Find the Cartesian equation of $\pi$.

The line ${l_3}$ passing through the point $(4,{\text{ }}0,{\text{ }}8)$ is perpendicular to $\pi$.

Find the coordinates of the point where ${l_3}$ meets $\pi$.

attempting to find a normal to $\pi {\text{ }}eg{\text{ }}\left( {\begin{array}{*{20}{c}} 3 \\ 2 \\ { - 2} \end{array}} \right) \times \left( {\begin{array}{*{20}{c}} 8 \\ {11} \\ 6 \end{array}} \right)$     (M1)

$\left( {\begin{array}{*{20}{c}} 3 \\ 2 \\ { - 2} \end{array}} \right) \times \left( {\begin{array}{*{20}{c}} 8 \\ {11} \\ 6 \end{array}} \right) = 17\left( {\begin{array}{*{20}{c}} 2 \\ { - 2} \\ 1 \end{array}} \right)$     (A1)

${{r}} \bullet \left( {\begin{array}{*{20}{c}} 2 \\ { - 2} \\ 1 \end{array}} \right) = \left( {\begin{array}{*{20}{c}} 1 \\ 5 \\ {12} \end{array}} \right) \bullet \left( {\begin{array}{*{20}{c}} 2 \\ { - 2} \\ 1 \end{array}} \right)$     M1

$2x - 2y + z = 4$ (or equivalent)     A1

[4 marks]

${l_3}:{{r}} = \left( {\begin{array}{*{20}{c}} 4 \\ 0 \\ 8 \end{array}} \right) + t\left( {\begin{array}{*{20}{c}} 2 \\ { - 2} \\ 1 \end{array}} \right),\;\;\;t \in \mathbb{R}$     (A1)

attempting to solve $\left( {\begin{array}{*{20}{c}} {4 + 2t} \\ { - 2t} \\ {8 + t} \end{array}} \right) \bullet \left( {\begin{array}{*{20}{c}} 2 \\ { - 2} \\ 1 \end{array}} \right) = 4\;\;\;{\text{for }}t\;\;\;ie{\text{ }}9t + 16 = 4\;\;\;{\text{for }}t$     M1

$t =  - \frac{4}{3}$     A1

$\left( {\frac{4}{3},{\text{ }}\frac{8}{3},{\text{ }}\frac{{20}}{3}} \right)$     A1

[4 marks]

Total [8 marks]

Part (a) was reasonably well done. Some candidates made numerical errors when attempting to find a normal to $\pi$.

In part (b), a number of candidates were awarded follow through marks from numerical errors committed in part (a).