Two lines with equations \({{\boldsymbol{r}}_1} = \left( {\begin{array}{*{20}{c}}
2\\
3\\
{ - 1}
\end{array}} \right) + s\left( {\begin{array}{*{20}{c}}
5\\
{ - 3}\\
2
\end{array}} \right)\) and \({{\boldsymbol{r}}_2} = \left( {\begin{array}{*{20}{c}}
9\\
2\\
2
\end{array}} \right) + t\left( {\begin{array}{*{20}{c}}
{ - 3}\\
5\\
{ - 1}
\end{array}} \right)\) intersect at the point P. Find the coordinates of P.

evidence of appropriate approach     (M1)

e.g. \(\left( {\begin{array}{*{20}{c}}
2\\
3\\
{ - 1}
\end{array}} \right) + s\left( {\begin{array}{*{20}{c}}
5\\
{ - 3}\\
2
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
9\\
2\\
2
\end{array}} \right) + t\left( {\begin{array}{*{20}{c}}
{ - 3}\\
5\\
{ - 1}
\end{array}} \right)\)

two correct equations     A1A1

e.g. \(2 + 5s = 9 - 3t\) ,  \(3 - 3s = 2 + 5t\) , \( - 1 + 2s = 2 - t\)

attempting to solve the equations     (M1)

one correct parameter \(s = 2\) , \(t =  - 1\)     A1

P is \((12, - 3,3)\) (accept \(\left( {\begin{array}{*{20}{c}}
{12}\\
{ - 3}\\
3
\end{array}} \right)\))      A1     N3

[6 marks] 

If this topic had been taught well then the candidates scored highly. The question was either well answered or not at all. Many candidates did not understand what was needed and tried to find the length of vectors or mid-points of lines. The other most common mistake was to use the values of the parameters to write the coordinates as \({\text{P}}(2{\text{, }} - 1)\).