strong><em><span style="font-family: times new roman,times; font-size: medium;">[2 marks] </span></em></strong></p>
<div class="question_part_label">b.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">Most candidates comfortably applied algebraic techniques to find new vectors. However, a significant number of candidates answered part (b) as the absolute numerical value of the vector components, which suggests a misunderstanding of the modulus notation. Those who understood the notation easily made the calculation.</span></p>
<div class="question_part_label">.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">Most candidates comfortably applied algebraic techniques to find new vectors. <br></span></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">Most candidates comfortably applied algebraic techniques to find new vectors. However, a significant number of candidates answered part (b) as the absolute numerical value of the vector components, which suggests a misunderstanding of the modulus notation. Those who understood the notation easily made the calculation.</span></p>
<div class="question_part_label">b.</div>
</div>
<br><hr><br><div class="specification">
<p><span style="font-family: times new roman,times; font-size: medium;">The vertices of the triangle PQR are defined by the position vectors</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(\overrightarrow {{\rm{OP}}} = \left( {\begin{array}{*{20}{c}}<br>4\\<br>{ - 3}\\<br>1<br>\end{array}} \right)\) , \(\overrightarrow {{\rm{OQ}}} = \left( {\begin{array}{*{20}{c}}<br>3\\<br>{ - 1}\\<br>2<br>\end{array}} \right)\) and \(\overrightarrow {{\rm{OR}}} = \left( {\begin{array}{*{20}{c}}<br>6\\<br>{ - 1}\\<br>5<br>\end{array}} \right)\) .</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;">Find</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">(i) \(\overrightarrow {{\rm{PQ}}} \) </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;">(ii) \(\overrightarrow {{\rm{PR}}} \) </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><span style="font-family: times new roman,times; font-size: medium;"> Show that \(\cos {\rm{R}}\widehat {\rm{P}}{\rm{Q}} = \frac{1}{2}\) .</span></p>
<div class="marks">[7]</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 \({\rm{sinR}}\widehat {\rm{P}}{\rm{Q}}\) .</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">(ii) Hence, find the area of triangle PQR, giving your answer in the form \(a\sqrt 3 \) .</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><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. \(\overrightarrow {{\rm{PQ}}} = \overrightarrow {{\rm{PO}}} + \overrightarrow {{\rm{OQ}}} \) , \({\rm{Q}} - {\rm{P}}\)</span></p>
<p><span style="font-family: Times New Roman; font-size: medium;">\(\overrightarrow {{\rm{PQ}}} = \left( {\begin{array}{*{20}{c}}<br>{ - 1}\\<br>2\\<br>1<br>\end{array}} \right)\) </span><em><strong><span style="font-family: times new roman,times; font-size: medium;">A1 N2 </span></strong></em></p>
<p><span style="font-family: times new roman,times; font-size: medium;">(ii) \(\overrightarrow {{\rm{PR}}} = \left( {\begin{array}{*{20}{c}}<br>2\\<br>2\\<br>4<br>\end{array}} \right)\) <em><strong>A1 N1</strong> </em></span></p>
<p><em><strong><span style="font-family: times new roman,times; font-size: medium;"> [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;"><strong>METHOD 1 </strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">choosing correct vectors \(\overrightarrow {{\rm{PQ}}} \) and \(\overrightarrow {{\rm{PR}}} \) <em><strong>(A1)(A1)</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">finding \(\overrightarrow {{\rm{PQ}}} \bullet \overrightarrow {{\rm{PR}}} \) , \(\left| {\overrightarrow {{\rm{PQ}}} } \right|\) , \(\left| {\overrightarrow {{\rm{PR}}} } \right|\) <em><strong>(A1) (A1)(A1)</strong> </em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(\overrightarrow {{\rm{PQ}}} \bullet \overrightarrow {{\rm{PR}}} = - 2 + 4 + 4( = 6)\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(\left| {\overrightarrow {{\rm{PQ}}} } \right| = \sqrt {{{( - 1)}^2} + {2^2} + {1^2}} \) \(\left( { = \sqrt 6 } \right)\) , \(\left| {\overrightarrow {{\rm{PR}}} } \right| = \sqrt {{2^2} + {2^2} + {4^2}} \) \(\left( { = \sqrt {24} } \right)\) </span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">substituting into formula for angle between two vectors <em><strong>M1 </strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g. \(\cos {\rm{R}}\widehat {\rm{P}}{\rm{Q}} = \frac{6}{{\sqrt 6 \times \sqrt {24} }}\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">simplifying to expression clearly leading to \(\frac{1}{2}\) <em><strong>A1 </strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g. \(\frac{6}{{\sqrt 6 \times 2\sqrt 6 }}\) , \(\frac{6}{{\sqrt {144} }}\) , \(\frac{6}{{12}}\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"><span style="font-family: times new roman,times; font-size: medium;">\(\cos {\rm{R}}\widehat {\rm{P}}{\rm{Q}} = \frac{1}{2}\)</span> <em><strong> AG N0</strong> </em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"><strong>METHOD 2</strong> </span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">evidence of choosing cosine rule (seen anywhere) <em><strong>(M1)</strong> </em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(\overrightarrow {{\rm{QR}}} = \left( {\begin{array}{*{20}{c}}<br>3\\<br>0\\<br>3<br>\end{array}} \right)\) <em><strong>A1 </strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(\left| {\overrightarrow {{\rm{QR}}} } \right| = \sqrt {18} \) , \(\left| {\overrightarrow {{\rm{PQ}}} } \right| = \sqrt 6 \) and \(\left| {\overrightarrow {{\rm{PR}}} } \right| = \sqrt {24} \) <em><strong>(A1)(A1)(A1) </strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(\cos {\rm{R}}\widehat {\rm{P}}{\rm{Q}} = \frac{{{{\left( {\sqrt 6 } \right)}^2} + {{\left( {\sqrt {24} } \right)}^2} - {{\left( {\sqrt {18} } \right)}^2}}}{{2\sqrt 6 \times \sqrt {24} }}\) <em><strong>A1</strong> </em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(\cos {\rm{R}}\widehat {\rm{P}}{\rm{Q}} = \frac{{6 + 24 - 18}}{{24}}\) \(\left( { = \frac{{12}}{{24}}} \right)\) <em><strong>A1 </strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(\cos {\rm{R}}\widehat {\rm{P}}{\rm{Q}} = \frac{1}{2}\) <em><strong> AG N0</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">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">(i) <strong>METHOD 1</strong> </span></p>
<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. using \({\rm{si}}{{\rm{n}}^{\rm{2}}}{\rm{R}}\widehat {\rm{P}}{\rm{Q + co}}{{\rm{s}}^{\rm{2}}}{\rm{R}}\widehat {\rm{P}}{\rm{Q}} = 1\) , diagram </span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">substituting correctly <em><strong>(A1) </strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g. \({\rm{sinR}}\widehat {\rm{P}}{\rm{Q}} = \sqrt {1 - {{\left( {\frac{1}{2}} \right)}^2}} \)</span></p>
<p><span style="font-family: Times New Roman; font-size: medium;">\({\rm{sinR}}\widehat {\rm{P}}{\rm{Q}} = \sqrt {\frac{3}{4}} \) \(\left( { = \frac{{\sqrt 3 }}{2}} \right)\) </span><em><span style="font-family: times new roman,times; font-size: medium;"><strong>A1 N3</strong> </span></em></p>
<p><span style="font-family: times new roman,times; font-size: medium;"><strong>METHOD 2</strong> </span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">since \(\cos \widehat {\rm{P}} = \frac{1}{2}\) , \(\widehat {\rm{P}} = 60^\circ \) <em><strong>(A1)</strong> </em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">evidence of approach </span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g. drawing a right triangle, finding the missing side <em><strong>(A1)</strong> </em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(\sin \widehat {\rm{P}} = \frac{{\sqrt 3 }}{2}\) <em><strong>A1 N3</strong> </em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">(ii) 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. attempt to substitute into \(\frac{1}{2}ab\sin C\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">correct substitution </span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g. area \( = \frac{1}{2}\sqrt 6 \times \sqrt {24} \times \frac{{\sqrt 3 }}{2}\) <em><strong>A1</strong> </em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">area \( = 3\sqrt 3 \) <em><strong> A1 N2</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">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;">Combining the vectors in (a) was generally well done, although some candidates reversed the subtraction, while others calculated the magnitudes. </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;">Many candidates successfully used scalar product and magnitude calculations to complete part (b). Alternatively, some used the cosine rule, and often achieved correct results. Some assumed the triangle was a right-angled triangle and thus did not earn full marks. Although PQR is indeed right-angled, in a “show that” question this attribute must be directly established. </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 attained the value for sine in (c) with little difficulty, some using the Pythagorean identity, while others knew the side relationships in a 30-60-90 triangle. Unfortunately, a good number of candidates then used the side values of \(1,2,\sqrt 3 \) to find the area of PQR , instead of the magnitudes of the vectors found in (a). Furthermore, the "hence" command was sometimes neglected as the value of sine was expected to be used in the approach. </span></p>
<div class="question_part_label">c.</div>
</div>
<br><hr><br><div class="specification">
<p><span style="font-family: TimesNewRomanPSMT;"><span style="font-family: TimesNewRomanPSMT;"></span></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">Let A and B be points such that \(\overrightarrow {{\rm{OA}}} = \left( {\begin{array}{*{20}{c}}<br>5\\<br>2\\<br>1<br>\end{array}} \right)\) and \(\overrightarrow {{\rm{OB}}} = \left( {\begin{array}{*{20}{c}}<br>6\\<br>0\\<br>3<br>\end{array}} \right)\) .</span></p>
<p align="LEFT"> </p>
<p align="LEFT"> </p>
<p><span style="font-family: TimesNewRomanPSMT;"><span style="font-family: TimesNewRomanPSMT;"></span></span></p>
<p align="LEFT"> </p>
<p><span style="font-family: TimesNewRomanPSMT;"><span style="font-family: TimesNewRomanPSMT;"></span></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;">Show that \(\overrightarrow {{\rm{AB}}} = \left( {\begin{array}{*{20}{c}}<br>1\\<br>{ - 2}\\<br>2<br>\end{array}} \right)\) .</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;">Let C and D be points such that ABCD is a <strong>rectangle</strong>.</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">Given that \(\overrightarrow {{\rm{AD}}} = \left( {\begin{array}{*{20}{c}}<br>4\\<br>p\\<br>1<br>\end{array}} \right)\) , show that \(p = 3\) .</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><span style="font-family: times new roman,times; font-size: medium;">Let C and D be points such that ABCD is a <strong>rectangle</strong>.</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">Find the coordinates of point C.</span></p>
<div class="marks">[4]</div>
<div class="question_part_label">c.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">Let C and D be points such that ABCD is a <strong>rectangle</strong>.</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">Find the area of rectangle ABCD.</span></p>
<div class="marks">[5]</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;">correct approach <strong>A1 </strong></span></p>
<p> </p>
<p><span style="font-family: times new roman,times; font-size: medium;"><em>e.g</em>. \(\overrightarrow {{\rm{AO}}} + \overrightarrow {{\rm{OB}}} ,\left( {\begin{array}{*{20}{c}}<br>6\\<br>0\\<br>3<br>\end{array}} \right) - \left( {\begin{array}{*{20}{c}}<br>5\\<br>2\\<br>1<br>\end{array}} \right)\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(\overrightarrow {{\rm{AB}}} = \left( {\begin{array}{*{20}{c}}<br>1\\<br>{ - 2}\\<br>2<br>\end{array}} \right)\) <em><strong>AG N0</strong></em></span></p>
<p> </p>
<p><strong><span style="font-family: times new roman,times; font-size: medium;"><em>[1 mark]</em> </span></strong></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;">recognizing \(\overrightarrow {{\rm{AD}}} \) is perpendicular to \(\overrightarrow {{\rm{AB}}} \) (may be seen in sketch) <em><strong>(R1)</strong></em> </span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"><em>e.g</em>. adjacent sides of rectangle are perpendicular </span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">recognizing dot product must be zero <em><strong> (R1) </strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"><em>e.g</em>. \(\overrightarrow {{\rm{AD}}} \bullet \overrightarrow {AB} = 0\)</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>e.g</em>. \((1 \times 4) + ( - 2 \times p) + (2 \times 1)\) , \(4 - 2p + 2 = 0\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">equation which clearly leads to \(p = 3\) <em><strong>A1</strong></em> </span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"><em>e.g</em>. \(6 - 2p = 0\) , \(2p = 6\) </span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(p = 3\) <em><strong>AG N0</strong> </em></span></p>
<p> </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 approach (seen anywhere including sketch) <em><strong>(A1)</strong> </em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"><em>e.g</em>. \(\overrightarrow {{\rm{OC}}} = \overrightarrow {{\rm{OB}}} + \overrightarrow {{\rm{BC}}} \) , \(\overrightarrow {{\rm{OD}}} + \overrightarrow {{\rm{DC}}} \)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">recognizing opposite sides are equal vectors (may be seen in sketch) <em><strong>(R1) </strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"><em>e.g</em>. \(\overrightarrow {{\rm{BC}}} = \overrightarrow {{\rm{AD}}} \) , \(\overrightarrow {{\rm{DC}}} = \overrightarrow {{\rm{AB}}} \) , \(\left( {\begin{array}{*{20}{c}}<br>6\\<br>0\\<br>3<br>\end{array}} \right) + \left( {\begin{array}{*{20}{c}}<br>4\\<br>3\\<br>1<br>\end{array}} \right)\) , \(\left( {\begin{array}{*{20}{c}}<br>9\\<br>5\\<br>2<br>\end{array}} \right) + \left( {\begin{array}{*{20}{c}}<br>1\\<br>{ - 2}\\<br>2<br>\end{array}} \right)\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">coordinates of point C are (10, 3, 4) (accept \(\left( {\begin{array}{*{20}{c}}<br>{10}\\<br>3\\<br>4<br>\end{array}} \right)\) ) <em><strong>A2 N4</strong></em> </span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"><strong>Note:</strong> Award <em><strong>A1</strong></em> for two correct values. </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">c.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">attempt to find one side of the rectangle <em><strong>(M1) </strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"><em>e.g</em>. substituting into magnitude formula </span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">two correct magnitudes <em><strong>A1A1 </strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"><em>e.g</em>. \(\sqrt {{{(1)}^2} + {{( - 2)}^2} + {2^2}} \) , 3 ; \(\sqrt {16 + 9 + 1} \) , \(\sqrt {26} \)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">multiplying magnitudes <em><strong>(M1) </strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"><em>e.g</em>. \(\sqrt {26} \times \sqrt 9 \) </span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\({\rm{area}} = \sqrt {234} ( = 3\sqrt {26} )\) (accept \(3 \times \sqrt {26} \) ) <em><strong>A1 N3</strong> </em></span></p>
<p><strong><em><span style="font-family: times new roman,times; font-size: medium;">[5 marks] </span></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><span style="font-family: times new roman,times; font-size: medium;">Part (a) was answered correctly by nearly every candidate. </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), the candidates who realized that the vectors must be perpendicular were successful using the scalar product to find <em>p</em> . Incorrect approaches included using magnitudes, or creating vector equations of lines for the sides and setting them equal to each other. In addition, there were a good number of candidates who worked backwards, using the given value of 3 for <em>p</em> to find the coordinates of point D. Candidates who work backwards on a "show that" question will earn no marks. </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;">Part (c) was more difficult for candidates, and was left blank by some. Some candidates found \(\overrightarrow {{\rm{AC}}} \) rather than \(\overrightarrow {{\rm{OC}}} \) , as required. Many candidates recognized that the opposite sides of the rectangle must be equal, but did not consider the directions of the vectors for those sides. There were also a good number of candidates who mislabelled the vertices of their rectangles, which led to them working with a rectangle ABDC, rather than ABCD. </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;">The majority of candidates who attempted part (d) were successful in multiplying the magnitudes of the sides. Unfortunately, there were some who set up their solutions correctly, then had arithmetic errors in their working. </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: 17.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;"><em>Distances in this question are in metres.</em></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 17.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">Ryan and Jack have model airplanes, which take off from level ground. Jack’s airplane takes off after Ryan’s.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 17.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">The position of Ryan’s airplane \(t\) seconds after it takes off is given by \(\boldsymbol{r}=\left( \begin{array}{c}5\\6\\0\end{array} \right) + t\left( \begin{array}{c} - 4\\2\\4\end{array} \right)\).</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 speed of Ryan’s airplane.</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 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 height of Ryan’s airplane after two seconds.</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 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 position of Jack’s airplane \(s\) seconds after <strong>it </strong>takes off is given by <strong><em>r</em></strong> = \(\left( \begin{array}{c} - 39\\44\\0\end{array} \right) + s\left( \begin{array}{c}4\\ - 6\\7\end{array} \right)\).</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">Show that the paths of the airplanes are perpendicular.</span></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 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 two airplanes collide at the point \((-23, 20, 28)\).</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;">How long after Ryan’s airplane takes off does Jack’s airplane take off?</span></p>
<div class="marks">[5]</div>
<div class="question_part_label">d.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">valid approach <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> magnitude of direction vector</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 working <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> \(\sqrt {{{( - 4)}^2} + {2^2} + {4^2}} ,{\text{ }}\sqrt { - {4^2} + {2^2} + {4^2}} \)</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;">\(6{\text{ (m}}{{\text{s}}^{ - 1}})\) <strong><em>A1 N2</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[3 marks]</em></strong></span></p>
<div class="question_part_label">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;">substituting \(2\) for \(t\)<em> </em><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> </span><span style="font-family: 'times new roman', times; font-size: medium;">\(0 + 2(4)\), <strong><em>r</em></strong> = \(\left( \begin{array}{c}5\\6\\0\end{array} \right) + 2\left( \begin{array}{c} - 4\\2\\4\end{array} \right),\left( \begin{array}{c} - 3\\10\\8\end{array} \right)\), \(y = 10\)</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;">\(8\) (metres) <strong><em>A1 N2</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>[2 marks]</em></strong></span></p>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px '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: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">choosing correct direction vectors \(\left( \begin{array}{c} - 4\\2\\4\end{array} \right)\) and \(\left( \begin{array}{c}4\\ - 6\\7\end{array} \right)\) <strong><em>(A1)(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;">evidence of scalar product <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> <strong><em>a</em></strong> \( \cdot \) <strong><em>b</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;">correct substitution into 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> \(( - 4 \times 4) + (2 \times - 6) + (4 \times 7)\)</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;">evidence of correct calculation of the scalar product as \(0\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;"><em>eg</em> \( - 16 - 12 + 28 = 0\)</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;">directions are perpendicular <strong><em>AG N0</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>METHOD 2</strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">choosing correct direction vectors \(\left( \begin{array}{c} - 4\\2\\4\end{array} \right)\) and \(\left( \begin{array}{c}4\\ - 6\\7\end{array} \right)\) <strong><em>(A1)(A1)</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">attempt to find angle between vectors <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;">correct substitution into numerator <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> \(\cos \theta = \frac{{ - 16 - 12 + 28}}{{\left| a \right|\left| b \right|}},{\text{ }}\cos \theta = 0\)</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;">\(\theta = 90^\circ \) <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;">directions are perpendicular <strong><em>AG N0</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[5 marks]</em></strong></span></p>
<div class="question_part_label">c.</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;"><strong>METHOD 1</strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>one</strong> correct equation for Ryan's airplane <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> \(5 - 4t = - 23,{\text{ }}6 + 2t = 20,{\text{ }}0 + 4t = 28\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">\(t = 7\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>one </strong>correct equation for Jack's airplane <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> \( - 39 + 4s = - 23,{\text{ }}44 - 6s = 20,{\text{ }}0 + 7s = 28\)</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;">\(s = 4\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">\(3\) (seconds later) <strong><em>A1 N2</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>METHOD 2</strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">valid approach <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> \(\left( \begin{array}{c}5\\6\\0\end{array} \right) + t\left( \begin{array}{c} - 4\\2\\4\end{array} \right) = \left( \begin{array}{c} - 39\\44\\0\end{array} \right) + s\left( \begin{array}{c}4\\ - 6\\7\end{array} \right)\), one correct equation</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>two </strong>correct equations <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> \(5 - 4t = - 39 + 4s,{\text{ }}6 + 2t = 44 - 6s,{\text{ }}4t = 7s\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">\(t = 7\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.5px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">\(s = 4\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">\(3\) (seconds later) <strong><em>A1 N2</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[5 marks]</em></strong></span></p>
<div class="question_part_label">d.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">c.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">d.</div>
</div>
<br><hr><br><div class="specification">
<p><span style="font-family: times new roman,times; font-size: medium;">The line \({L_1}\) passes through the points P(2, 4, 8) and Q(4, 5, 4) .</span></p>
</div>
<div class="specification">
<p><span style="font-family: times new roman,times; font-size: medium;">The line \({L_2}\) is perpendicular to \({L_1}\) , and parallel to \(\left( {\begin{array}{*{20}{c}}<br>{3p}\\<br>{2p}\\<br>4<br>\end{array}} \right)\) , where \(p \in \mathbb{Z}\) .</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) Find \(\overrightarrow {{\rm{PQ}}} \) </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 write down a vector equation for \({L_1}\) in the form \({\boldsymbol{r}} = {\boldsymbol{a}} + s{\boldsymbol{b}}\) .</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;">(i) Find the value of <em>p</em> .</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">(ii) Given that \({L_2}\) passes through \({\text{R}}(10{\text{, }}6{\text{, }}- 40)\) , write down a vector equation </span><span style="font-family: times new roman,times; font-size: medium;">for \({L_2}\) .</span></p>
<div class="marks">[7]</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;">The lines \({L_1}\) and \({L_2}\) intersect at the point A. Find the <em>x</em>-coordinate of A.</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) evidence of approach <em><strong>(M1)</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g. \(\overrightarrow {{\rm{PO}}} + \overrightarrow {{\rm{OQ}}} \) , \({\rm{P}} - {\rm{Q}}\)</span></p>
<p><span style="font-family: Times New Roman; font-size: medium;">\(\overrightarrow {{\rm{PQ}}} = \left( {\begin{array}{*{20}{c}}<br>2\\<br>1\\<br>{ - 4}<br>\end{array}} \right)\) </span><em><strong><span style="font-family: times new roman,times; font-size: medium;">A1 N2</span></strong></em></p>
<p><span style="font-family: times new roman,times; font-size: medium;">(ii) <strong>any</strong> correct equation in the form \({\boldsymbol{r}} = {\boldsymbol{a}} + s{\boldsymbol{b}}\) (accept any parameter for <em>s</em>)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">where <em><strong>a</strong></em> is \(\left( {\begin{array}{*{20}{c}}<br>2\\<br>4\\<br>8<br>\end{array}} \right)\) or \(\left( {\begin{array}{*{20}{c}}<br>4\\<br>5\\<br>4<br>\end{array}} \right)\) </span><span style="font-family: times new roman,times; font-size: medium;">, and <em><strong>b</strong></em> is a scalar multiple of \(\left( {\begin{array}{*{20}{c}}<br>2\\<br>1\\<br>{ - 4}<br>\end{array}} \right)\) </span><em><strong><span style="font-family: times new roman,times; font-size: medium;">A2 N2</span></strong></em></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g. \({\boldsymbol{r}} = \left( {\begin{array}{*{20}{c}}<br>2\\<br>4\\<br>8<br>\end{array}} \right) + s\left( {\begin{array}{*{20}{c}}<br>2\\<br>1\\<br>{ - 4}<br>\end{array}} \right)\) , \({\boldsymbol{r}} = \left( {\begin{array}{*{20}{c}}<br>{4 + 2s}\\<br>{5 + 1s}\\<br>{4 - 4s}<br>\end{array}} \right)\) , \({\boldsymbol{r}} = 2{\boldsymbol{i}} + 4{\boldsymbol{j}} + 8{\boldsymbol{k}} + s(2{\boldsymbol{i}} + 1{\boldsymbol{j}} - 4{\boldsymbol{k}})\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"><strong>Note</strong>: Award <em><strong>A1</strong></em> for the form \({\boldsymbol{a}} + s{\boldsymbol{b}}\) , <em><strong>A1</strong></em> for \({\boldsymbol{L}} = {\boldsymbol{a}} + s{\boldsymbol{b}}\) , <em><strong>A0</strong></em> for \({\boldsymbol{r}} = {\boldsymbol{b}} + s{\boldsymbol{a}}\) .</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) choosing correct direction vectors for \({L_1}\) and \({L_2}\) <em><strong>(A1) (A1)</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g. \(\left( {\begin{array}{*{20}{c}}<br>2\\<br>1\\<br>{ - 4}<br>\end{array}} \right)\) , \(\left( {\begin{array}{*{20}{c}}<br>{3p}\\<br>{2p}\\<br>4<br>\end{array}} \right)\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">evidence of equating scalar product to 0 <em><strong>(M1)</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">correct calculation of scalar product <em><strong>A1</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g. \(2 \times 3p + 1 \times 2p + ( - 4) \times 4\) , \(8p - 16 = 0\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(p = 2\) <em><strong>A1 N3</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">(ii) <strong>any</strong> correct expression in the form \({\boldsymbol{r}} = {\boldsymbol{a}} + t{\boldsymbol{b}}\) (accept any parameter for <em>t</em>)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">where <em><strong>a</strong></em> is \(\left( {\begin{array}{*{20}{c}}<br>{10}\\<br>6\\<br>{ - 40}<br>\end{array}} \right)\) </span><span style="font-family: times new roman,times; font-size: medium;">, and <em><strong>b</strong></em> is a scalar multiple of \(\left( {\begin{array}{*{20}{c}}<br>6\\<br>4\\<br>4<br>\end{array}} \right)\) </span><em><strong><span style="font-family: times new roman,times; font-size: medium;">A2 N2</span></strong></em></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g. \({\boldsymbol{r}} = \left( {\begin{array}{*{20}{c}}<br>{10}\\<br>6\\<br>{ - 40}<br>\end{array}} \right) + t\left( {\begin{array}{*{20}{c}}<br>6\\<br>4\\<br>4<br>\end{array}} \right)\) , \({\boldsymbol{r}} = \left( {\begin{array}{*{20}{c}}<br>{10 + 6s}\\<br>{6 + 4s}\\<br>{ - 40 + 4s}<br>\end{array}} \right)\) , \({\boldsymbol{r}} = 10{\boldsymbol{i}} + 6{\boldsymbol{j}} - 40{\boldsymbol{k}} + s(6{\boldsymbol{i}} + 4{\boldsymbol{j}} + 4{\boldsymbol{k}})\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"><strong>Note</strong>: Award <em><strong>A1</strong></em> for the form \({\boldsymbol{a}} + t{\boldsymbol{b}}\) , <em><strong>A1</strong></em> for \({\boldsymbol{L}} = {\boldsymbol{a}} + t{\boldsymbol{b}}\) (unless they </span><span style="font-family: times new roman,times; font-size: medium;">have been penalised for \({\boldsymbol{L}} = {\boldsymbol{a}} + s{\boldsymbol{b}}\) in part (a)), <em><strong>A0</strong></em> for \({\boldsymbol{r}} = {\boldsymbol{b}} + t{\boldsymbol{a}}\) .</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">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;">appropriate approach <em><strong>(M1)</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g. \(\left( {\begin{array}{*{20}{c}}<br>2\\<br>4\\<br>8<br>\end{array}} \right) + s\left( {\begin{array}{*{20}{c}}<br>2\\<br>1\\<br>{ - 4}<br>\end{array}} \right) = \left( {\begin{array}{*{20}{c}}<br>{10}\\<br>6\\<br>{ - 40}<br>\end{array}} \right) + t\left( {\begin{array}{*{20}{c}}<br>6\\<br>4\\<br>4<br>\end{array}} \right)\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">any two correct equations with <strong>different</strong> parameters <em><strong>A1A1</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g. \(2 + 2s = 10 + 6t\) , \(4 + s = 6 + 4t\) , \(8 - 4s = - 40 + 4t\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">attempt to solve simultaneous equations <em><strong>(M1)</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">correct working <em><strong>(A1)</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g. \( - 6 = - 2 - 2t\) , \(4 = 2t\) , \( - 4 + 5s = 46\) , \(5s = 50\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"><strong>one</strong> correct parameter \(s = 10\) , \(t = 2\) <em><strong>A1</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(x = 22\) (accept (22, 14, −32)) <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;">In part (a), nearly all the candidates correctly found the vector PQ, and the majority went onto find the correct vector equation of the line. There are still many candidates who do not write this equation in the correct form, using "<strong><em>r </em></strong>= ", and these candidates were penalized one mark.</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;">In part (b), the majority of candidates knew to set the scalar product equal to zero for the perpendicular vectors, and were able to find the correct value of <em>p</em>. </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;">A good number of candidates used the correct method to find the intersection of the two lines, though some algebraic and arithmetic errors kept some from finding the correct final answer. </span></p>
<div class="question_part_label">c.</div>
</div>
<br><hr><br><div class="specification">
<p>A line \(L\) passes through points \({\text{A}}( - 3,{\text{ }}4,{\text{ }}2)\) and \({\text{B}}( - 1,{\text{ }}3,{\text{ }}3)\).</p>
</div>
<div class="specification">
<p>The line \(L\) also passes through the point \({\text{C}}(3,{\text{ }}1,{\text{ }}p)\).</p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Show that \(\overrightarrow {{\text{AB}}} = \left( {\begin{array}{*{20}{c}} 2 \\ { - 1} \\ 1 \end{array}} \right)\).</p>
<div class="marks">[1]</div>
<div class="question_part_label">a.i.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Find a vector equation for \(L\).</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 value of \(p\).</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>The point D has coordinates \(({q^2},{\text{ }}0,{\text{ }}q)\). Given that \(\overrightarrow {{\text{DC}}} \) is perpendicular to \(L\), find the possible values of \(q\).</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>correct approach <strong><em>A1</em></strong></p>
<p> </p>
<p><em>eg</em>\(\,\,\,\,\,\)\(\left( {\begin{array}{*{20}{c}} { - 1} \\ 3 \\ 3 \end{array}} \right) - \left( {\begin{array}{*{20}{c}} { - 3} \\ 4 \\ 2 \end{array}} \right),{\text{ }}\left( {\begin{array}{*{20}{c}} 3 \\ { - 4} \\ { - 2} \end{array}} \right) + \left( {\begin{array}{*{20}{c}} { - 1} \\ 3 \\ 3 \end{array}} \right)\)</p>
<p> </p>
<p>\(\overrightarrow {{\text{AB}}} = \left( {\begin{array}{*{20}{c}} 2 \\ { - 1} \\ 1 \end{array}} \right)\) <strong><em>AG N0</em></strong></p>
<p><strong><em>[1 mark]</em></strong></p>
<div class="question_part_label">a.i.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p>any correct equation in the form \(r = a + tb\) (any parameter for \(t\))</p>
<p> </p>
<p>where \(a\) is \(\left( {\begin{array}{*{20}{c}} { - 3} \\ 4 \\ 2 \end{array}} \right)\) or \(\left( {\begin{array}{*{20}{c}} { - 1} \\ 3 \\ 3 \end{array}} \right)\) and \(b\) is a scalar multiple of \(\left( {\begin{array}{*{20}{c}} 2 \\ { - 1} \\ 1 \end{array}} \right)\) <strong><em>A2 N2</em></strong></p>
<p> </p>
<p><em>eg</em>\(\,\,\,\,\,\)\(r = \left( {\begin{array}{*{20}{c}} { - 3} \\ 4 \\ 2 \end{array}} \right) + t\left( {\begin{array}{*{20}{c}} 2 \\ { - 1} \\ 1 \end{array}} \right),{\text{ }}(x,{\text{ }}y,{\text{ }}z) = ( - 1,{\text{ }}3,{\text{ }}3) + s( - 2,{\text{ }}1,{\text{ }} - 1),{\text{ }}r = \left( {\begin{array}{*{20}{c}} { - 3 + 2t} \\ {4 - t} \\ {2 + t} \end{array}} \right)\)</p>
<p> </p>
<p><strong>Note:</strong> Award <strong><em>A1 </em></strong>for the form \(a + tb\), <strong>A1</strong> for the form \(L = a + tb\), <strong>A0</strong> for the form \(r = b + ta\).</p>
<p> </p>
<p><strong><em>[2 marks]</em></strong></p>
<div class="question_part_label">a.ii.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><strong>METHOD 1 – finding value of parameter</strong></p>
<p>valid approach <strong><em>(M1)</em></strong></p>
<p> </p>
<p><em>eg</em>\(\,\,\,\,\,\)\(\left( {\begin{array}{*{20}{c}} { - 3} \\ 4 \\ 2 \end{array}} \right) + t\left( {\begin{array}{*{20}{c}} 2 \\ { - 1} \\ 1 \end{array}} \right) = \left( {\begin{array}{*{20}{c}} 3 \\ 1 \\ p \end{array}} \right),{\text{ }}( - 1,{\text{ }}3,{\text{ }}3) + s( - 2,{\text{ }}1,{\text{ }} - 1) = (3,{\text{ }}1,{\text{ }}p)\)</p>
<p> </p>
<p>one correct equation (not involving \(p\)) <strong><em>(A1)</em></strong></p>
<p><em>eg</em>\(\,\,\,\,\,\)\( - 3 + 2t = 3,{\text{ }} - 1 - 2s = 3,{\text{ }}4 - t = 1,{\text{ }}3 + s = 1\)</p>
<p>correct parameter from their equation (may be seen in substitution) <strong><em>A1</em></strong></p>
<p><em>eg</em>\(\,\,\,\,\,\)\(t = 3,{\text{ }}s = - 2\)</p>
<p>correct substitution <strong><em>(A1)</em></strong></p>
<p> </p>
<p><em>eg</em>\(\,\,\,\,\,\)\(\left( {\begin{array}{*{20}{c}} { - 3} \\ 4 \\ 2 \end{array}} \right) + 3\left( {\begin{array}{*{20}{c}} 2 \\ { - 1} \\ 1 \end{array}} \right) = \left( {\begin{array}{*{20}{c}} 3 \\ 1 \\ p \end{array}} \right),{\text{ }}3 - ( - 2)\)</p>
<p> </p>
<p>\(p = 5\,\,\,\,\,\left( {{\text{accept }}\left( {\begin{array}{*{20}{c}} 3 \\ 1 \\ 5 \end{array}} \right)} \right)\) <strong><em>A1 N2</em></strong></p>
<p> </p>
<p><strong>METHOD 2 – eliminating parameter</strong></p>
<p>valid approach <strong><em>(M1)</em></strong></p>
<p> </p>
<p><em>eg</em>\(\,\,\,\,\,\)\(\left( {\begin{array}{*{20}{c}} { - 3} \\ 4 \\ 2 \end{array}} \right) + t\left( {\begin{array}{*{20}{c}} 2 \\ { - 1} \\ 1 \end{array}} \right) = \left( {\begin{array}{*{20}{c}} 3 \\ 1 \\ p \end{array}} \right),{\text{ }}( - 1,{\text{ }}3,{\text{ }}3) + s( - 2,{\text{ }}1,{\text{ }} - 1) = (3,{\text{ }}1,{\text{ }}p)\)</p>
<p> </p>
<p>one correct equation (not involving \(p\)) <strong><em>(A1)</em></strong></p>
<p><em>eg</em>\(\,\,\,\,\,\)\( - 3 + 2t = 3,{\text{ }} - 1 - 2s = 3,{\text{ }}4 - t = 1,{\text{ }}3 + s = 1\)</p>
<p>correct equation (with \(p\)) <strong><em>A1</em></strong></p>
<p><em>eg</em>\(\,\,\,\,\,\)\(2 + t = p,{\text{ }}3 - s = p\)</p>
<p>correct working to solve for \(p\) <strong><em>(A1)</em></strong></p>
<p><em>eg</em>\(\,\,\,\,\,\)\(7 = 2p - 3,{\text{ }}6 = 1 + p\)</p>
<p> </p>
<p>\(p = 5\,\,\,\,\,\left( {{\text{accept }}\left( {\begin{array}{*{20}{c}} 3 \\ 1 \\ 5 \end{array}} \right)} \right)\) <strong><em>A1 N2</em></strong></p>
<p> </p>
<p><strong><em>[5 marks]</em></strong></p>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p>valid approach to find \(\overrightarrow {{\text{DC}}} \) or \(\overrightarrow {{\text{CD}}} \) <strong><em>(M1)</em></strong></p>
<p> </p>
<p><em>eg</em>\(\,\,\,\,\,\)\(\left( {\begin{array}{*{20}{c}} 3 \\ 1 \\ 5 \end{array}} \right) - \left( {\begin{array}{*{20}{c}} {{q^2}} \\ 0 \\ q \end{array}} \right),{\text{ }}\left( {\begin{array}{*{20}{c}} {{q^2}} \\ 0 \\ q \end{array}} \right) - \left( {\begin{array}{*{20}{c}} 3 \\ 1 \\ 5 \end{array}} \right),{\text{ }}\left( {\begin{array}{*{20}{c}} {{q^2}} \\ 0 \\ q \end{array}} \right) - \left( {\begin{array}{*{20}{c}} 3 \\ 1 \\ p \end{array}} \right)\)</p>
<p> </p>
<p>correct vector for \(\overrightarrow {{\text{DC}}} \) or \(\overrightarrow {{\text{CD}}} \) (may be seen in scalar product) <strong><em>A1</em></strong></p>
<p> </p>
<p><em>eg</em>\(\,\,\,\,\,\)\(\left( {\begin{array}{*{20}{c}} {3 - {q^2}} \\ 1 \\ {5 - q} \end{array}} \right),{\text{ }}\left( {\begin{array}{*{20}{c}} {{q^2} - 3} \\ { - 1} \\ {q - 5} \end{array}} \right),{\text{ }}\left( {\begin{array}{*{20}{c}} {3 - {q^2}} \\ 1 \\ {p - q} \end{array}} \right)\)</p>
<p> </p>
<p>recognizing scalar product of \(\overrightarrow {{\text{DC}}} \) or \(\overrightarrow {{\text{CD}}} \) with direction vector of \(L\) is zero (seen anywhere) <strong><em>(M1)</em></strong></p>
<p> </p>
<p><em>eg</em>\(\,\,\,\,\,\)\(\left( {\begin{array}{*{20}{c}} {3 - {q^2}} \\ 1 \\ {p - q} \end{array}} \right) \bullet \left( {\begin{array}{*{20}{c}} 2 \\ { - 1} \\ 1 \end{array}} \right) = 0,{\text{ }}\overrightarrow {{\text{DC}}} \bullet \overrightarrow {{\text{AC}}} = 0,{\text{ }}\left( {\begin{array}{*{20}{c}} {3 - {q^2}} \\ 1 \\ {5 - q} \end{array}} \right) \bullet \left( {\begin{array}{*{20}{c}} 2 \\ { - 1} \\ 1 \end{array}} \right) = 0\)</p>
<p> </p>
<p>correct scalar product in terms of only \(q\) <strong><em>A1</em></strong></p>
<p><em>eg</em>\(\,\,\,\,\,\)\(6 - 2{q^2} - 1 + 5 - q,{\text{ }}2{q^2} + q - 10 = 0,{\text{ }}2(3 - {q^2}) - 1 + 5 - q\)</p>
<p>correct working to solve quadratic <strong><em>(A1)</em></strong></p>
<p><em>eg</em>\(\,\,\,\,\,\)\((2q + 5)(q - 2),{\text{ }}\frac{{ - 1 \pm \sqrt {1 - 4(2)( - 10)} }}{{2(2)}}\)</p>
<p>\(q = - \frac{5}{2},{\text{ }}2\) <strong><em>A1A1 N3</em></strong></p>
<p> </p>
<p><strong><em>[7 marks]</em></strong></p>
<div class="question_part_label">c.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
[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>
<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 line \(L\) is parallel to the vector \(\left( \begin{array}{l}3\\2\end{array} \right)\).</span></p>
</div>
<div class="specification">
<p><span style="font-family: 'times new roman', times; font-size: medium;">The line \(L\) passes through the point \((9, 4)\).</span></p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">Find the gradient of the line \(L\).</span></p>
<div class="marks">[2]</div>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">Find the equation of the line \(L\) in the form \(y = ax + b\).</span></p>
<div class="marks">[3]</div>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">Write down a vector equation for the line \(L\).</span></p>
<div class="marks">[2]</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: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">attempt to find gradient <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> reference to change in \(x\) is \(3\) and/or \(y\) is \(2\), \(\frac{3}{2}\)</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;">gradient \( = \frac{2}{3}\) <strong><em>A1 N2</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>[2 marks]</em></strong></span></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 16.0px 'Times New Roman'; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;">attempt to substitute coordinates and/or gradient into Cartesian equation</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 16.0px 'Times New Roman'; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;">for a line <strong><em>(M1)</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 16.0px 'Times New Roman'; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;"><em>eg</em> \(y - 4 = m(x - 9),{\text{ }}y = \frac{2}{3}x + b,{\text{ }}9 = a(4) + c\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 16.0px 'Times New Roman'; color: #3f3f3f;"><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: 16.0px 'Times New Roman'; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;"><em>eg</em> \(4 = \frac{2}{3}(9) + c,{\text{ }}y - 4 = \frac{2}{3}(x - 9)\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 16.0px 'Times New Roman'; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;">\(y = \frac{2}{3}x - 2{\text{ }}\left( {{\text{accept }}a = \frac{2}{3},{\text{ }}b = - 2} \right)\) <strong><em>A1 N2</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 16.0px 'Times New Roman'; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[3 marks]</em></strong></span></p>
<div class="question_part_label">b.</div>
</div>
<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;"><strong>any</strong> correct equation in the form <strong><em>r</em></strong> = <strong><em>a</em></strong> + <em>t</em><strong><em>b</em></strong> (any parameter for <em>t</em>), where <strong><em>a</em></strong> indicates position <em>eg</em> \(\left( \begin{array}{l}9\\4\end{array} \right)\) or \(\left( \begin{array}{c}0\\ - 2\end{array} \right)\), and <strong><em>b</em></strong> is a scalar multiple of </span><span style="font-family: 'times new roman', times; font-size: medium; background-color: #f7f7f7;">\(\left( \begin{array}{l}3\\2\end{array} \right)\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;"><em>eg </em><strong><em>r </em></strong>= \(\left( \begin{array}{c}9\\4\end{array} \right) + t\left( \begin{array}{c}3\\2\end{array} \right),\left( \begin{array}{c}x\\y\end{array} \right) = \left( \begin{array}{c}3t + 9\\2t + 4\end{array} \right)\), <strong><em>r </em></strong>= 0<em>i </em>− 2 <em>j </em>+ <em>s</em>(3<em>i </em>+ 2 <em>j</em>)<span style="font: 24.5px 'Times New Roman';"> </span><strong><em>A2 N2</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman'; min-height: 25.0px;"> </p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>Note: </strong>Award <strong><em>A1 </em></strong>for <strong><em>a</em></strong> + <em>t</em><strong><em>b</em></strong>, <strong><em>A1 </em></strong>for <em>L</em> = <strong><em>a</em></strong> + <em>t</em><strong><em>b</em></strong>, <strong><em>A0 </em></strong>for <strong><em>r</em></strong> = <strong><em>b</em></strong> + <em>t</em><strong><em>a</em></strong>.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman'; min-height: 25.0px;"> </p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[2 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><span style="font-family: times new roman,times; font-size: medium;">The following diagram shows the obtuse-angled triangle ABC such that \(\overrightarrow {{\rm{AB}}} = \left( {\begin{array}{*{20}{c}}<br>{ - 3}\\<br>0\\<br>{ - 4}<br>\end{array}} \right)\) and \(\overrightarrow {{\rm{AC}}} = \left( {\begin{array}{*{20}{c}}<br>{ - 2}\\<br>2\\<br>{ - 6}<br>\end{array}} \right)\) .</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"><br><img src="images/lbd.png" alt></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 \(\overrightarrow {{\rm{BA}}} \) </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;">(ii) Find \(\overrightarrow {{\rm{BC}}} \) </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(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;">(i) Find \(\cos {\rm{A}}\widehat {\rm{B}}{\rm{C}}\) .</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">(ii) Hence, find \({\rm{sinA}}\widehat {\rm{B}}{\rm{C}}\) .</span></p>
<div class="marks">[7]</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;">The point D is such that \(\overrightarrow {{\rm{CD}}} = \left( {\begin{array}{*{20}{c}}<br>{ - 4}\\<br>5\\<br>p<br>\end{array}} \right)\) , where \(p > 0\) .</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">(i) Given that \(\overrightarrow {|{\rm{CD}}|} = \sqrt {50} \)</span><span style="font-family: times new roman,times; font-size: medium;"> , show that \(p = 3\) .</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">(ii) Hence, show that \(\overrightarrow {{\rm{CD}}} \) </span><span style="font-family: times new roman,times; font-size: medium;">is perpendicular to \(\overrightarrow {{\rm{BC}}} \) </span><span style="font-family: times new roman,times; font-size: medium;">.</span></p>
<p> </p>
<div class="marks">[6]</div>
<div class="question_part_label">c(i) and (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) \(\overrightarrow {{\rm{BA}}} = \left( {\begin{array}{*{20}{c}}<br>3\\<br>0\\<br>4<br>\end{array}} \right)\) <em><strong>A1 N1</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">(ii) evidence of combining vectors <em><strong>(M1)</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g. \(\overrightarrow {{\rm{AB}}} + \overrightarrow {{\rm{BC}}} = \overrightarrow {{\rm{AC}}} \) , \(\overrightarrow {{\rm{BA}}} + \overrightarrow {{\rm{AC}}} \) , \(\left( {\begin{array}{*{20}{c}}<br>{ - 2}\\<br>2\\<br>{ - 6}<br>\end{array}} \right) - \left( {\begin{array}{*{20}{c}}<br>{ - 3}\\<br>0\\<br>{ - 4}<br>\end{array}} \right)\)</span></p>
<p><span style="font-family: Times New Roman; font-size: medium;">\(\overrightarrow {{\rm{BC}}} = \left( {\begin{array}{*{20}{c}}<br>1\\<br>2\\<br>{ - 2}<br>\end{array}} \right)\) </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;"> [3 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) <strong>METHOD 1</strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">finding \(\overrightarrow {{\rm{BA}}} \bullet \overrightarrow {{\rm{BC}}} \) , \(|\overrightarrow {{\rm{BA}}} |\) , \(|\overrightarrow {{\rm{BC}}} |\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g. \(\overrightarrow {{\rm{BA}}} \bullet \overrightarrow {{\rm{BC}}} = 3 \times 1 + 0 + 4 \times - 2\) , \(\overrightarrow {|{\rm{BA}}} | = \sqrt {{3^2} + {4^2}} \) , \(\overrightarrow {|{\rm{BC}}} | = 3\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">substituting into formula for \(\cos \theta \) <span lang="EN-US"> </span><em><strong>M1</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g. \(\frac{{3 \times 1 + 0 + 4 \times - 2}}{{3\sqrt {{3^2} + 0 + {4^2}} }}\) , \(\frac{{ - 5}}{{5 \times 3}}\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(cos {\rm{A}}\widehat {\rm{B}}{\rm{C}} = - \frac{5}{{15}}\) \(\left( { = - \frac{1}{3}} \right)\) <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;">finding \(|\overrightarrow {{\rm{AC}}} |\) , \(|\overrightarrow {{\rm{BA}}} |\) , \(|\overrightarrow {{\rm{BC}}} |\) </span><span style="font-family: times new roman,times; font-size: medium;"> <em><strong>(A1)(A1)(A1)</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g. \(|\overrightarrow {{\rm{AC}}} | = \sqrt {{2^2} + {2^2} + {6^2}} \) , \(|\overrightarrow {{\rm{BA}}} | = \sqrt {{3^2} + {4^2}} \) , \(|\overrightarrow {{\rm{BC}}} | = 3\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">substituting into cosine rule <em><strong>M1</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g. \(\frac{{{5^2} + {3^2} - {{\left( {\sqrt {44} } \right)}^2}}}{{2 \times 5 \times 3}}\) , \(\frac{{25 + 9 - 44}}{{30}}\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(\cos {\rm{A}}\widehat {\rm{B}}{\rm{C}} = - \frac{{10}}{{30}}\) \(\left( { = - \frac{1}{3}} \right)\) <em><strong>A1 N3</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">(ii) evidence of using Pythagoras <em><strong>(M1)</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g. right-angled triangle with values, \({\sin ^2}x + {\cos ^2}x = 1\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(\sin {\rm{A}}\widehat {\rm{B}}{\rm{C}} = \frac{{\sqrt 8 }}{3}\) \(\left( { = \frac{{2\sqrt 2 }}{3}} \right)\) <em><strong>A1 N2</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">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;">(i) attempt to find an expression for \(|\overrightarrow {{\rm{CD}}} |\) </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. \(\sqrt {{{( - 4)}^2} + {5^2} + {p^2}} \) , \(|\overrightarrow {{\rm{CD}}} {|^2} = {4^2} + {5^2} + {p^2}\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">correct equation <em><strong>A1</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g. \(\sqrt {{{( - 4)}^2} + {5^2} + {p^2}} = \sqrt {50} \) , \({4^2} + {5^2} + {p^2} = 50\)</span></p>
<p><span style="font-family: Times New Roman; font-size: medium;">\({p^2} = 9\) </span><span style="font-family: times new roman,times; font-size: medium;"> <em><strong>A1</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(p = 3\) <em><strong>AG N0</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">(ii) evidence of scalar product <em><strong>(M1)</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g. \(\left( {\begin{array}{*{20}{c}}<br>{ - 4}\\<br>5\\<br>3<br>\end{array}} \right) \bullet \left( {\begin{array}{*{20}{c}}<br>1\\<br>2\\<br>{ - 2}<br>\end{array}} \right)\) , \(\overrightarrow {{\rm{CD}}} \bullet \overrightarrow {{\rm{BC}}} \)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">correct substitution</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g. \( - 4 \times 1 + 5 \times 2 + 3 \times - 2\) , \( - 4 + 10 - 6\) </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;"> \(\overrightarrow {{\rm{CD}}} \bullet \overrightarrow {{\rm{BC}}} = 0\) <em><strong>A1</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(\overrightarrow {{\rm{CD}}} \) is perpendicular to \(\overrightarrow {{\rm{BC}}} \) <em><strong>AG N0</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">c(i) and (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;">Many candidates answered (a) correctly, although some reversed the vectors when finding \(\overrightarrow {{\rm{BC}}} \) , while others miscopied the vectors from the question paper. </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;">Students had no difficulty finding the scalar product and magnitudes of the vectors used in finding the cosine. However, few recognized that \(\overrightarrow {{\rm{BA}}} \) is the vector to apply in the formula to find the cosine value. Most used \(\overrightarrow {{\rm{AB}}} \) to obtain a positive cosine, which neglects that the angle is obtuse and thus has a negative cosine. Surprisingly few students could then take a value for cosine and use it to find a value for sine. Most left (bii) blank entirely. </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;">Part (c) proved accessible for many candidates. Some created an expression for \(|\overrightarrow {{\rm{CD}}} |\) and then substituted the given \(p = 3\) to obtain \(\sqrt {50} \) , which does not satisfy the "show that" instruction. Many students recognized that the scalar product must be zero for vectors to be perpendicular, and most provided the supporting calculations. </span></p>
<div class="question_part_label">c(i) and (ii).</div>
</div>
<br><hr><br><div class="specification">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;"><span style="font-family: TimesNewRomanPSMT;">The line </span><em><span style="font-family: TimesNewRomanPS-ItalicMT;">L </span></em><span style="font-family: TimesNewRomanPSMT;">passes through the point \((5, - 4,10)\)</span><span style="font-family: TimesNewRomanPSMT;"> and is parallel to the vector \(\left( {\begin{array}{*{20}{c}}<br>4\\<br>{ - 2}\\<br>5<br>\end{array}} \right)\) .</span></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;"><span style="font-family: TimesNewRomanPSMT;">Write down a vector equation for line </span><em><span style="font-family: TimesNewRomanPS-ItalicMT;">L </span></em><span style="font-family: TimesNewRomanPSMT;">.</span></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;"><span style="font-family: TimesNewRomanPSMT;">The line </span><em><span style="font-family: TimesNewRomanPS-ItalicMT;">L </span></em><span style="font-family: TimesNewRomanPSMT;">intersects the </span><em><span style="font-family: TimesNewRomanPS-ItalicMT;">x</span></em><span style="font-family: TimesNewRomanPSMT;">-axis at the point P. Find the </span><em><span style="font-family: TimesNewRomanPS-ItalicMT;">x</span></em><span style="font-family: TimesNewRomanPSMT;">-coordinate of P.</span></span></p>
<div class="marks">[6]</div>
<div class="question_part_label">b.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;"><span style="font-size: medium;"><strong>any </strong></span><span style="font-family: Times New Roman,Times New Roman; font-size: medium;"><span style="font-family: Times New Roman,Times New Roman; font-size: medium;">correct equation in the form \({\boldsymbol{r}} = {\boldsymbol{a}} + t{\boldsymbol{b}}\) (accept any parameter for </span></span><em><span style="font-family: Times New Roman,Times New Roman; font-size: medium;"><span style="font-family: Times New Roman,Times New Roman; font-size: medium;">t</span></span></em><span style="font-family: Times New Roman,Times New Roman; font-size: medium;"><span style="font-family: Times New Roman,Times New Roman; font-size: medium;">) </span></span></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"><span style="font-family: Times New Roman,Times New Roman; font-size: medium;"><span style="font-family: Times New Roman,Times New Roman; font-size: medium;">where </span></span><strong><em><span style="font-family: Times New Roman,Times New Roman; font-size: medium;"><span style="font-family: Times New Roman,Times New Roman; font-size: medium;">a </span></span></em></strong><span style="font-family: Times New Roman,Times New Roman; font-size: medium;"><span style="font-family: Times New Roman,Times New Roman; font-size: medium;">is \(\left( {\begin{array}{*{20}{c}}<br>5\\<br>{ - 4}\\<br>{10}<br>\end{array}} \right)\) , and </span></span><strong><em><span style="font-family: Times New Roman,Times New Roman; font-size: medium;"><span style="font-family: Times New Roman,Times New Roman; font-size: medium;">b </span></span></em></strong><span style="font-family: Times New Roman,Times New Roman; font-size: medium;"><span style="font-family: Times New Roman,Times New Roman; font-size: medium;">is a scalar multiple of \(\left( {\begin{array}{*{20}{c}}<br>4\\<br>{ - 2}\\<br>5<br>\end{array}} \right)\) </span></span><strong><em><span style="font-family: Times New Roman,Times New Roman; font-size: medium;"><span style="font-family: Times New Roman,Times New Roman; font-size: medium;">A2 N2 </span></span></em></strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"><em><span style="font-family: Times New Roman,Times New Roman; font-size: medium;"><span style="font-family: Times New Roman,Times New Roman; font-size: medium;">e.g.</span></span></em> \({\boldsymbol{r}} = \left( {\begin{array}{*{20}{c}}<br> 5 \\ <br> { - 4} \\ <br> {10} <br>\end{array}} \right) + t\left( {\begin{array}{*{20}{c}}<br> 4 \\ <br> { - 2} \\ <br> 5 <br>\end{array}} \right){\text{, }}{\boldsymbol{r}} = 5{\boldsymbol{i}} - 4{\boldsymbol{j}} + 10{\boldsymbol{k}} + t( - 8{\boldsymbol{i}} + 4{\boldsymbol{j}} - 10{\boldsymbol{k}})\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"><span style="font-size: medium;"><strong>Note</strong></span><span style="font-family: Times New Roman,Times New Roman; font-size: medium;"><span style="font-family: Times New Roman,Times New Roman; font-size: medium;">: Award </span></span><strong><em><span style="font-family: Times New Roman,Times New Roman; font-size: medium;"><span style="font-family: Times New Roman,Times New Roman; font-size: medium;">A1 </span></span></em></strong><span style="font-family: Times New Roman,Times New Roman; font-size: medium;"><span style="font-family: Times New Roman,Times New Roman; font-size: medium;">for the form \({\boldsymbol{a}} + t{\boldsymbol{b}}\) , </span></span><strong><em><span style="font-family: Times New Roman,Times New Roman; font-size: medium;"><span style="font-family: Times New Roman,Times New Roman; font-size: medium;">A1 </span></span></em></strong><span style="font-family: Times New Roman,Times New Roman; font-size: medium;"><span style="font-family: Times New Roman,Times New Roman; font-size: medium;">for \(L = {\boldsymbol{a}} + t{\boldsymbol{b}}\) , </span></span><strong><em><span style="font-family: Times New Roman,Times New Roman; font-size: medium;"><span style="font-family: Times New Roman,Times New Roman; font-size: medium;">A0 </span></span></em></strong><span style="font-family: Times New Roman,Times New Roman; font-size: medium;"><span style="font-family: Times New Roman,Times New Roman; font-size: medium;">for \({\boldsymbol{r}} = {\boldsymbol{b}} + t{\boldsymbol{a}}\) . </span></span></span></p>
<p> </p>
<p><em><strong><span style="font-family: times new roman,times; font-size: medium;">[2 marks] </span></strong></em></p>
<p> </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;"><span style="font-size: medium;">recognizing that \(y = 0\) or \(z = 0\) at </span><em><span style="font-family: Times New Roman,Times New Roman; font-size: medium;"><span style="font-family: Times New Roman,Times New Roman; font-size: medium;">x</span></span></em><span style="font-size: medium;">-intercept (seen anywhere) </span><strong><em><span style="font-family: Times New Roman,Times New Roman; font-size: medium;"><span style="font-family: Times New Roman,Times New Roman; font-size: medium;">(R1) </span></span></em></strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"><span style="font-size: medium;">attempt to set up equation for </span><em><span style="font-family: Times New Roman,Times New Roman; font-size: medium;"><span style="font-family: Times New Roman,Times New Roman; font-size: medium;">x</span></span></em><span style="font-size: medium;">-intercept (must suggest \(x \ne 0\) ) </span><strong><em><span style="font-family: Times New Roman,Times New Roman; font-size: medium;"><span style="font-family: Times New Roman,Times New Roman; font-size: medium;">(M1) </span></span></em></strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"><span style="font-family: Times New Roman,Times New Roman; font-size: medium;"><span style="font-family: Times New Roman,Times New Roman; font-size: medium;"><em>e.g. </em>\(L = \left( {\begin{array}{*{20}{c}}<br>x\\<br>0\\<br>0<br>\end{array}} \right)\) </span></span><span style="font-size: medium;">, \(5 + 4t = x\) , \(r = \left( {\begin{array}{*{20}{c}}<br>1\\<br>0\\<br>0<br>\end{array}} \right)\)</span></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"><span style="font-size: medium;">one correct equation in one variable </span><strong><em><span style="font-family: Times New Roman,Times New Roman; font-size: medium;"><span style="font-family: Times New Roman,Times New Roman; font-size: medium;">(A1) </span></span></em></strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"><span style="font-family: Times New Roman,Times New Roman; font-size: medium;"><span style="font-family: Times New Roman,Times New Roman; font-size: medium;"><em>e.g.</em> \( - 4 - 2t = 0\) </span></span><span style="font-size: medium;">, \(10 + 5t = 0\) </span></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"><span style="font-size: medium;">finding \(t = - 2\) </span><strong><em><span style="font-family: Times New Roman,Times New Roman; font-size: medium;"><span style="font-family: Times New Roman,Times New Roman; font-size: medium;">A1 </span></span></em></strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"><span style="font-size: medium;">correct working </span><strong><em><span style="font-family: Times New Roman,Times New Roman; font-size: medium;"><span style="font-family: Times New Roman,Times New Roman; font-size: medium;">(A1) </span></span></em></strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"><span style="font-family: Times New Roman,Times New Roman; font-size: medium;"><span style="font-family: Times New Roman,Times New Roman; font-size: medium;"><em>e.g. </em>\(x = 5 + ( - 2)(4)\)</span></span></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"><span style="font-family: Times New Roman,Times New Roman; font-size: medium;"><span style="font-family: Times New Roman,Times New Roman; font-size: medium;">\(x = - 3\) (accept \(( - 3{\text{, }}0{\text{, }}0)\)) </span></span></span><span style="font-family: times new roman,times; font-size: medium;"><span style="font-family: Times New Roman,Times New Roman; font-size: medium;"><span style="font-family: Times New Roman,Times New Roman; font-size: medium;"><em><strong>A1 N3 </strong></em></span></span></span></p>
<p> </p>
<p><strong><span style="font-family: times new roman,times; font-size: medium;"><em>[6 marks]</em> </span></strong></p>
<p> </p>
<div class="question_part_label">b.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
<p><span style="font-size: medium; font-family: times new roman,times;">In part (a), the majority of candidates correctly recognized the equation that contains the position and direction vectors of a line. However, we saw a large number of candidates who continue to write their equations using "\(L = \) ", rather than the mathematically correct "\({\boldsymbol{r}} = \) " or "\(\left( {\begin{array}{*{20}{c}}<br>x\\<br>y\\<br>z<br>\end{array}} \right) = \)". <strong><em>r</em></strong> and \(\left( {\begin{array}{*{20}{c}}<br> x \\ <br> y \\ <br> z <br>\end{array}} \right)\) represent vectors, whereas <em>L</em><em> </em>is simply the name of the line. For part (b), very few candidates recognized that a general point on the <em>x</em>-axis will be given by the vector \(\left( {\begin{array}{*{20}{c}}<br>x\\<br>0\\<br>0<br>\end{array}} \right)\) . Common errors included candidates setting their equation equal to\(\left( {\begin{array}{*{20}{c}}<br>0\\<br>0\\<br>0<br>\end{array}} \right)\) , or \(\left( {\begin{array}{*{20}{c}}<br>1\\<br>0\\<br>0<br>\end{array}} \right)\) , or even just the number \(0\).</span></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-size: medium; font-family: times new roman,times;">In part (a), the majority of candidates correctly recognized the equation that contains the position and direction vectors of a line. However, we saw a large number of candidates who continue to write their equations using "\(L = \) ", rather than the mathematically correct "\({\boldsymbol{r}} = \) " or "\(\left( {\begin{array}{*{20}{c}}<br>x\\<br>y\\<br>z<br>\end{array}} \right) = \)". <strong><em>r</em></strong> and \(\left( {\begin{array}{*{20}{c}}<br> x \\ <br> y \\ <br> z <br>\end{array}} \right)\) represent vectors, whereas <em>L</em><em> </em>is simply the name of the line. For part (b), very few candidates recognized that a general point on the <em>x</em>-axis will be given by the vector \(\left( {\begin{array}{*{20}{c}}<br>x\\<br>0\\<br>0<br>\end{array}} \right)\) . Common errors included candidates setting their equation equal to\(\left( {\begin{array}{*{20}{c}}<br>0\\<br>0\\<br>0<br>\end{array}} \right)\) , or \(\left( {\begin{array}{*{20}{c}}<br>1\\<br>0\\<br>0<br>\end{array}} \right)\) , or even just the number \(0\).</span></p>
<div class="question_part_label">b.</div>
</div>
<br><hr><br><div class="specification">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">A particle is moving with a constant velocity along line <em>L</em> . Its initial position </span><span style="font-family: times new roman,times; font-size: medium;">is A(6 , −2 , 10) . After one second the particle has moved to B( 9, −6 , 15) .</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) Find the velocity vector, \(\overrightarrow {{\rm{AB}}} \) </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) Find the speed of the particle.</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><span style="font-family: times new roman,times; font-size: medium;">Write down an equation of the line <em>L</em> .</span></p>
<div class="marks">[2]</div>
<div class="question_part_label">b.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">(i) evidence of approach <strong><em>(M1) </em></strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g. \(\overrightarrow {{\rm{AO}}} + \overrightarrow {{\rm{OB}}} \) , \({\rm{B}} - {\rm{A}}\) , \(\left( {\begin{array}{*{20}{c}}<br>{9 - 6}\\<br>{ - 6 + 2}\\<br>{15 - 10}<br>\end{array}} \right)\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(\overrightarrow {{\rm{AB}}} = \left( {\begin{array}{*{20}{c}}<br>3\\<br>{ - 4}\\<br>5<br>\end{array}} \right)\) (accept \(\left( {3, - 4, 5} \right)\) ) <em><strong>A1 N2</strong> </em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">(ii) evidence of finding the magnitude of the velocity vector <strong><em>M1</em> </strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g. \({\text{speed}} = \sqrt {{3^2} + {4^2} + {5^2}} \)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\({\text{speed}} = \sqrt {50} \) \(\left( { = 5\sqrt 2 } \right)\) <strong><em>A1 N1</em></strong></span></p>
<p><strong><em><span style="font-family: times new roman,times; font-size: medium;">[4 marks]</span></em></strong></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;">correct <strong>equation</strong> (accept Cartesian and parametric forms) <strong><em>A2 N2</em> </strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g. \({\boldsymbol{r}} = \left( {\begin{array}{*{20}{c}}<br>6\\<br>{ - 2}\\<br>{10}<br>\end{array}} \right) + t\left( {\begin{array}{*{20}{c}}<br>3\\<br>{ - 4}\\<br>5<br>\end{array}} \right)\) , \({\boldsymbol{r}} = \left( {\begin{array}{*{20}{c}}<br>9\\<br>{ - 6}\\<br>{15}<br>\end{array}} \right) + t\left( {\begin{array}{*{20}{c}}<br>3\\<br>{ - 4}\\<br>5<br>\end{array}} \right)\)</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>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">This question was quite well done. Marks were lost when candidates found the vector \(\overrightarrow {{\text{BA}}} \) instead of \(\overrightarrow {{\text{AB}}} \) in part (a) and for not writing their vector equation as an equation.</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;">In part (b), a few candidates switched the position and velocity vectors or used the vectors \(\overrightarrow {{\text{OA}}} \) and \(\overrightarrow {{\text{OB}}} \) to incorrectly write the vector equation.</span></p>
<div class="question_part_label">b.</div>
</div>
<br><hr><br><div class="specification">
<p><span style="font-family: times new roman,times; font-size: medium;">The line \({L_1}\) is represented by the vector equation \({\boldsymbol{r}} = \left( {\begin{array}{*{20}{c}}<br>{ - 3}\\<br>{ - 1}\\<br>{ - 25}<br>\end{array}} \right) + p\left( {\begin{array}{*{20}{c}}<br>2\\<br>1\\<br>{ - 8}<br>\end{array}} \right)\) .</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">A second line \({L_2}\) is parallel to \({L_1}\) and passes through the point B(\( - 8\), \( - 5\), \(25\)) .</span></p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">Write down a vector equation for \({L_2}\) in the form \({\boldsymbol{r}} = {\boldsymbol{a}} + t{\boldsymbol{b}}\) .</span></p>
<div class="marks">[2]</div>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">A third line \({L_3}\) is perpendicular to \({L_1}\) and is represented by \({\boldsymbol{r}} = \left( {\begin{array}{*{20}{c}}<br>5\\<br>0\\<br>3<br>\end{array}} \right) + q\left( {\begin{array}{*{20}{c}}<br>{ - 7}\\<br>{ - 2}\\<br>k<br>\end{array}} \right)\) .</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">Show that \(k = - 2\) .</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><span style="font-family: times new roman,times; font-size: medium;">The lines \({L_1}\) and \({L_3}\) intersect at the point A.</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">Find the coordinates of A.</span></p>
<div class="marks">[6]</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;"><span>The lines </span><span>\({L_2}\)</span><span>and </span><span>\({L_3}\)</span><span>intersect at point C where \(\overrightarrow {{\rm{BC}}} = \left( {\begin{array}{*{20}{c}}<br>6\\<br>3\\<br>{ - 24}<br>\end{array}} \right)\) .</span></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">(i) Find \(\overrightarrow {{\rm{AB}}} \) </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 \(|\overrightarrow {{\rm{AC}}} |\) .</span></p>
<p> </p>
<p> </p>
<div class="marks">[5]</div>
<div class="question_part_label">d(i) and (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;">any correct equation in the form \({\boldsymbol{r}} = {\boldsymbol{a}} + t{\boldsymbol{b}}\) (accept any parameter) <em><strong>A2 N2</strong> </em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g. \({\boldsymbol{r}} = \left( {\begin{array}{*{20}{c}}<br>{ - 8}\\<br>{ - 5}\\<br>{25}<br>\end{array}} \right) + t\left( {\begin{array}{*{20}{c}}<br>2\\<br>1\\<br>{ - 8}<br>\end{array}} \right)\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"><strong>Note</strong>: Award <em><strong>A1</strong></em> for \({\boldsymbol{a}} + t{\boldsymbol{b}}\) , <em><strong>A1</strong></em> for \(L = {\boldsymbol{a}} + t{\boldsymbol{b}}\) , <em><strong>A0</strong></em> for \({\boldsymbol{r}} = {\boldsymbol{b}} + t{\boldsymbol{a}}\) .</span></p>
<p><em><strong><span style="font-family: times new roman,times; font-size: medium;"> [2 marks]</span></strong></em></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">recognizing scalar product must be zero (seen anywhere) <em><strong>R1</strong> </em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g. \({\boldsymbol{a}} \bullet {\boldsymbol{b}} = 0\) </span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">evidence of choosing direction vectors \(\left( {\begin{array}{*{20}{c}}<br>2\\<br>1\\<br>{ - 8}<br>\end{array}} \right),\left( {\begin{array}{*{20}{c}}<br>{ - 7}\\<br>{ - 2}\\<br>k<br>\end{array}} \right)\) <em><strong>(A1)(A1) </strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">correct calculation of scalar product <em><strong>(A1) </strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g. \(2( - 7) + 1( - 2) - 8k\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">simplification that clearly leads to solution <em><strong>A1</strong> </em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g. \( - 16 - 8k\) , \( - 16 - 8k = 0\)</span></p>
<p><span style="font-family: Times New Roman; font-size: medium;">\(k = - 2\) </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;"> [5 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;">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. \({L_1} = {L_3}\) , \(\left( {\begin{array}{*{20}{c}}<br>{ - 3}\\<br>{ - 1}\\<br>{ - 25}<br>\end{array}} \right) + p\left( {\begin{array}{*{20}{c}}<br>2\\<br>1\\<br>{ - 8}<br>\end{array}} \right) = \left( {\begin{array}{*{20}{c}}<br>5\\<br>0\\<br>3<br>\end{array}} \right) + q\left( {\begin{array}{*{20}{c}}<br>{ - 7}\\<br>{ - 2}\\<br>{ - 2}<br>\end{array}} \right)\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">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. \( - 3 + 2p = 5 - 7q\) , \( - 1 + p = - 2q\) , \(- 25 - 8p = 3 - 2q\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">attempting to solve 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 (\(p = - 3\) , \(q = 2\) ) <em><strong> A1</strong> </em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">the coordinates of A are \(( - 9, - 4, - 1)\) <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">c.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">(i) 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. \(\overrightarrow {{\rm{OA}}} + \overrightarrow {{\rm{AB}}} = \overrightarrow {{\rm{OB}}} \) , \(\overrightarrow {{\rm{AB}}} = \left( {\begin{array}{*{20}{c}}<br>{ - 8}\\<br>{ - 5}\\<br>{25}<br>\end{array}} \right) - \left( {\begin{array}{*{20}{c}}<br>{ - 9}\\<br>{ - 4}\\<br>{ - 1}<br>\end{array}} \right)\)</span></p>
<p><span style="font-family: Times New Roman; font-size: medium;">\(\overrightarrow {{\rm{AB}}} = \left( {\begin{array}{*{20}{c}}<br>1\\<br>{ - 1}\\<br>{26}<br>\end{array}} \right)\) </span><em><span style="font-family: times new roman,times; font-size: medium;"><strong>A1 N2</strong> </span></em></p>
<p><span style="font-family: times new roman,times; font-size: medium;">(ii) finding \(\overrightarrow {{\rm{AC}}} = \left( {\begin{array}{*{20}{c}}<br>7\\<br>2\\<br>2<br>\end{array}} \right)\) <em><strong>A1 </strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">evidence of finding magnitude <em><strong>(M1)</strong> </em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g. \(|\overrightarrow {{\rm{AC}}} | = \sqrt {{7^2} + {2^2} + {2^2}} \)</span></p>
<p><span style="font-family: Times New Roman; font-size: medium;">\(|\overrightarrow {{\rm{AC}}} | = \sqrt {57} \) </span><em><strong><span style="font-family: times new roman,times; font-size: medium;">A1 N3</span></strong></em></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">d(i) and (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;">Many candidates gave a correct vector equation for the line.</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;">A common error was to misplace </span><span style="font-family: times new roman,times; font-size: medium;">the initial position and direction vectors. Those who set the scalar product of the direction </span><span style="font-family: times new roman,times; font-size: medium;">vectors to zero typically solved for <em>k</em> successfully. Those who substituted \(k = - 2\) earned fewer </span><span style="font-family: times new roman,times; font-size: medium;">marks for working backwards in a "show that" question.</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;">Many went on to find the coordinates </span><span style="font-family: times new roman,times; font-size: medium;">of point A, however some used the same letter, say <em>p</em>, for each parameter and thus could not </span><span style="font-family: times new roman,times; font-size: medium;">solve the system.</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;">Part (d) proved challenging as many candidates did not consider that \(\overrightarrow {{\rm{AB}}} + \overrightarrow {{\rm{BC}}} = \overrightarrow {{\rm{AC}}} \) . Rather, many attempted to find the coordinates of point C, which became a more arduous and error-prone task.</span></p>
<div class="question_part_label">d(i) and (ii).</div>
</div>
<br><hr><br><div class="specification">
<p class="p1">The position vectors of points <span class="s1">P </span>and <span class="s1">Q </span>are <span class="s1"><strong><em>i</em></strong> \( + \) 2 <strong><em>j</em></strong> \( - \) <strong><em>k </em></strong></span>and <span class="s1">7<strong><em>i</em></strong> \( + \) 3<strong><em>j</em></strong> \( - \) 4<strong><em>k </em></strong></span>respectively.</p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p class="p1">Find a vector equation of the line that passes through <span class="s1">P </span><span class="s2">and </span><span class="s1">Q</span><span class="s2">.</span></p>
<div class="marks">[4]</div>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p class="p1"><span class="s1">The line through </span><span class="s2">P </span>and <span class="s2">Q </span>is perpendicular to the vector <span class="s2">2<strong><em>i </em></strong>\( + \) <em>n</em><strong><em>k</em></strong></span>. Find the value of \(n\).</p>
<div class="marks">[3]</div>
<div class="question_part_label">b.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p class="p1">valid attempt to find direction vector <span class="Apple-converted-space"> </span><span class="s1"><strong><em>(M1)</em></strong></span></p>
<p class="p2"><em>eg</em>\(\,\,\,\,\,\)\(\overrightarrow {{\text{PQ}}} ,{\text{ }}\overrightarrow {{\text{QP}}} \)</p>
<p class="p3">correct direction vector (or multiple of) <span class="s1"><strong><em>(A1)</em></strong></span></p>
<p class="p4"><span class="s1"><em>eg</em>\(\,\,\,\,\,\)</span><span class="s2">6</span><strong><em>i</em></strong> \( + \) <strong><em>j</em></strong> \( - \) <span class="s2">3</span><strong><em>k</em></strong></p>
<p class="p1"><strong>any </strong>correct equation in the form <span class="s3"><strong><em>r</em></strong> \( = \) <strong><em>a</em></strong> \( + \) <em>t</em><strong><em>b </em></strong></span>(any parameter for \(t\)) <span class="Apple-converted-space"> </span><span class="s1"><strong><em>A2 <span class="Apple-converted-space"> </span>N3</em></strong></span></p>
<p class="p4"><span class="s4">where </span><strong><em>a </em></strong><span class="s4">is </span><strong><em>i</em></strong> \( + \) <span class="s2">2</span><strong><em>j</em></strong> \( - \) <strong><em>k</em></strong> <span class="s2">or 7</span><strong><em>i</em></strong> \( + \) <span class="s2">3</span><strong><em>j</em></strong> \( - \) <span class="s2">4</span><strong><em>k </em></strong><span class="s4">, and </span><strong><em>b </em></strong><span class="s2">is a scalar multiple of 6</span><strong><em>i</em></strong> \( + \) <strong><em>j</em></strong> \( - \) <span class="s2">3</span><strong><em>k</em></strong></p>
<p class="p4"><span class="s1"><em>eg</em>\(\,\,\,\,\,\)</span><strong><em>r</em></strong> \( = \) <span class="s2">7</span><span class="s1"><strong><em>i</em></strong></span> \( + \) <span class="s2">3</span><strong><em>j</em></strong> \( - \) <span class="s2">4</span><strong><em>k</em></strong> \( + \) <em>t</em><span class="s2">(6</span><span class="s1"><strong><em>i</em></strong></span> \( + \) <strong><em>j</em></strong> \( - \) <span class="s2">3</span><strong><em>k</em></strong><span class="s2">), </span><strong><em>r</em></strong> \( = \left( {\begin{array}{*{20}{c}} {1 + 6s} \\ {2 + 1s} \\ { - 1 - 3s} \end{array}} \right),{\text{ }}r = \left( {\begin{array}{*{20}{c}} 1 \\ 2 \\ { - 1} \end{array}} \right) + t\left( {\begin{array}{*{20}{c}} { - 6} \\ { - 1} \\ 3 \end{array}} \right)\)</p>
<p class="p5"> </p>
<p class="p1"><strong>Notes: </strong>Award <span class="s1"><strong><em>A1 </em></strong></span>for the form <span class="s3"><strong><em>a</em></strong> \( + \) <em>t</em><strong><em>b</em></strong></span>, <span class="s1"><strong><em>A1 </em></strong></span>for the form <span class="s3"><strong><em>L</em></strong> \( = \) <strong><em>a</em></strong> \( + \) <em>t</em><strong><em>b</em></strong></span>, <span class="s1"><strong><em>A0 </em></strong></span>for the form <span class="s3"><strong><em>r</em></strong> \( = \) <strong><em>b</em></strong> \( + \) <em>t</em><strong><em>a</em></strong></span>.</p>
<p class="p6"> </p>
<p class="p2"><strong><em>[4 marks]</em></strong></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">correct expression for scalar product <span class="Apple-converted-space"> </span><span class="s1"><strong><em>(A1)</em></strong></span></p>
<p class="p2"><em>eg</em>\(\,\,\,\,\,\)\(6 \times 2 + 1 \times 0 + ( - 3) \times n,{\text{ }} - 3n + 12\)</p>
<p class="p1">setting scalar product equal to zero (seen anywhere) <span class="Apple-converted-space"> </span><span class="s1"><strong><em>(M1)</em></strong></span></p>
<p class="p2"><em>eg</em>\(\,\,\,\,\,\)<span class="s2"><strong><em>u</em></strong></span> \( \bullet \) <span class="s2"><strong><em>v</em></strong></span> \( = 0,{\text{ }} - 3n + 12 = 0\)</p>
<p class="p2"><span class="Apple-converted-space">\(n = 4\) </span><strong><em>A1 <span class="Apple-converted-space"> </span>N2</em></strong></p>
<p class="p2"><strong><em>[3 marks]</em></strong></p>
<div class="question_part_label">b.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">b.</div>
</div>
<br><hr><br><div class="specification">
<p><span style="font-family: times new roman,times; font-size: medium;">A line <em>L</em> passes through \({\text{A}}(1{\text{, }} - 1{\text{, }}2)\) and is parallel to the line \({\boldsymbol{r}} = \left( {\begin{array}{*{20}{c}}<br>{ - 2}\\<br>1\\<br>5<br>\end{array}} \right) + s\left( {\begin{array}{*{20}{c}}<br>1\\<br>3\\<br>{ - 2}<br>\end{array}} \right)\) .</span></p>
</div>
<div class="specification">
<p><span style="font-family: times new roman,times; font-size: medium;">The line <em>L</em> passes through point P when \(t = 2\) .</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;">Write down a vector equation for <em>L</em> in the form \({\boldsymbol{r}} = {\boldsymbol{a}} + t{\boldsymbol{b}}\) .</span></p>
<div class="marks">[2]</div>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">Find</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">(i) \(\overrightarrow {{\rm{OP}}} \) </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;">(ii) \(|\overrightarrow {{\rm{OP}}} |\) .</span></p>
<div class="marks">[4]</div>
<div class="question_part_label">b(i) and (ii).</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;">correct equation in the form \({\boldsymbol{r}} = {\boldsymbol{a}} + t{\boldsymbol{b}}\) <em><strong>A2 N2</strong></em></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\({\boldsymbol{r}} = \left( {\begin{array}{*{20}{c}}<br>1\\<br>{ - 1}\\<br>2<br>\end{array}} \right) + t\left( {\begin{array}{*{20}{c}}<br>1\\<br>3\\<br>{ - 2}<br>\end{array}} \right)\)</span></p>
<p><em><strong><span style="font-family: times new roman,times; font-size: medium;">[2 marks]</span></strong></em></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">(i) attempt to substitute \(t = 2\) into the equation <em><strong>(M1)</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g. \(\left( {\begin{array}{*{20}{c}}<br>2\\<br>6\\<br>{ - 4}<br>\end{array}} \right)\) , \(\left( {\begin{array}{*{20}{c}}<br>1\\<br>{ - 1}\\<br>2<br>\end{array}} \right) + 2\left( {\begin{array}{*{20}{c}}<br>1\\<br>3\\<br>{ - 2}<br>\end{array}} \right)\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(\overrightarrow {{\rm{OP}}} = \left( {\begin{array}{*{20}{c}}<br>3\\<br>5\\<br>{ - 2}<br>\end{array}} \right)\) <em><strong> A1 N2</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">(ii) correct substitution into formula for magnitude <em><strong>A1</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g. \(\sqrt {{3^2} + {5^2} + - {2^2}} \) , \(\sqrt {{3^2} + {5^2} + {2^2}} \)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(|\overrightarrow {{\rm{OP}}}| = \sqrt {38} \) <em><strong>A1 N1</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(i) and (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;">Many candidates answered this question well. Some continue to write the vector equation in (a) using "<em>L</em> =", which does not earn full marks. </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;">Part (b) proved accessible for most, although small arithmetic errors were not uncommon. Some candidates substituted \(t = 2\) into the original equation, and a few answered \(\overrightarrow {{\rm{OP}}} = \left( {\begin{array}{*{20}{c}}<br>2\\<br>6\\<br>{ - 4}<br>\end{array}} \right)\) . A small but surprising number of candidates left this question blank, suggesting the topic was not given adequate attention in course preparation. </span></p>
<p> </p>
<p> </p>
<div class="question_part_label">b(i) and (ii).</div>
</div>
<br><hr><br><div class="specification">
<p class="p1">Let \({L_x}\) be a family of lines with equation given by \(r\) \( = \left( {\begin{array}{*{20}{c}} x \\ {\frac{2}{x}} \end{array}} \right) + t\left( {\begin{array}{*{20}{c}} {{x^2}} \\ { - 2} \end{array}} \right)\), where \(x > 0\).</p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p class="p1">Write down the equation of \({L_1}\).</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">A line \({L_a}\) crosses the \(y\)-axis at a point \(P\).</p>
<p class="p1">Show that \(P\) has coordinates \(\left( {0,{\text{ }}\frac{4}{a}} \right)\).</p>
<div class="marks">[6]</div>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p class="p1">The line \({L_a}\) crosses the \(x\)-axis at \({\text{Q}}(2a,{\text{ }}0)\). Let \(d = {\text{P}}{{\text{Q}}^2}\).</p>
<p class="p1">Show that \(d = 4{a^2} + \frac{{16}}{{{a^2}}}\).</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">There is a minimum value for \(d\). Find the value of \(a\) that gives this minimum value.</p>
<div class="marks">[7]</div>
<div class="question_part_label">d.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p>attempt to substitute \(x = 1\) <strong><em>(M1)</em></strong></p>
<p><em>eg</em>\(\;\;\;\)<strong><em>r </em></strong>\( = \left( {\begin{array}{*{20}{c}} 1 \\ {\frac{2}{1}} \end{array}} \right) + t\left( {\begin{array}{*{20}{c}} {{1^2}} \\ { - 2} \end{array}} \right),{\text{ }}{L_1} = \left( {\begin{array}{*{20}{c}} 1 \\ 2 \end{array}} \right) + t\left( {\begin{array}{*{20}{c}} 1 \\ { - 2} \end{array}} \right)\)</p>
<p>correct equation (vector or Cartesian, but do not accept “\({L_1}\)”)</p>
<p><em>eg</em>\(\;\;\;\)<strong><em>r </em></strong>\( = \left( {\begin{array}{*{20}{c}} 1 \\ 2 \end{array}} \right) + t\left( {\begin{array}{*{20}{c}} 1 \\ { - 2} \end{array}} \right),{\text{ }}y = - 2x + 4\;\;\;\)(must be an equation) <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 class="p1">appropriate approach <span class="Apple-converted-space"> </span><strong><em>(M1)</em></strong></p>
<p class="p1"><em>eg</em>\(\;\;\;\left( {\begin{array}{*{20}{c}} 0 \\ y \end{array}} \right) = \left( {\begin{array}{*{20}{c}} a \\ {\frac{2}{a}} \end{array}} \right) + t\left( {\begin{array}{*{20}{c}} {{a^2}} \\ { - 2} \end{array}} \right)\)</p>
<p class="p1">correct equation for \(x\)-coordinate <span class="Apple-converted-space"> </span><strong><em>A1</em></strong></p>
<p class="p1"><em>eg</em>\(\;\;\;0 = a + t{a^2}\)</p>
<p class="p1">\(t = \frac{{ - 1}}{a}\) <span class="Apple-converted-space"> </span><strong><em>A1</em></strong></p>
<p class="p1">substituting <strong>their </strong>parameter to find \(y\) <span class="Apple-converted-space"> </span><strong><em>(M1)</em></strong></p>
<p class="p1"><em>eg</em>\(\;\;\;y = \frac{2}{a} - 2\left( {\frac{{ - 1}}{a}} \right),{\text{ }}\left( {\begin{array}{*{20}{c}} a \\ {\frac{2}{a}} \end{array}} \right) - \frac{1}{a}\left( {\begin{array}{*{20}{c}} {{a^2}} \\ { - 2} \end{array}} \right)\)</p>
<p class="p1">correct working <span class="Apple-converted-space"> </span><strong><em>A1</em></strong></p>
<p class="p1"><em>eg</em>\(\;\;\;y = \frac{2}{a} + \frac{2}{a},{\text{ }}\left( {\begin{array}{*{20}{c}} a \\ {\frac{2}{a}} \end{array}} \right) - \left( {\begin{array}{*{20}{c}} a \\ { - \frac{2}{a}} \end{array}} \right)\)</p>
<p class="p1">finding correct expression for \(y\) <span class="Apple-converted-space"> </span><strong><em>A1</em></strong></p>
<p class="p1"><em>eg</em>\(\;\;\;y = \frac{4}{a},{\text{ }}\left( {\begin{array}{*{20}{c}} 0 \\ {\frac{4}{a}} \end{array}} \right)\) \({\text{P}}\left( {0,{\text{ }}\frac{4}{a}} \right)\) <span class="Apple-converted-space"> </span><strong><em>AG <span class="Apple-converted-space"> </span>N0</em></strong></p>
<p class="p1"><strong><em>[6 marks]</em></strong></p>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p>valid approach <strong><em>M1</em></strong></p>
<p><em>eg</em>\(\;\;\;\)distance formula, Pythagorean Theorem, \(\overrightarrow {{\text{PQ}}} = \left( {\begin{array}{*{20}{c}} {2a} \\ { - \frac{4}{a}} \end{array}} \right)\)</p>
<p>correct simplification <strong><em>A1</em></strong></p>
<p><em>eg</em>\(\;\;\;{(2a)^2} + {\left( {\frac{4}{a}} \right)^2}\)</p>
<p>\(d = 4{a^2} + \frac{{16}}{{{a^2}}}\) <strong><em>AG N0</em></strong></p>
<p><strong><em>[2 marks]</em></strong></p>
<div class="question_part_label">c.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p>recognizing need to find derivative <strong><em>(M1)</em></strong></p>
<p><em>eg</em>\(\;\;\;d',{\text{ }}d'(a)\)</p>
<p>correct derivative <strong><em>A2</em></strong></p>
<p><em>eg</em>\(\;\;\;8a - \frac{{32}}{{{a^3}}},{\text{ }}8x - \frac{{32}}{{{x^3}}}\)</p>
<p>setting <strong>their </strong>derivative equal to \(0\) <strong><em>(M1)</em></strong></p>
<p><em>eg</em>\(\;\;\;8a - \frac{{32}}{{{a^3}}} = 0\)</p>
<p>correct working <strong><em>(A1)</em></strong></p>
<p><em>eg</em>\(\;\;\;8a = \frac{{32}}{{{a^3}}},{\text{ }}8{a^4} - 32 = 0\)</p>
<p>working towards solution <strong><em>(A1)</em></strong></p>
<p><em>eg</em>\(\;\;\;{a^4} = 4,{\text{ }}{a^2} = 2,{\text{ }}a = \pm \sqrt 2 \)</p>
<p>\(a = \sqrt[4]{4}\;\;\;(a = \sqrt 2 )\;\;\;({\text{do not accept }} \pm \sqrt 2 )\)<strong><em> A1 N3</em></strong></p>
<p><strong><em>[7 marks]</em></strong></p>
<p><strong><em>Total [17 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">In part (a), most candidates correctly substituted 1 for \(x\), although many of them did not earn full marks for their work here, as they wrote their vector equation using \({L_1} = \), not understanding that \({L_1}\) is the name of the line, and not a vector.</p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">Very few candidates answered parts (b) and (c) correctly, often working backwards from the given answer, which is not appropriate in "show that" questions. In these types of questions, candidates are required to clearly show their working and reasoning, which will hopefully lead them to the given answer.</p>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">Very few candidates answered parts (b) and (c) correctly, often working backwards from the given answer, which is not appropriate in "show that" questions. In these types of questions, candidates are required to clearly show their working and reasoning, which will hopefully lead them to the given answer.</p>
<div class="question_part_label">c.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">Fortunately, a good number of candidates recognized the need to find the derivative of the given expression for \(d\) in part (d) of the question, and so were able to earn at least some of the available marks in the final part.</p>
<div class="question_part_label">d.</div>
</div>
<br><hr><br><div class="question">
<p class="p1">The following diagram shows triangle \(ABC\).</p>
<p class="p1" style="text-align: center;"><img src="images/Schermafbeelding_2015-12-14_om_04.52.14.png" alt></p>
<p>Let \(\overrightarrow {{\text{AB}}} \bullet \overrightarrow {{\text{AC}}} = - 5\sqrt 3 \) and \(\left| {\overrightarrow {{\text{AB}}} } \right|\left| {\overrightarrow {{\text{AC}}} } \right| = 10\). Find the area of triangle \(ABC\).</p>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question">
<p>attempt to find \(\cos {\rm{C\hat AB}}\) (seen anywhere) <strong><em>(M1)</em></strong></p>
<p><em>eg</em>\(\;\;\;\cos \theta = \frac{{\overrightarrow {{\text{AB}}} \bullet \overrightarrow {{\text{AC}}} }}{{\left| {\overrightarrow {{\text{AB}}} } \right|\left| {\overrightarrow {{\text{AC}}} } \right|}}\)</p>
<p>\(\cos {\rm{C\hat AB}} = \frac{{ - 5\sqrt 3 }}{{10}}\;\;\;\left( { = - \frac{{\sqrt 3 }}{2}} \right)\) <strong><em>A1</em></strong></p>
<p>valid attempt to find \(\sin {\rm{C\hat AB}}\) <strong><em>(M1)</em></strong></p>
<p><em>eg</em>\(\;\;\;\)triangle, Pythagorean identity, \({\rm{C\hat AB}} = \frac{{5\pi }}{6},{\text{ }}150^\circ \)</p>
<p>\(\sin {\rm{C\hat AB}} = \frac{1}{2}\) <strong><em>(A1)</em></strong></p>
<p>correct substitution into formula for area <strong><em>(A1)</em></strong></p>
<p><em>eg</em>\(\;\;\;\frac{1}{2} \times 10 \times \frac{1}{2},{\text{ }}\frac{1}{2} \times 10 \times \sin \frac{\pi }{6}\)</p>
<p>\({\text{area}} = \frac{{10}}{4}\;\;\;\left( { = \frac{5}{2}} \right)\) <strong><em>A1 N3</em></strong></p>
<p><strong><em>[6 marks]</em></strong></p>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question">
<p class="p1">The large majority of candidates were able to find the correct expression for \(\cos {\rm{C\hat AB}}\), but few recognized that an angle with a negative cosine will be obtuse, rather than acute, and many stated that \({\rm{C\hat AB}} = 30^\circ \). When substituting into the triangle area formula, a common error was to substitute \(5\sqrt 3 \) rather than 10, as many did not understand the relationship between the magnitude of a vector and the length of a line segment in the triangle formula.</p>
<p class="p1">Some of the G2 comments from schools suggested that it might have been easier for their students if this question were split into two parts. While we do tend to provide more support on the earlier questions in the paper, questions 6 and 7 are usually presented with little or no scaffolding. On these later questions, the candidates are often required to use knowledge from different areas of the syllabus within a single question.</p>
</div>
<br><hr><br><div class="question" style="padding-left: 20px; padding-right: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">Let <em><strong>u </strong></em>\( = \left( {\begin{array}{*{20}{c}}<br>2\\<br>3\\<br>{ - 1}<br>\end{array}} \right)\) and <em><strong>w</strong></em> \( = \left( {\begin{array}{*{20}{c}}<br>3\\<br>{ - 1}\\<br>p<br>\end{array}} \right)\) . Given that <strong><em>u</em></strong> is perpendicular to <strong><em>w</em></strong> , find the value of <em>p</em> .</span></p>
<div class="marks">[3]</div>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">Let \({\boldsymbol{v}} = \left( {\begin{array}{*{20}{c}}<br> 1 \\ <br> q \\ <br> 5 <br>\end{array}} \right)\)</span><span style="font-family: times new roman,times; font-size: medium;"> . Given that \(\left| {\boldsymbol{v}} \right| = \sqrt {42} \), find the possible values of \(q\) .</span></p>
<div class="marks">[3]</div>
<div class="question_part_label">b.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">evidence of equating scalar product to 0 <em><strong>(M1)</strong> </em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(2 \times 3 + 3 \times ( - 1) + ( - 1) \times p = 0\) (\(6 - 3 - p = 0\), \(3 - p = 0\)) <em><strong>A1</strong> </em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(p = 3\) <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">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">evidence of substituting into magnitude formula <em><strong>(M1)</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"><em>e.g.</em> \(\sqrt {1 + {q^2} + 25} \) , \(1 + {q^2} + 25\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">setting up a correct equation <em><strong>A1</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"><em>e.g.</em> \(\sqrt {1 + {q^2} + 25} = \sqrt {42} \) , \(1 + {q^2} + 25 = 42\) , \({q^2} = 16\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(q = \pm 4\) <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">b.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">Most candidates knew to set the scalar product equal to zero. </span></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">Most candidates knew to set the scalar product equal to zero. A pleasing number found both answers for \(q\), although some often neglected to provide both solutions.</span></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;">Find the cosine of the angle between the two vectors \(3{\boldsymbol{i}} + 4{\boldsymbol{j}} + 5{\boldsymbol{k}}\) and \(4{\boldsymbol{i}} - 5{\boldsymbol{j}} - 3{\boldsymbol{k}}\) .</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;">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 \( = 12 - 20 - 15\) (\( = - 23\)) </span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">magnitudes \( = \sqrt {{3^2} + {4^2} + {5^2}} \) , \( = \sqrt {{4^2} + {{( - 5)}^2} + {{( - 3)}^2}} \) , \(\left( {\sqrt {50} ,\sqrt {50} } \right)\)</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;">e.g. \(\cos \theta = \frac{{12 - 20 - 15}}{{\left( {\sqrt {{3^2} + {4^2} + {5^2}} } \right) \times \left( {\sqrt {{4^2} + {{( - 5)}^2} + {{( - 3)}^2}} } \right)}}\) </span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(\cos \theta = - \frac{{23}}{{50}}\) \(( = - 0.46)\) <em><strong>A2 N4</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><span style="font-family: times new roman,times; font-size: medium;">Many candidates performed well in finding the magnitudes and scalar product to use the formula for angle between vectors. Some experienced trouble with the arithmetic to obtain the required result. A significant number of candidates isolated the theta answering with \({\text{arccos}}\left( {\frac{{ - 23}}{{50}}} \right)\) .</span></p>
</div>
<br><hr><br><div class="question">
<p>Six equilateral triangles, each with side length 3 cm, are arranged to form a hexagon.<br>This is shown in the following diagram.</p>
<p><img src="data:image/png;base64,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"></p>
<p>The vectors <em><strong>p</strong></em> , <em><strong>q</strong></em> and <em><strong>r</strong></em> are shown on the diagram.</p>
<p>Find <em><strong>p</strong></em>•(<em><strong>p</strong></em> + <em><strong>q</strong></em> + <em><strong>r</strong></em>).</p>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question">
<p><strong>METHOD 1 </strong>(using |<em><strong>p</strong></em>| |2<em><strong>q</strong></em>| cos<em>θ</em>)</p>
<p>finding <em><strong>p</strong></em> + <em><strong>q</strong></em> + <em><strong>r (A1)</strong></em></p>
<p><em>eg </em> 2<em><strong>q</strong></em>, <img src="data:image/png;base64,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"></p>
<p>| <em><strong>p</strong></em> + <em><strong>q</strong></em> + <em><strong>r </strong></em>| = 2 × 3 (= 6) (seen anywhere) <em><strong>A1</strong></em></p>
<p>correct angle between <em><strong>p</strong></em> and <em><strong>q</strong></em> (seen anywhere) <em><strong>(A1)</strong></em></p>
<p>\(\frac{\pi }{3}\) (accept 60°)</p>
<p>substitution of <strong>their</strong> values <em><strong>(M1)</strong></em></p>
<p><em>eg</em> 3 × 6 × cos\(\left( {\frac{\pi }{3}} \right)\)</p>
<p>correct value for cos\(\left( {\frac{\pi }{3}} \right)\) (seen anywhere) <em><strong>(A1)</strong></em></p>
<p><em>eg</em> \(\frac{1}{2},\,\,\,3 \times 6 \times \frac{1}{2}\)</p>
<p><em><strong>p</strong></em>•(<em><strong>p</strong></em> + <em><strong>q</strong></em> + <em><strong>r</strong></em>) = 9 <em><strong> A1 N3</strong></em></p>
<p> </p>
<p><strong>METHOD 2</strong> (scalar product using distributive law)</p>
<p>correct expression for scalar distribution <em><strong>(A1)</strong></em></p>
<p>eg <em><strong>p</strong></em>• <em><strong>p</strong></em> + <em><strong>p</strong></em>•<em><strong>q</strong></em> + <em><strong>p</strong></em>•<em><strong>r</strong></em></p>
<p>three correct angles between the vector pairs (seen anywhere) <em><strong>(A2)</strong></em></p>
<p><em>eg </em> 0° between <em><strong>p</strong></em> and <em><strong>p</strong></em>, \(\frac{\pi }{3}\) between <em><strong>p</strong></em> and <em><strong>q</strong></em>, \(\frac{{2\pi }}{3}\) between <em><strong>p</strong></em> and <em><strong>r</strong></em></p>
<p><strong>Note:</strong> Award <em><strong>A1</strong> </em>for only two correct angles.</p>
<p>substitution of <strong>their</strong> values <em><strong>(M1)</strong></em></p>
<p><em>eg </em> 3.3.cos0 +3.3.cos\(\frac{\pi }{3}\) + 3.3.cos120</p>
<p>one correct value for cos0, cos\(\left( {\frac{\pi }{3}} \right)\) or cos\(\left( {\frac{2\pi }{3}} \right)\) (seen anywhere) <em><strong>A1</strong></em></p>
<p><em>eg </em>\(\frac{1}{2},\,\,\,3 \times 6 \times \frac{1}{2}\)</p>
<p><em><strong>p</strong></em>•(<em><strong>p</strong></em> + <em><strong>q</strong></em> + <em><strong>r</strong></em>) = 9 <em><strong> A1 N3</strong></em></p>
<p> </p>
<p><strong>METHOD 3</strong> (scalar product using relative position vectors)</p>
<p>valid attempt to find one component of <em><strong>p</strong></em> or <em><strong>r</strong></em> <em><strong>(M1)</strong></em></p>
<p><em>eg </em> sin 60 = \(\frac{x}{3}\), cos 60 = \(\frac{x}{3}\), one correct value \(\frac{3}{2},\,\,\frac{{3\sqrt 3 }}{2},\,\,\frac{{ - 3\sqrt 3 }}{2}\)</p>
<p>one correct vector (two or three dimensions) (seen anywhere) <em><strong>A1</strong></em></p>
<p><em>eg </em>\(p = \left( \begin{gathered}<br> \,\,\,\frac{3}{2} \hfill \\<br> \frac{{3\sqrt 3 }}{2} \hfill \\ <br>\end{gathered} \right),\,\,q = \left( \begin{gathered}<br> 3 \hfill \\<br> 0 \hfill \\ <br>\end{gathered} \right),\,\,r = \left( \begin{gathered}<br> \,\,\,\,\frac{3}{2} \hfill \\<br> - \frac{{3\sqrt 3 }}{2} \hfill \\<br> \,\,\,\,0 \hfill \\ <br>\end{gathered} \right)\)</p>
<p>three correct vectors <em><strong>p</strong></em> + <em><strong>q</strong></em> + <em><strong>r </strong></em>= 2<em><strong>q</strong></em> <em><strong>(A1)</strong></em></p>
<p><em><strong>p</strong></em> + <em><strong>q</strong></em> + <em><strong>r </strong></em>= \(\left( \begin{gathered}<br> 6 \hfill \\<br> 0 \hfill \\ <br>\end{gathered} \right)\) or \(\left( \begin{gathered}<br> 6 \hfill \\<br> 0 \hfill \\<br> 0 \hfill \\ <br>\end{gathered} \right)\) (seen anywhere, including scalar product) <em><strong>(A1)</strong></em></p>
<p>correct working <em><strong>(A1)</strong></em><br><em>eg </em>\(\left( {\frac{3}{2} \times 6} \right) + \left( {\frac{{3\sqrt 3 }}{2} \times 0} \right),\,\,9 + 0 + 0\)</p>
<p><em><strong>p</strong></em>•(<em><strong>p</strong></em> + <em><strong>q</strong></em> + <em><strong>r</strong></em>) = 9 <em><strong> A1 N3</strong></em></p>
<p><strong><em>[6 marks]</em></strong></p>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question">
[N/A]
</div>
<br><hr><br><div class="specification">
<p 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;">In the following diagram, \(\overrightarrow {{\text{OP}}} \) = <strong><em>p</em></strong>, \(\overrightarrow {{\text{OQ}}} \) = <strong><em>q</em></strong> and \(\overrightarrow {{\text{PT}}} = \frac{1}{2}\overrightarrow {{\text{PQ}}} \).</span></p>
<p style="font: normal normal normal 21px/normal 'Times New Roman'; min-height: 25px; text-align: center; margin: 0px;"><img src="images/maths_1.png" alt></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;">Express each of the following vectors in terms of <strong><em>p </em></strong>and <strong><em>q</em></strong>,</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;">\(\overrightarrow {{\text{QP}}} \);</span></p>
<div class="marks">[2]</div>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">\(\overrightarrow {{\text{OT}}} \).</span></p>
<div class="marks">[3]</div>
<div class="question_part_label">b.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 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> \(\overrightarrow {{\text{QP}}} = \overrightarrow {{\text{QO}}} + \overrightarrow {{\text{OP}}} ,{\text{ P}} - {\text{Q}}\)</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;">\(\overrightarrow {{\text{QP}}} \) = <strong><em>p</em></strong> – <strong><em>q</em></strong> <strong><em>A1 N2</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>[2 marks]</em></strong></span></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 20.5px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">recognizing correct vector for \(\overrightarrow {{\text{QT}}} \) or \(\overrightarrow {{\text{PT}}} \) <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> \(\overrightarrow {{\text{QT}}} = \frac{1}{2}\)(<strong><em>p</em></strong> – <strong><em>q</em></strong>), \(\overrightarrow {{\text{PT}}} = \frac{1}{2}\)(<strong><em>q</em></strong> – <strong><em>p</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;">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> \(\overrightarrow {{\text{OT}}} = \overrightarrow {{\text{OP}}} + \overrightarrow {{\text{PT}}} ,{\text{ }}\overrightarrow {{\text{OQ}}} + \overrightarrow {{\text{QT}}} ,{\text{ }}\overrightarrow {{\text{OP}}} + \frac{1}{2}\overrightarrow {{\text{PQ}}} \)</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;">\(\overrightarrow {{\text{OT}}} = \frac{1}{2}\)(<strong><em>p</em></strong> + <strong><em>q</em></strong>) \(\left( {{\text{accept }}\frac{{p + q}}{2}} \right)\) <strong><em>A1 N2</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>[3 marks]</em></strong></span></p>
<div class="question_part_label">b.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
[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>Point A has coordinates (−4, −12, 1) and point B has coordinates (2, −4, −4).</p>
</div>
<div class="specification">
<p>The line <em>L</em> passes through A and B.</p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Show that \(\mathop {{\text{AB}}}\limits^ \to = \left( \begin{gathered}<br> \,6 \hfill \\<br> \,8 \hfill \\<br> - 5 \hfill \\ <br>\end{gathered} \right)\)</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>Find a vector equation for <em>L</em>.</p>
<div class="marks">[2]</div>
<div class="question_part_label">b.i.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Point <em>C</em> (<em>k</em> , 12 , −<em>k</em>) is on <em>L</em>. Show that <em>k</em> = 14.</p>
<div class="marks">[4]</div>
<div class="question_part_label">b.ii.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Find \(\mathop {{\text{OB}}}\limits^ \to \, \bullet \mathop {{\text{AB}}}\limits^ \to \).</p>
<div class="marks">[2]</div>
<div class="question_part_label">c.i.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Write down the value of angle OBA.</p>
<div class="marks">[1]</div>
<div class="question_part_label">c.ii.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Point D is also on <em>L</em> and has coordinates (8, 4, −9).</p>
<p>Find the area of triangle OCD.</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>correct approach <em><strong>A1</strong></em></p>
<p><em>eg</em> \(\mathop {{\text{AO}}}\limits^ \to \,\, + \,\,\mathop {{\text{OB}}}\limits^ \to ,\,\,\,{\text{B}} - {\text{A}}\,{\text{, }}\,\left( \begin{gathered}<br> \,\,2 \hfill \\<br> - 4 \hfill \\<br> - 4 \hfill \\ <br>\end{gathered} \right) - \left( \begin{gathered}<br> \, - 4 \hfill \\<br> - 12 \hfill \\<br> \,\,\,1 \hfill \\ <br>\end{gathered} \right)\)</p>
<p>\(\mathop {{\text{AB}}}\limits^ \to = \left( \begin{gathered}<br>\,6 \hfill \\<br>\,8 \hfill \\<br>- 5 \hfill \\ <br>\end{gathered} \right)\) <em><strong>AG N0</strong></em></p>
<p><em><strong>[1 mark]</strong></em></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p>any correct equation in the form <em><strong>r</strong></em> = <em><strong>a</strong></em> + <em>t<strong>b</strong></em> (any parameter for <em>t</em>) <em><strong>A2 N2</strong></em></p>
<p>where <strong><em>a</em></strong> is \(\left( \begin{gathered}<br> \,\,2 \hfill \\<br> - 4 \hfill \\<br> - 4 \hfill \\ <br>\end{gathered} \right)\) or \(\left( \begin{gathered}<br> \, - 4 \hfill \\<br> - 12 \hfill \\<br> \,\,\,1 \hfill \\ <br>\end{gathered} \right)\) and <em><strong>b</strong></em> is a scalar multiple of \(\left( \begin{gathered}<br> \,6 \hfill \\<br> \,8 \hfill \\<br> - 5 \hfill \\ <br>\end{gathered} \right)\)</p>
<p><em>eg</em> <em><strong>r</strong></em> \( = \left( \begin{gathered}<br> \, - 4 \hfill \\<br> - 12 \hfill \\<br> \,\,\,1 \hfill \\ <br>\end{gathered} \right) + t\left( \begin{gathered}<br> \,6 \hfill \\<br> \,8 \hfill \\<br> - 5 \hfill \\ <br>\end{gathered} \right),\,\,\left( {x,\,\,y,\,\,z} \right) = \left( {2,\,\, - 4,\,\, - 4} \right) + t\left( {6,\,\,8,\,\, - 5} \right),\) <em><strong>r </strong></em>\( = \left( \begin{gathered}<br> \, - 4 + 6t \hfill \\<br> - 12 + 8t \hfill \\<br> \,\,\,1 - 5t \hfill \\ <br>\end{gathered} \right)\)</p>
<p><strong>Note:</strong> Award <em><strong>A1</strong></em> for the form <em><strong>a</strong></em> + <em>t<strong>b</strong></em>, <em><strong>A1</strong></em> for the form <em><strong>L</strong></em> = <em><strong>a</strong></em> + <em>t<strong>b</strong></em>, <em><strong>A0</strong></em> for the form <em><strong>r</strong></em> = <em><strong>b</strong></em> + <em>t<strong>a</strong></em>.</p>
<p><em><strong>[2 marks]</strong></em></p>
<div class="question_part_label">b.i.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><strong>METHOD 1</strong> (solving for <em>t</em>)</p>
<p>valid approach <em><strong>(M1)</strong></em></p>
<p><em>eg</em> \(\left( \begin{gathered}<br> \,k \hfill \\<br> 12 \hfill \\<br> - k \hfill \\ <br>\end{gathered} \right) = \left( \begin{gathered}<br> \,\,2 \hfill \\<br> - 4 \hfill \\<br> - 4 \hfill \\ <br>\end{gathered} \right) + t\left( \begin{gathered}<br> \,6 \hfill \\<br> \,8 \hfill \\<br> - 5 \hfill \\ <br>\end{gathered} \right),\,\,\left( \begin{gathered}<br> \,k \hfill \\<br> 12 \hfill \\<br> - k \hfill \\ <br>\end{gathered} \right) = \left( \begin{gathered}<br> \, - 4 \hfill \\<br> - 12 \hfill \\<br> \,\,\,1 \hfill \\ <br>\end{gathered} \right) + t\left( \begin{gathered}<br> \,6 \hfill \\<br> \,8 \hfill \\<br> - 5 \hfill \\ <br>\end{gathered} \right)\)</p>
<p>one correct equation <em><strong>A1</strong></em></p>
<p>eg −4 + 8<em>t</em> = 12, −12 + 8<em>t</em> = 12</p>
<p>correct value for <em>t <strong>(A1)</strong></em></p>
<p><em>eg t</em> = 2 or 3</p>
<p>correct substitution <em><strong>A1</strong></em></p>
<p><em>eg </em> 2 + 6(2), −4 + 6(3), −[1 + 3(−5)]</p>
<p><em>k</em> = 14 <em><strong>AG N0</strong></em></p>
<p> </p>
<p><strong>METHOD 2</strong> (solving simultaneously)</p>
<p>valid approach <em><strong>(M1)</strong></em></p>
<p><em>eg </em>\(\left( \begin{gathered}<br>\,k \hfill \\<br>12 \hfill \\<br>- k \hfill \\ <br>\end{gathered} \right) = \left( \begin{gathered}<br>\,\,2 \hfill \\<br>- 4 \hfill \\<br>- 4 \hfill \\ <br>\end{gathered} \right) + t\left( \begin{gathered}<br>\,6 \hfill \\<br>\,8 \hfill \\<br>- 5 \hfill \\ <br>\end{gathered} \right),\,\,\left( \begin{gathered}<br>\,k \hfill \\<br>12 \hfill \\<br>- k \hfill \\ <br>\end{gathered} \right) = \left( \begin{gathered}<br>\, - 4 \hfill \\<br>- 12 \hfill \\<br>\,\,\,1 \hfill \\ <br>\end{gathered} \right) + t\left( \begin{gathered}<br>\,6 \hfill \\<br>\,8 \hfill \\<br>- 5 \hfill \\ <br>\end{gathered} \right)\)</p>
<p>two correct equations in <em><strong>A1</strong></em></p>
<p><em>eg k</em> = −4 + 6<em>t,</em> −<em>k</em> = 1 −5<em>t</em></p>
<p><strong>EITHER</strong> (eliminating <em>k</em>)</p>
<p>correct value for <em>t</em> <em><strong>(A1)</strong></em></p>
<p><em>eg t</em> = 2 or 3</p>
<p>correct substitution <em><strong>A1</strong></em></p>
<p><em>eg </em> 2 + 6(2), −4 + 6(3)</p>
<p><strong>OR</strong> (eliminating <em>t</em>)</p>
<p>correct equation(s) <em><strong>(A1)</strong></em></p>
<p><em>eg </em> 5<em>k</em> + 20 = 30<em>t</em> <strong>and </strong>−6<em>k</em> − 6 = 30<em>t</em>, −<em>k</em> = 1 − 5\(\left( {\frac{{k + 4}}{6}} \right)\)</p>
<p>correct working clearly leading to <em>k</em> = 14 <em><strong>A1</strong></em></p>
<p><em>eg </em>−<em>k</em> + 14 = 0, −6<em>k</em> = 6 −5<em>k</em> − 20, 5<em>k</em> = −20 + 6(1 + <em>k</em>)</p>
<p><strong>THEN </strong></p>
<p><em>k</em> = 14 <em><strong>AG N0</strong></em></p>
<p><em><strong>[4 marks]</strong></em></p>
<p> </p>
<div class="question_part_label">b.ii.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p>correct substitution into scalar product <em><strong>A1</strong></em></p>
<p><em>eg </em>(2)(6) − (4)(8) − (4)(−5), 12 − 32 + 20</p>
<p>\(\mathop {{\text{OB}}}\limits^ \to \, \bullet \mathop {{\text{AB}}}\limits^ \to \) = 0 <em><strong>A1 N0</strong></em></p>
<p><em><strong>[2 marks]</strong></em></p>
<p> </p>
<div class="question_part_label">c.i.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p>\({\text{O}}\mathop {\text{B}}\limits^ \wedge {\text{A}} = \frac{\pi }{2},\,\,90^\circ \,\,\,\,\,\left( {{\text{accept}}\,\frac{{3\pi }}{2},\,\,270^\circ } \right)\,\) <strong><em>A1 N1</em></strong></p>
<p><strong><em>[1 marks]</em></strong></p>
<div class="question_part_label">c.ii.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><strong>METHOD 1</strong> (\(\frac{1}{2}\) × height × CD)</p>
<p>recognizing that OB is altitude of triangle with base CD (seen anywhere) <em><strong> M1</strong></em></p>
<p><em>eg </em>\(\frac{1}{2} \times \left| {\mathop {{\text{OB}}}\limits^ \to } \right| \times \left| {\mathop {{\text{CD}}}\limits^ \to } \right|,\,\,{\text{OB}} \bot {\text{CD}},\) sketch showing right angle at B</p>
<p><img 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"></p>
<p>\(\mathop {{\text{CD}}}\limits^ \to = \left( \begin{gathered}<br> - 6 \hfill \\<br> - 8 \hfill \\<br> \,5 \hfill \\ <br>\end{gathered} \right)\) or \(\mathop {{\text{DC}}}\limits^ \to = \left( \begin{gathered}<br> \,6 \hfill \\<br> \,8 \hfill \\<br> - 5 \hfill \\ <br>\end{gathered} \right)\) (seen anywhere) <em><strong>(A1)</strong></em></p>
<p>correct magnitudes (seen anywhere) <em><strong>(A1)(A1)</strong></em></p>
<p>\(\left| {\mathop {{\text{OB}}}\limits^ \to } \right| = \sqrt {{{\left( 2 \right)}^2} + {{\left( { - 4} \right)}^2} + {{\left( { - 4} \right)}^2}} = \left( {\sqrt {36} } \right)\)</p>
<p>\(\left| {\mathop {{\text{CD}}}\limits^ \to } \right| = \sqrt {{{\left( { - 6} \right)}^2} + {{\left( { - 8} \right)}^2} + {{\left( 5 \right)}^2}} = \left( {\sqrt {125} } \right)\)</p>
<p>correct substitution into \(\frac{1}{2}bh\) <em><strong>A1</strong></em></p>
<p><em>eg</em> \(\frac{1}{2} \times 6 \times \sqrt {125} \) </p>
<p>area \( = 3\sqrt {125} ,\,\,15\sqrt 5 \) <em><strong>A1 N3</strong></em></p>
<p> </p>
<p><strong>METHOD 2</strong> (subtracting triangles)</p>
<p>recognizing that OB is altitude of either ΔOBD or ΔOBC(seen anywhere) <em><strong>M1</strong></em></p>
<p>eg \(\frac{1}{2} \times \left| {\mathop {{\text{OB}}}\limits^ \to } \right| \times \left| {\mathop {{\text{BD}}}\limits^ \to } \right|,\,\,{\text{OB}} \bot {\text{BC}},\) sketch of triangle showing right angle at B</p>
<p><img 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"></p>
<p>one correct vector \(\mathop {{\text{BD}}}\limits^ \to \) or \(\mathop {{\text{DB}}}\limits^ \to \) or \(\mathop {{\text{BC}}}\limits^ \to \) or \(\mathop {{\text{CB}}}\limits^ \to \) (seen anywhere) <em><strong>(A1)</strong></em></p>
<p>eg \(\mathop {{\text{BD}}}\limits^ \to = \left( \begin{gathered}<br> \,6 \hfill \\<br> \,8 \hfill \\<br> - 5 \hfill \\ <br>\end{gathered} \right)\), \(\mathop {{\text{CB}}}\limits^ \to = \left( \begin{gathered}<br> - 12 \hfill \\<br> - 16 \hfill \\<br> \,10 \hfill \\ <br>\end{gathered} \right)\)</p>
<p>\(\left| {\mathop {{\text{OB}}}\limits^ \to } \right| = \sqrt {{{\left( 2 \right)}^2} + {{\left( { - 4} \right)}^2} + {{\left( { - 4} \right)}^2}} = \left( {\sqrt {36} } \right)\) (seen anywhere) <em><strong>(A1)</strong></em></p>
<p>one correct magnitude of a base (seen anywhere)<em><strong> (A1)</strong></em></p>
<p>\(\left| {\mathop {{\text{BD}}}\limits^ \to } \right| = \sqrt {{{\left( 6 \right)}^2} + {{\left( 8 \right)}^2} + {{\left( 5 \right)}^2}} = \left( {\sqrt {125} } \right),\,\,\left| {\mathop {{\text{BC}}}\limits^ \to } \right| = \sqrt {144 + 256 + 100} = \left( {\sqrt {500} } \right)\)</p>
<p>correct working <strong><em>A1</em></strong></p>
<p><em>eg </em>\(\frac{1}{2} \times 6 \times \sqrt {500} - \frac{1}{2} \times 6 \times 5\sqrt 5 ,\,\,\frac{1}{2} \times 6 \times \sqrt {500} \times {\text{sin}}90 - \frac{1}{2} \times 6 \times 5\sqrt 5 \times {\text{sin}}90\)</p>
<p>area \( = 3\sqrt {125} ,\,\,15\sqrt 5 \) <em><strong>A1 N3</strong></em></p>
<p> </p>
<p><strong>METHOD 3</strong> (using \(\frac{1}{2}\)<em>ab</em> sin <em>C</em> with ΔOCD)</p>
<p>two correct side lengths (seen anywhere) <em><strong>(A1)(A1)</strong></em></p>
<p>\(\left| {\mathop {{\text{OD}}}\limits^ \to } \right| = \sqrt {{{\left( 8 \right)}^2} + {{\left( 4 \right)}^2} + {{\left( { - 9} \right)}^2}} = \left( {\sqrt {161} } \right),\,\,\left| {\mathop {{\text{CD}}}\limits^ \to } \right| = \sqrt {{{\left( { - 6} \right)}^2} + {{\left( { - 8} \right)}^2} + {{\left( 5 \right)}^2}} = \left( {\sqrt {125} } \right),\,\) \(\left| {\mathop {{\text{OC}}}\limits^ \to } \right| = \sqrt {{{\left( {14} \right)}^2} + {{\left( {12} \right)}^2} + {{\left( { - 14} \right)}^2}} = \left( {\sqrt {536} } \right)\)</p>
<p>attempt to find cosine ratio (seen anywhere) <em><strong>M1</strong></em><br><em>eg</em> \(\frac{{536 - 286}}{{ - 2\sqrt {161} \sqrt {125} }},\,\,\frac{{{\text{OD}} \bullet {\text{DC}}}}{{\left| {OD} \right|\left| {DC} \right|}}\)</p>
<p>correct working for sine ratio <em><strong>A1</strong></em></p>
<p><em>eg </em>\(\frac{{{{\left( {125} \right)}^2}}}{{161 \times 125}} + {\text{si}}{{\text{n}}^2}\,D = 1\)</p>
<p>correct substitution into \(\frac{1}{2}ab\,\,{\text{sin}}\,C\) <em><strong>A1</strong></em></p>
<p><em>eg</em> \(0.5 \times \sqrt {161} \times \sqrt {125} \times \frac{6}{{\sqrt {161} }}\)</p>
<p>area \( = 3\sqrt {125} ,\,\,15\sqrt 5 \) <em><strong>A1 N3</strong></em></p>
<p><em><strong>[6 marks]</strong></em></p>
<div class="question_part_label">d.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">b.i.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">b.ii.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">c.i.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">c.ii.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">d.</div>
</div>
<br><hr><br>