• This sample question was produced by experienced DP mathematics senior examiners to aid teachers in preparing for external assessment in the new MAA course. There may be minor differences in formatting compared to formal exam papers.

Attempts to calculate $2{{ 1 _m}} \cdot 2{{ -5 -m}}$ (M1)

$= -1 - m^2$ A1

since $m^2 \geq 0, -1 - m^2 < 0$ for $m \in \mathbb{R}$R1

so $l_1$ and $l_2$ are never perpendicular to each otherAG

[3 marks]

(since $l_1$ is parallel to $\Pi$, $l_1$ is perpendicular to the normal of $\Pi$ and so) $\begin{pmatrix} 2 \\ 1 \\ m \end{pmatrix} \cdot \begin{pmatrix} 1 \\ 4 \\ -1 \end{pmatrix} = 0$R1 $2 + 4 - m = 0$ $m = 6$ A1

[2 marks]

since there are no points in common, $3, -2, 0$ does not lie in $\Pi$

EITHER substitutes $3, -2, 0$ into $x+4y-z\neq p$ (M1)

OR $\begin{bmatrix} 3\\ -2\\ 0 \end{bmatrix} \cdot \begin{bmatrix} 1\\ 4\\ -1 \end{bmatrix} \neq p$ (M1)

THEN $p\neq -5$A1

[2 marks]