(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 $OAB$.

(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]

Finding $\overrightarrow {{\text{AB}}}$ and its magnitude were mostly well done.

Mostly correct answers with common errors being using both position vectors or writing it as “$L =$” instead of “${\mathbf{r}} =$”.

Many candidates assumed that $\overrightarrow {{\text{AB}}}  = \overrightarrow {{\text{BC}}}$, although this was not indicated on the diagram nor given in the question.

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.

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.