Line \({L_1}\) has equation \({\boldsymbol{r}_1} = \left( {\begin{array}{*{20}{c}}
{10}\\
6\\
{ - 1}
\end{array}} \right) + s\left( {\begin{array}{*{20}{c}}
2\\
{ - 5}\\
{ - 2}
\end{array}} \right)\) and line \({L_2}\) has equation \({\boldsymbol{r}_2} = \left( {\begin{array}{*{20}{c}}
2\\
1\\
{ - 3}
\end{array}} \right) + t\left( {\begin{array}{*{20}{c}}
3\\
5\\
2
\end{array}} \right)\) .

Lines \({L_1}\) and \({L_2}\) intersect at point A. Find the coordinates of A.

appropriate approach     (M1)

eg   \(\left( {\begin{array}{*{20}{c}}
{10}\\
6\\
{ - 1}
\end{array}} \right) + s\left( {\begin{array}{*{20}{c}}
2\\
{ - 5}\\
{ - 2}
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
2\\
1\\
{ - 3}
\end{array}} \right) + t\left( {\begin{array}{*{20}{c}}
3\\
5\\
2
\end{array}} \right)\) , \({L_1} = {L_2}\)

any two correct equations     A1A1

eg   \(10 + 2s = 2 + 3t\) , \(6 - 5s = 1 + 5t\) , \( - 1 - 2s =  - 3 + 2t\)

attempt to solve     (M1)

eg substituting one equation into another

one correct parameter     A1

eg   \(s = - 1\) , \(t = 2\)

correct substitution     (A1)

eg   \(2 + 3(2)\) , \(1 + 5(2)\) , \( - 3 + 2(2)\)

A \( = \) (\(8\), \(11\), \(1\)) (accept column vector)     A1     N4

[7 marks]

Most students were able to set up one or more equations, but few chose to use their GDCs to solve the resulting system. Algebraic errors prevented many of these candidates from obtaining the final three marks. Some candidates stopped after finding the value of s and/or \(t\).