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Question
A line \(L\) passes through points \({\text{A}}( – 2,{\text{ }}4,{\text{ }}3)\) and \({\text{B}}( – 1,{\text{ }}3,{\text{ }}1)\).
(i) Show that \(\overrightarrow {{\text{AB}}} = \left( {\begin{array}{*{20}{c}} 1 \\ { – 1} \\ { – 2} \end{array}} \right)\).
(ii) Find \(\left| {\overrightarrow {{\text{AB}}} } \right|\).
Find a vector equation for \(L\).
The following diagram shows the line \(L\) and the origin \(O\). The point \(C\) also lies on \(L\).
Point \(C\) has position vector \(\left( {\begin{array}{*{20}{c}} 0 \\ y \\ { – 1} \end{array}} \right)\).
Show that \(y = 2\).
(i) Find \(\overrightarrow {{\text{OC}}} \bullet \overrightarrow {{\text{AB}}} \).
(ii) Hence, write down the size of the angle between \(C\) and \(L\).
Hence or otherwise, find the area of triangle \(OAB\).
Answer/Explanation
Markscheme
(i) correct approach A1
eg\(\;\;\;{\text{B}} – {\text{A, AO}} + {\text{OB}}\)
\(\overrightarrow {{\text{AB}}} = \left( {\begin{array}{*{20}{c}} 1 \\ { – 1} \\ { – 2} \end{array}} \right)\) AG N0
(ii) correct substitution (A1)
eg\(\;\;\;\sqrt {{{(1)}^2} + {{( – 1)}^2} + {{( – 2)}^2}} ,{\text{ }}\sqrt {1 + 1 + 4} \)
\(\left| {\overrightarrow {{\text{AB}}} } \right| = \sqrt 6 \) A1 N2
[3 marks]
any correct equation in the form \(r = a + tb\) (any parameter for \(t\))
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 \(\left( {\begin{array}{*{20}{c}} 1 \\ { – 1} \\ { – 2} \end{array}} \right)\) A2 N2
eg\(\;\;\;\(r\) = \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)\)
Note: Award A1 for the form \({\mathbf{a}} + t{\mathbf{b}}\), A1 for the form \(L = {\mathbf{a}} + t{\mathbf{b}}\), A0 for the form \({\mathbf{r}} = {\mathbf{b}} + t{\mathbf{a}}\).
METHOD 1
valid approach (M1)
eg\(\;\;\;\left( {\begin{array}{*{20}{c}} { – 1} \\ 3 \\ 1 \end{array}} \right) + t\left( {\begin{array}{*{20}{c}} 1 \\ { – 1} \\ { – 2} \end{array}} \right) = \left( {\begin{array}{*{20}{c}} 0 \\ y \\ { – 1} \end{array}} \right),{\text{ }}\left( {\begin{array}{*{20}{c}} 0 \\ y \\ { – 1} \end{array}} \right) = \left( {\begin{array}{*{20}{c}} { – 2} \\ 4 \\ 3 \end{array}} \right) + s\left( {\begin{array}{*{20}{c}} 1 \\ { – 1} \\ { – 2} \end{array}} \right)\)
one correct equation from their approach A1
eg\(\;\;\; – 1 + t = 0,{\text{ }}1 – 2t = – 1,{\text{ }} – 2 + s = 0,{\text{ }}3 – 2s = – 1\)
one correct value for their parameter and equation A1
eg\(\;\;\;t = 1,{\text{ }}s = 2\)
correct substitution A1
eg\(\;\;\;3 + 1( – 1),{\text{ }}4 + 2( – 1)\)
\(y = 2\) AG N0
METHOD 2
valid approach (M1)
eg\(\;\;\;\overrightarrow {{\text{AC}}} = k\overrightarrow {{\text{AB}}} \)
correct working A1
eg\(\;\;\;\overrightarrow {{\text{AC}}} = \left( {\begin{array}{*{20}{c}} 2 \\ {y – 4} \\ { – 4} \end{array}} \right),{\text{ }}\left( {\begin{array}{*{20}{c}} 2 \\ {y – 4} \\ { – 4} \end{array}} \right) = k\left( {\begin{array}{*{20}{c}} 1 \\ { – 1} \\ { – 2} \end{array}} \right)\)
\(k = 2\) A1
correct substitution A1
eg\(\;\;\;y – 4 = – 2\)
\(y = 2\) AG N0
[4 marks]
(i) correct substitution A1
eg\(\;\;\;0(1) + 2( – 1) – 1( – 2),{\text{ }}0 – 2 + 2\)
\(\overrightarrow {{\text{OC}}} \bullet \overrightarrow {{\text{AB}}} = 0\) A1 N1
(ii) \(9{0^ \circ }\) or \(\frac{\pi }{2}\) A1 N1
[3 marks]
METHOD 1 \({\text{(area}} = 0.5 \times {\text{height}} \times {\text{base)}}\)
\(\left| {\overrightarrow {{\text{OC}}} } \right| = \sqrt {0 + {2^2} + {{( – 1)}^2}} \;\;\;\left( { = \sqrt 5 } \right)\;\;\;\)(seen anywhere) A1
valid approach (M1)
eg\(\;\;\;\frac{1}{2} \times \left| {\overrightarrow {{\text{AB}}} } \right| \times \left| {\overrightarrow {{\text{OC}}} } \right|,{\text{ }}\left| {\overrightarrow {{\text{OC}}} } \right|\) is height of triangle
correct substitution A1
eg\(\;\;\;\frac{1}{2} \times \sqrt 6 \times \sqrt {0 + {{(2)}^2} + {{( – 1)}^2}} ,{\text{ }}\frac{1}{2} \times \sqrt 6 \times \sqrt 5 \)
area is \(\frac{{\sqrt {30} }}{2}\) A1 N2
METHOD 2 (difference of two areas)
one correct magnitude (seen anywhere) A1
eg\(\;\;\;\left| {\overrightarrow {{\text{OC}}} } \right| = \sqrt {{2^2} + {{( – 1)}^2}} \;\;\;\left( { = \sqrt 5 } \right),\;\;\;\left| {\overrightarrow {{\text{AC}}} } \right| = \sqrt {4 + 4 + 16} \;\;\;\left( { = \sqrt {24} } \right),\;\;\;\left| {\overrightarrow {{\text{BC}}} } \right| = \sqrt 6 \)
valid approach (M1)
eg\(\;\;\;\Delta {\text{OAC}} – \Delta {\text{OBC}}\)
correct substitution A1
eg\(\;\;\;\frac{1}{2} \times \sqrt {24} \times \sqrt 5 – \frac{1}{2} \times \sqrt 5 \times \sqrt 6 \)
area is \(\frac{{\sqrt {30} }}{2}\) A1 N2
METHOD 3 \({\text{(area}} = \frac{1}{2}ab\sin C{\text{ for }}\Delta {\text{OAB)}}\)
one correct magnitude of \(\overrightarrow {{\text{OA}}} \) or \(\overrightarrow {{\text{OB}}} \) (seen anywhere) A1
eg\(\;\;\;\left| {\overrightarrow {{\text{OA}}} } \right| = \sqrt {{{( – 2)}^2} + {4^2} + {3^2}} \;\;\;\left( { = \sqrt {29} } \right),\;\;\;\left| {\overrightarrow {{\text{OB}}} } \right| = \sqrt {1 + 9 + 1} \;\;\;\left( { = \sqrt {11} } \right)\)
valid attempt to find \(\cos \theta \) or \(\sin \theta \) (M1)
eg\(\;\;\;\cos {\text{C}} = \frac{{ – 1 – 3 – 2}}{{\sqrt 6 \times \sqrt {11} }}\;\;\;\left( { = \frac{{ – 6}}{{\sqrt {66} }}} \right),\;\;\;29 = 6 + 11 – 2\sqrt 6 \sqrt {11} \cos \theta ,{\text{ }}\frac{{\sin \theta }}{{\sqrt 5 }} = \frac{{\sin 90}}{{\sqrt {29} }}\)
correct substitution into \(\frac{1}{2}ab\sin {\text{C}}\) A1
eg\(\;\;\;\frac{1}{2} \times \sqrt 6 \times \sqrt {11} \times \sqrt {1 – \frac{{36}}{{66}}} ,{\text{ }}0.5 \times \sqrt 6 \times \sqrt {29} \times \frac{{\sqrt 5 }}{{\sqrt {29} }}\)
area is \(\frac{{\sqrt {30} }}{2}\) A1 N2
[4 marks]
Total [16 marks]