(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]