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

Hence, write down a vector equation for \({L_1}\) .

Given that \({L_1}\) is perpendicular to \({L_2}\) , show that \(p = - 6\) .

The line \({L_1}\) intersects the line \({L_2}\) at point Q. Find the \(x\)-coordinate of Q.

valid approach     (M1)

eg   \(\left( {\begin{array}{*{20}{c}}
7\\
{ - 4}\\
3
\end{array}} \right) - \left( {\begin{array}{*{20}{c}}
1\\
{ - 2}\\
{ - 1}
\end{array}} \right)\) , \({\rm{A}} - {\rm{B}}\) , \(\overrightarrow {{\rm{AB}}}  = \overrightarrow {{\rm{AO}}}  + \overrightarrow {{\rm{OB}}} \)

\(\overrightarrow {{\rm{AB}}}  = \left( {\begin{array}{*{20}{c}}
6\\
{ - 2}\\
4
\end{array}} \right)\)     A1     N2

[2 marks]

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

where \(\boldsymbol{a} = \left( {\begin{array}{*{20}{c}}
1\\
{ - 2}\\
3
\end{array}} \right)\) and \(\boldsymbol{b}\) is a scalar multiple of \(\overrightarrow {{\rm{AB}}} \)     A2     N2

eg   \(\boldsymbol{r} = \left( {\begin{array}{*{20}{c}}
1\\
{ - 2}\\
3
\end{array}} \right) + t\left( {\begin{array}{*{20}{c}}
6\\
{ - 2}\\
4
\end{array}} \right)\) , \((x,y,z) = (1, - 2,3) + t(3, - 1,2)\) , \(\boldsymbol{r} = \left( {\begin{array}{*{20}{c}}
{1 + 6t}\\
{ - 2 - 2t}\\
{3 + 4t}
\end{array}} \right)\)

Note: Award A1 for \(\boldsymbol{a} + t\boldsymbol{b}\) , A1 for \({L_1} = \boldsymbol{a} + t\boldsymbol{b}\) , A0 for \(\boldsymbol{r} = \boldsymbol{b} + t\boldsymbol{a}\) .

[2 marks]

recognizing that scalar product \( = 0\) (seen anywhere)     R1

correct calculation of scalar product     (A1)

eg   \(6(3) - 2( - 3) + 4p\) , \(18 + 6 + 4p\)

correct working     A1

eg   \(24 + 4p = 0\) , \(4p =  - 24\)

\(p = - 6\)     AG     N0

[3 marks]

setting lines equal     (M1)

eg   \({L_1} = {L_2}\) , \(\left( {\begin{array}{*{20}{c}}
1\\
{ - 2}\\
3
\end{array}} \right) + t\left( {\begin{array}{*{20}{c}}
6\\
{ - 2}\\
4
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
{ - 1}\\
2\\
{15}
\end{array}} \right) + s\left( {\begin{array}{*{20}{c}}
3\\
{ - 3}\\
{ - 6}
\end{array}} \right)\)

any two correct equations with different parameters     A1A1

eg   \(1 + 6t = 1 + 3s\) , \( - 2 - 2t = 2 - 3s\) , \(3 + 4t = 15 - 6s\)

attempt to solve their simultaneous equations     (M1)

one correct parameter     A1

eg   \(t = \frac{1}{2}\) , \(s = \frac{5}{3}\)

attempt to substitute parameter into vector equation     (M1)

eg   \(\left( {\begin{array}{*{20}{c}}
1\\
{ - 2}\\
3
\end{array}} \right) + \frac{1}{2}\left( {\begin{array}{*{20}{c}}
6\\
{ - 2}\\
4
\end{array}} \right)\) , \(1 + \frac{1}{2} \times 6\)

\(x = 4\) (accept (4, -3, 5), ignore incorrect values for \(y\) and \(z\))     A1     N3

[7 marks]

While many candidates can find a vector given two points, few could write down a fully correct vector equation of a line.

While many candidates can find a vector given two points, few could write down a fully correct vector equation of a line. Most candidates wrote their equation as “ \({L_1} = \) ”, which misrepresents that the resulting equation must still be a vector.

Those who recognized that vector perpendicularity means the scalar product is zero found little difficulty answering part (b). Occasionally a candidate would use the given \(p = 6\) to show the scalar product is zero. However, working backward from the given answer earns no marks in a question that requires candidates to show that this value is achieved.

While many candidates knew to set the lines equal to find an intersection point, a surprising number could not carry the process to correct completion. Some could not solve a simultaneous pair of equations, and for those who did, some did not know what to do with the parameter value. Another common error was to set the vector equations equal using the same parameter, from which the candidates did not recognize a system to solve. Furthermore, it is interesting to note that while only one parameter value is needed to answer the question, most candidates find or attempt to find both, presumably out of habit in the algorithm.