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

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

(i)     Find the value of p .

(ii)    Given that \({L_2}\) passes through \({\text{R}}(10{\text{, }}6{\text{, }}- 40)\) , write down a vector equation for \({L_2}\) .

The lines \({L_1}\) and \({L_2}\) intersect at the point A. Find the x-coordinate of A.

(i) evidence of approach     (M1)

e.g. \(\overrightarrow {{\rm{PO}}}  + \overrightarrow {{\rm{OQ}}} \) , \({\rm{P}} - {\rm{Q}}\)

\(\overrightarrow {{\rm{PQ}}}  = \left( {\begin{array}{*{20}{c}}
2\\
1\\
{ - 4}
\end{array}} \right)\)     A1     N2

(ii) any correct equation in the form \({\boldsymbol{r}} = {\boldsymbol{a}} + s{\boldsymbol{b}}\) (accept any parameter for s)

where a is \(\left( {\begin{array}{*{20}{c}}
2\\
4\\
8
\end{array}} \right)\) or \(\left( {\begin{array}{*{20}{c}}
4\\
5\\
4
\end{array}} \right)\) , and b is a scalar multiple of \(\left( {\begin{array}{*{20}{c}}
2\\
1\\
{ - 4}
\end{array}} \right)\)     A2     N2

e.g. \({\boldsymbol{r}} = \left( {\begin{array}{*{20}{c}}
2\\
4\\
8
\end{array}} \right) + s\left( {\begin{array}{*{20}{c}}
2\\
1\\
{ - 4}
\end{array}} \right)\) , \({\boldsymbol{r}} = \left( {\begin{array}{*{20}{c}}
{4 + 2s}\\
{5 + 1s}\\
{4 - 4s}
\end{array}} \right)\) , \({\boldsymbol{r}} = 2{\boldsymbol{i}} + 4{\boldsymbol{j}} + 8{\boldsymbol{k}} + s(2{\boldsymbol{i}} + 1{\boldsymbol{j}} - 4{\boldsymbol{k}})\)

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

[4 marks]

(i) choosing correct direction vectors for \({L_1}\) and \({L_2}\)     (A1)    (A1)

e.g. \(\left( {\begin{array}{*{20}{c}}
2\\
1\\
{ - 4}
\end{array}} \right)\) , \(\left( {\begin{array}{*{20}{c}}
{3p}\\
{2p}\\
4
\end{array}} \right)\)

evidence of equating scalar product to 0     (M1)

correct calculation of scalar product     A1

e.g. \(2 \times 3p + 1 \times 2p + ( - 4) \times 4\)  , \(8p - 16 = 0\)

\(p = 2\)     A1     N3

(ii) any correct expression in the form \({\boldsymbol{r}} = {\boldsymbol{a}} + t{\boldsymbol{b}}\) (accept any parameter for t)

where a is \(\left( {\begin{array}{*{20}{c}}
{10}\\
6\\
{ - 40}
\end{array}} \right)\) , and b is a scalar multiple of \(\left( {\begin{array}{*{20}{c}}
6\\
4\\
4
\end{array}} \right)\)     A2     N2

e.g. \({\boldsymbol{r}} = \left( {\begin{array}{*{20}{c}}
{10}\\
6\\
{ - 40}
\end{array}} \right) + t\left( {\begin{array}{*{20}{c}}
6\\
4\\
4
\end{array}} \right)\) , \({\boldsymbol{r}} = \left( {\begin{array}{*{20}{c}}
{10 + 6s}\\
{6 + 4s}\\
{ - 40 + 4s}
\end{array}} \right)\) , \({\boldsymbol{r}} = 10{\boldsymbol{i}} + 6{\boldsymbol{j}} - 40{\boldsymbol{k}} + s(6{\boldsymbol{i}} + 4{\boldsymbol{j}} + 4{\boldsymbol{k}})\)

Note: Award A1 for the form \({\boldsymbol{a}} + t{\boldsymbol{b}}\) , A1 for \({\boldsymbol{L}} = {\boldsymbol{a}} + t{\boldsymbol{b}}\) (unless they have been penalised for \({\boldsymbol{L}} = {\boldsymbol{a}} + s{\boldsymbol{b}}\) in part (a)), A0 for \({\boldsymbol{r}} = {\boldsymbol{b}} + t{\boldsymbol{a}}\) .

[7 marks]

appropriate approach     (M1)

e.g. \(\left( {\begin{array}{*{20}{c}}
2\\
4\\
8
\end{array}} \right) + s\left( {\begin{array}{*{20}{c}}
2\\
1\\
{ - 4}
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
{10}\\
6\\
{ - 40}
\end{array}} \right) + t\left( {\begin{array}{*{20}{c}}
6\\
4\\
4
\end{array}} \right)\)

any two correct equations with different parameters     A1A1

e.g. \(2 + 2s = 10 + 6t\) , \(4 + s = 6 + 4t\) , \(8 - 4s = - 40 + 4t\)

attempt to solve simultaneous equations     (M1)

correct working (A1)

e.g. \( - 6 = - 2 - 2t\) , \(4 = 2t\) , \( - 4 + 5s = 46\) , \(5s = 50\)

one correct parameter \(s = 10\) , \(t = 2\)     A1

\(x = 22\) (accept (22, 14, −32))     A1     N4

[7 marks]

In part (a), nearly all the candidates correctly found the vector PQ, and the majority went onto find the correct vector equation of the line. There are still many candidates who do not write this equation in the correct form, using "r = ", and these candidates were penalized one mark.

In part (b), the majority of candidates knew to set the scalar product equal to zero for the perpendicular vectors, and were able to find the correct value of p.

A good number of candidates used the correct method to find the intersection of the two lines, though some algebraic and arithmetic errors kept some from finding the correct final answer.