(i)     Show that \(\overrightarrow {{\rm{PQ}}}  = \left( {\begin{array}{*{20}{c}}
1\\
{ - 2}\\
2
\end{array}} \right)\) .

(ii)    Hence, write down an equation for \({L_1}\) in the form \({\boldsymbol{r}} = {\boldsymbol{a}} + t{\boldsymbol{b}}\) .

Find the cosine of the angle between \(\overrightarrow {{\rm{PQ}}} \) and \({L_2}\) .

The lines \({L_1}\) and \({L_2}\) intersect at the point R. Find the coordinates of R.

(i) evidence of correct approach     A1

e.g. \(\overrightarrow {{\rm{PQ}}}  = \overrightarrow {{\rm{OQ}}}  - \overrightarrow {{\rm{OP}}} \) , \(Q - P\)

\(\overrightarrow {{\rm{PQ}}}  = \left( {\begin{array}{*{20}{c}}
1\\
{ - 2}\\
2
\end{array}} \right)\)     AG     N0

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

where a is either \(\overrightarrow {{\rm{OP}}} \) or \(\overrightarrow {{\rm{OQ}}} \) and b is a scalar multiple of \(\overrightarrow {{\rm{PQ}}} \) 

e.g. \({\boldsymbol{r}} = \left( {\begin{array}{*{20}{c}}
{ - 1}\\
6\\
{ - 1}
\end{array}} \right) + t\left( {\begin{array}{*{20}{c}}
1\\
{ - 2}\\
2
\end{array}} \right)\) , \({\boldsymbol{r}} = \left( {\begin{array}{*{20}{c}}
  t \\
  {4 - 2t} \\
  {1 + 2t}
\end{array}} \right)\), \({\boldsymbol{r}} = 4{\boldsymbol{j}} + {\boldsymbol{k}} + t({\boldsymbol{i}} - 2{\boldsymbol{j}} + 2{\boldsymbol{k}})\)

[3 marks]

choosing a correct direction vector for \({L_2}\)     (A1)

e.g. \(\left( {\begin{array}{*{20}{c}}
3\\
0\\
{ - 4}
\end{array}} \right)\)

finding scalar products and magnitudes     (A1)(A1)(A1)

scalar product \( = 1(3) - 2(0) + 2( - 4)\) \(( = - 5)\)

magnitudes \( = \sqrt {{1^2} + {{( - 2)}^2} + {2^2}} \) \(( = 3)\) , \(\sqrt {{3^2} + {0^2} + {{( - 4)}^2}} \) \(( = 5)\)

substitution into formula     M1

e.g. \(\cos \theta  = \frac{{ - 5}}{{\sqrt 9  \times \sqrt {25} }}\)

\(\cos \theta  = - \frac{1}{3}\)     A2     N5

[7 marks]

evidence of valid approach     (M1)

e.g. equating lines, \({L_1} = {L_2}\)

EITHER

one correct equation in one variable     A2

e.g. \(6 - 2t = 2\)

OR

two correct equations in two variables     A1A1

e.g. \(2t + 4s = 0\) , \(t - 3s = 5\)

THEN

attempt to solve     (M1)

one correct parameter     A1

e.g. \(t = 2\) , \(s = - 1\)

correct substitution of either parameter     (A1)

e.g. \({\boldsymbol{r}} = \left( {\begin{array}{*{20}{c}}
4\\
2\\
{ - 1}
\end{array}} \right) + ( - 1)\left( {\begin{array}{*{20}{c}}
3\\
0\\
{ - 4}
\end{array}} \right)\) , \({\boldsymbol{r}} = \left( {\begin{array}{*{20}{c}}
{ - 1}\\
6\\
{ - 1}
\end{array}} \right) + ( + 2)\left( {\begin{array}{*{20}{c}}
1\\
{ - 2}\\
2
\end{array}} \right)\)

coordinates \({\text{R}}(1{\text{, }}2{\text{, }}3)\)     A1     N3

[7 marks]

A pleasing number of candidates were successful on this straightforward vector and line question. Part (a) was generally well answered, although a few candidates still labelled their line \(L =\) or used a position vector for the direction vector. Follow-through marking allowed full recovery from the latter error. 

Few candidates wrote down their direction vector in part (b) which led to lost follow-through marks, and a common error was finding an incorrect scalar product due to difficulty multiplying by zero.

Part (c) was generally well understood with some candidates realizing that the equation in just one variable led to the correct parameter more quickly than solving a system of two equations to find both parameters. Some candidates gave the answer as (s, t) instead of substituting those parameters, indicating a more rote understanding of the problem. Another common error was using the same parameter for both lines.

There were an alarming number of misreads of negative signs from the question or from the candidate working.