EXN.1.AHL.TZ0.8

pestleMathematicsAAHLPaper 1EXN· ahl-3-13-scalar-(dot)-product, ahl-3-17-vector-equations-of-a-planesource ↗

The lines $l_{1}$ and $l_{2}$ have the following vector equations where $λ, μ \in ℝ$ and $m \in ℝ$ .

$l_{1} : r_{1} = (\begin{matrix} 3 \\ - 2 \\ 0 \end{matrix}) + λ (\begin{matrix} 2 \\ 1 \\ m \end{matrix}) l_{2} : r_{2} = (\begin{matrix} - 1 \\ - 4 \\ - 2 m \end{matrix}) + μ (\begin{matrix} 2 \\ - 5 \\ - m \end{matrix})$

The plane $Π$ has Cartesian equation $x + 4 y - z = p$ where $p \in ℝ$ .

Given that $l_{1}$ and $Π$ have no points in common, find

Show that $l_{1}$ and $l_{2}$ are never perpendicular to each other.

[3]

the value of $m$ .

[2]

b.i.

the condition on the value of $p$ .

[2]

b.ii.

Markscheme / solution

* 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 $(\begin{matrix} 2 \\ 1 \\ m \end{matrix}) \cdot (\begin{matrix} 2 \\ - 5 \\ - m \end{matrix})$ (M1)

$= - 1 - m^{2}$ A1

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

so $l_{1}$ and $l_{2}$ are never perpendicular to each other AG

[3 marks]

(since $l_{1}$ is parallel to $Π$ , $l_{1}$ is perpendicular to the normal of $Π$ and so)

$(\begin{matrix} 2 \\ 1 \\ m \end{matrix}) \cdot (\begin{matrix} 1 \\ 4 \\ - 1 \end{matrix}) = 0$ R1

$2 + 4 - m = 0$

$m = 6$ A1

[2 marks]

b.i.

since there are no points in common, $(3, - 2, 0)$ does not lie in $Π$

EITHER

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

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

THEN

$p \neq - 5$ A1

[2 marks]

b.ii.

Examiners’ report

[N/A]

b.i.

[N/A]

b.ii.