(i)     Show that $\overrightarrow {{\text{AB}}}  = \left( {\begin{array}{*{20}{c}} { - 2} \\ 8 \\ 2 \end{array}} \right)$.

(ii)     Write down a vector equation for ${L_1}$.

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 ${L_2}$ intersect at a point $C$.

Show that the coordinates of $C$ are $( - 1,{\text{ }}1,{\text{ }}2)$.

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 ${\rm{A\hat CD}}$.

(i)     correct approach     A1

eg$\;\;\;{\text{OB}} - {\text{OA, }}\left( {\begin{array}{*{20}{c}} { - 2} \\ 5 \\ 3 \end{array}} \right) - \left( {\begin{array}{*{20}{c}} 0 \\ { - 3} \\ 1 \end{array}} \right),{\text{ B}} - {\text{A}}$

$\overrightarrow {{\text{AB}}}  = \left( {\begin{array}{*{20}{c}} { - 2} \\ 8 \\ 2 \end{array}} \right)$     AG     N0

(ii)     any correct equation in the form $r = a +$ t$b$ (accept any parameter for $t$)

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 $\left( {\begin{array}{*{20}{c}} { - 2} \\ 8 \\ 2 \end{array}} \right)$     A2     N2

eg$r$ = \left( { $$\begin{array}{*{20}{c}} 0 \\ { - 3} \\ 1 \end{array}$$ } \right) + t\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),$r = 2i + 5j + 3k +$ t$( - 2i + 8j + 2k)$

Note:     Award A1 for the form $a$ + t$b$, A1 for the form $L = \(a$ + t$b$,

A0 for the form $r$ = $b$ + t$a$.

[3 marks]

valid approach     (M1)

eg$\;\;\;$equating lines, ${L_1} = {L_2}$

one correct equation in one variable     A1

eg$\;\;\; - 2t =  - 1,{\text{ }} - 2 - 2t =  - 1$

valid attempt to solve     (M1)

eg$\;\;\;2t = 1,{\text{ }} - 2t = 1$

one correct parameter     A1

eg$\;\;\;t = \frac{1}{2},{\text{ }}t =  - \frac{1}{2},{\text{ }}s =  - 6$

correct substitution of either parameter     A1

eg$\;\;\;r = \left( {\begin{array}{*{20}{c}} 0 \\ { - 3} \\ 1 \end{array}} \right) + \frac{1}{2}\left( {\begin{array}{*{20}{c}} { - 2} \\ 8 \\ 2 \end{array}} \right),{\text{ }}r = \left( {\begin{array}{*{20}{c}} { - 2} \\ 5 \\ 3 \end{array}} \right) - \frac{1}{2}\left( {\begin{array}{*{20}{c}} { - 2} \\ 8 \\ 2 \end{array}} \right),{\text{ }}r = \left( {\begin{array}{*{20}{c}} { - 1} \\ 7 \\ { - 4} \end{array}} \right) - 6\left( {\begin{array}{*{20}{c}} 0 \\ 1 \\ { - 1} \end{array}} \right)$

the coordinates of $C$ are $( - 1,{\text{ }}1,{\text{ }}2)$, or position vector of $C$ is $\left( {\begin{array}{*{20}{c}} { - 1} \\ 1 \\ 2 \end{array}} \right)$     AG     N0

Note:     If candidate uses the same parameter in both vector equations and working shown, award M1A1M1A0A0.

[5 marks]

valid approach     (M1)

eg$\;\;\;$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 $\overrightarrow {{\text{CD}}}$

$\overrightarrow {{\text{CA}}}  = \left( {\begin{array}{*{20}{c}} 1 \\ { - 4} \\ { - 1} \end{array}} \right)$     (A1)

Notes:     Exceptions to FT:

1 if candidate indicates that they are finding $\overrightarrow {{\text{CA}}}$, but makes an error, award M1A0;

2 if candidate finds an incorrect vector (including $\overrightarrow {{\text{AC}}}$), award M0A0.

In both cases, if working shown, full FT may be awarded for subsequent correct FT work.

Award the final (A1) for simplification of their value for ${\rm{A\hat CD}}$.

Award the final A2 for finding their arc cos. If their value of cos does not allow them to find an angle, they cannot be awarded this A2.

finding $\left| {\overrightarrow {{\text{CA}}} } \right|$ (may be seen in cosine formula)     A1

eg$\;\;\;\sqrt {{1^2} + {{( - 4)}^2} + {{( - 1)}^2}} ,{\text{ }}\sqrt {18}$

correct substitution into cosine formula     (A1)

eg$\;\;\;\frac{{ - 9}}{{\sqrt {18} \sqrt {18} }}$

finding $\cos {\rm{A\hat CD}} - \frac{1}{2}$     (A1)

${\rm{A\hat CD}} = \frac{{2\pi }}{3}\;\;\;(120^\circ )$     A2     N2

Notes:     Award A1 if additional answers are given.

Award A1 for answer $\frac{\pi }{3}{\rm{ (60^\circ )}}$.

[7 marks]

Total [15 marks]