(i)     Show that \(\overrightarrow {{\rm{AB}}} = \left( {\begin{array}{*{20}{c}}
{ - 4}\\
6\\
{ - 1}
\end{array}} \right)\) .

(ii)    Find \({\rm{B}}\widehat {\rm{A}}{\rm{O}}\) .

The line \({L_1}\) has equation \(\left( {\begin{array}{*{20}{c}}
x\\
y\\
z
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
{ - 3}\\
4\\
2
\end{array}} \right) + s\left( {\begin{array}{*{20}{c}}
{ - 4}\\
6\\
{ - 1}
\end{array}} \right)\) .

Write down the coordinates of two points on \({L_1}\) .

The line \({L_2}\) passes through A and is parallel to \(\overrightarrow {{\rm{OB}}} \) .

(i)     Find a vector equation for \({L_2}\) , giving your answer in the form \({\boldsymbol{r}} = {\boldsymbol{a}} + t{\boldsymbol{b}}\) . 

(ii)    Point \(C(k, - k,5)\) is on  \({L_2}\) . Find the coordinates of C.

The line \({L_3}\) has equation \(\left( {\begin{array}{*{20}{c}}
x\\
y\\
z
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
3\\
{ - 8}\\
0
\end{array}} \right) + p\left( {\begin{array}{*{20}{c}}
1\\
{ - 2}\\
{ - 1}
\end{array}} \right)\) and passes through the point C. 

 Find the value of p at C.

(i) evidence of approach     M1

e.g. \(\overrightarrow {{\rm{AO}}} + \overrightarrow {{\rm{OB}}} = \overrightarrow {{\rm{AB}}} \)

\(\overrightarrow {{\rm{AB}}} = \left( {\begin{array}{*{20}{c}}
{ - 4}\\
6\\
{ - 1}
\end{array}} \right)\)    AG     N0

(ii) for choosing correct vectors, (\(\overrightarrow {{\rm{AO}}} \) with \(\overrightarrow {{\rm{AB}}} \) or \(\overrightarrow {{\rm{OA}}} \) with \(\overrightarrow {{\rm{BA}}} \) )     (A1)(A1)

Note: Using \(\overrightarrow {{\rm{AO}}} \) with \(\overrightarrow {{\rm{BA}}} \) will lead to \(\pi  - 0.799\) . If they then say \({\rm{B}}\widehat {\rm{A}}{\rm{O}} = 0.799\) , this is a correct solution.

calculating \(\overrightarrow {{\rm{AO}}} \bullet \overrightarrow {{\rm{AB}}} \) , \(\left| {\overrightarrow {{\rm{AO}}} } \right|\) , \(\left| {\overrightarrow {{\rm{AB}}} } \right|\)     (A1)(A1)(A1)

e.g. \({d_1} \bullet {d_2} = ( - 1)( - 4) + (2)(6) + ( - 3)( - 1)( = 19)\)

\(\left| {{d_1}} \right| = \sqrt {{{( - 1)}^2} + {2^2} + {{( - 3)}^2}} ( = \sqrt {14} )\) , \(\left| {{d_2}} \right| = \sqrt {{{( - 4)}^2} + {6^2} + {{( - 1)}^2}} ( = \sqrt {53} )\)

evidence of using the formula to find the angle     M1

e.g. \(\cos \theta = \frac{{( - 1)( - 4) + (2)(6) + ( - 3)( - 1)}}{{\sqrt {{{( - 1)}^2} + {2^2} + {{( - 3)}^2}} \sqrt {{{( - 4)}^2} + {6^2} + {{( - 1)}^2}} }}\) , \(\frac{{19}}{{\sqrt {14} \sqrt {53} }}\) , \(0.69751 \ldots \)

\({\rm{B}}\widehat {\rm{A}}{\rm{O}} = 0.799\) radians (accept \(45.8^\circ \))     A1     N3

[8 marks]

two correct answers     A1A1

e.g. (1, \( - 2\), 3) , (\( - 3\), 4, 2) , (\( - 7\), 10, 1), (\( - 11\), 16, 0)     N2

[2 marks]

(i) \({\boldsymbol{r}} = \left( {\begin{array}{*{20}{c}}
1\\
{ - 2}\\
3
\end{array}} \right) + t\left( {\begin{array}{*{20}{c}}
{ - 3}\\
4\\
2
\end{array}} \right)\)     A2     N2

(ii) C on \({L_2}\) , so \(\left( {\begin{array}{*{20}{c}}
k\\
{ - k}\\
5
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
1\\
{ - 2}\\
3
\end{array}} \right) + t\left( {\begin{array}{*{20}{c}}
{ - 3}\\
4\\
2
\end{array}} \right)\)     (M1)

evidence of equating components (A1)

e.g. \(1 - 3t = k\) , \( - 2 + 4t = - k\) , \(5 = 3 + 2t\)

one correct value \(t = 1\) , \(k = - 2\) (seen anywhere)     (A1)

coordinates of C are \(( - 2{\text{, }}2{\text{, }}5)\)     A1     N3

[6 marks]

for setting up one (or more) correct equation using \(\left( {\begin{array}{*{20}{c}}
{ - 2}\\
2\\
5
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
3\\
{ - 8}\\
0
\end{array}} \right) + p\left( {\begin{array}{*{20}{c}}
1\\
{ - 2}\\
{ - 1}
\end{array}} \right)\)     (M1)

e.g. \(3 + p = - 2\) , \( - 8 - 2p = 2\) , \( - p = 5\)

\(p = - 5\)     A1     N2

[2 marks]

Part (ai) was done well by most students. Most knew how to approach finding the angle in part (aii). The problems occurred when the incorrect vectors were chosen. If the vectors being used were stated, then follow through marks could be given.

Part (b) was well done.

In part (ci), the error that occurred most often was the incorrect choice for the direction vector.

Those that were able to find the coordinates in part (cii) were also able to be successful in part (d).