Write down a vector equation for \({L_2}\) in the form \({\boldsymbol{r}} = {\boldsymbol{a}} + t{\boldsymbol{b}}\) .

A third line \({L_3}\) is perpendicular to \({L_1}\) and is represented by \({\boldsymbol{r}} = \left( {\begin{array}{*{20}{c}}
5\\
0\\
3
\end{array}} \right) + q\left( {\begin{array}{*{20}{c}}
{ - 7}\\
{ - 2}\\
k
\end{array}} \right)\) .

Show that \(k = - 2\) .

The lines \({L_1}\) and \({L_3}\) intersect at the point A.

Find the coordinates of A.

The lines \({L_2}\)and \({L_3}\)intersect at point C where \(\overrightarrow {{\rm{BC}}} = \left( {\begin{array}{*{20}{c}}
6\\
3\\
{ - 24}
\end{array}} \right)\) .

(i)     Find \(\overrightarrow {{\rm{AB}}} \) .

(ii)    Hence, find \(|\overrightarrow {{\rm{AC}}} |\) .

any correct equation in the form \({\boldsymbol{r}} = {\boldsymbol{a}} + t{\boldsymbol{b}}\) (accept any parameter)     A2     N2

e.g. \({\boldsymbol{r}} = \left( {\begin{array}{*{20}{c}}
{ - 8}\\
{ - 5}\\
{25}
\end{array}} \right) + t\left( {\begin{array}{*{20}{c}}
2\\
1\\
{ - 8}
\end{array}} \right)\)

Note: Award A1 for \({\boldsymbol{a}} + t{\boldsymbol{b}}\) , A1 for \(L = {\boldsymbol{a}} + t{\boldsymbol{b}}\) , A0 for \({\boldsymbol{r}} = {\boldsymbol{b}} + t{\boldsymbol{a}}\) .

[2 marks]

recognizing scalar product must be zero (seen anywhere)     R1

e.g. \({\boldsymbol{a}} \bullet {\boldsymbol{b}} = 0\)

evidence of choosing direction vectors \(\left( {\begin{array}{*{20}{c}}
2\\
1\\
{ - 8}
\end{array}} \right),\left( {\begin{array}{*{20}{c}}
{ - 7}\\
{ - 2}\\
k
\end{array}} \right)\)     (A1)(A1)

correct calculation of scalar product     (A1)

e.g. \(2( - 7) + 1( - 2) - 8k\)

simplification that clearly leads to solution     A1

e.g. \( - 16 - 8k\) , \( - 16 - 8k = 0\)

\(k = - 2\)     AG     N0

[5 marks]

evidence of equating vectors     (M1)

e.g. \({L_1} = {L_3}\) , \(\left( {\begin{array}{*{20}{c}}
{ - 3}\\
{ - 1}\\
{ - 25}
\end{array}} \right) + p\left( {\begin{array}{*{20}{c}}
2\\
1\\
{ - 8}
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
5\\
0\\
3
\end{array}} \right) + q\left( {\begin{array}{*{20}{c}}
{ - 7}\\
{ - 2}\\
{ - 2}
\end{array}} \right)\)

any two correct equations     A1A1

e.g. \( - 3 + 2p = 5 - 7q\) ,  \( - 1 + p = - 2q\) , \(- 25 - 8p = 3 - 2q\)

attempting to solve equations     (M1)

finding one correct parameter (\(p = - 3\) , \(q = 2\) )     A1

the coordinates of A are \(( - 9, - 4, - 1)\)     A1     N3

[6 marks]

(i) evidence of appropriate approach     (M1)

e.g. \(\overrightarrow {{\rm{OA}}}  + \overrightarrow {{\rm{AB}}} = \overrightarrow {{\rm{OB}}} \) , \(\overrightarrow {{\rm{AB}}}  = \left( {\begin{array}{*{20}{c}}
{ - 8}\\
{ - 5}\\
{25}
\end{array}} \right) - \left( {\begin{array}{*{20}{c}}
{ - 9}\\
{ - 4}\\
{ - 1}
\end{array}} \right)\)

\(\overrightarrow {{\rm{AB}}} = \left( {\begin{array}{*{20}{c}}
1\\
{ - 1}\\
{26}
\end{array}} \right)\)     A1     N2

(ii) finding \(\overrightarrow {{\rm{AC}}} = \left( {\begin{array}{*{20}{c}}
7\\
2\\
2
\end{array}} \right)\)     A1

evidence of finding magnitude     (M1)

e.g. \(|\overrightarrow {{\rm{AC}}} | = \sqrt {{7^2} + {2^2} + {2^2}} \)

\(|\overrightarrow {{\rm{AC}}} | = \sqrt {57} \)     A1     N3

[5 marks]

Many candidates gave a correct vector equation for the line.

A common error was to misplace the initial position and direction vectors. Those who set the scalar product of the direction vectors to zero typically solved for k successfully. Those who substituted \(k = - 2\) earned fewer marks for working backwards in a "show that" question.

Many went on to find the coordinates of point A, however some used the same letter, say p, for each parameter and thus could not solve the system.

Part (d) proved challenging as many candidates did not consider that \(\overrightarrow {{\rm{AB}}} + \overrightarrow {{\rm{BC}}} = \overrightarrow {{\rm{AC}}} \) . Rather, many attempted to find the coordinates of point C, which became a more arduous and error-prone task.