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Date	November 2021	Marks available	2	Reference code	21N.2.AHL.TZ0.11
Level	Additional Higher Level	Paper	Paper 2	Time zone	Time zone 0
Command term	Show that	Question number	11	Adapted from	N/A

Question

Three points $A (3, 0, 0), B (0, - 2, 0)$ and $C (1, 1, - 7)$ lie on the plane $Π_{1}$ .

Plane $Π_{2}$ has equation $3 x - y + 2 z = 2$ .

The plane $Π_{3}$ is given by $2 x - 2 z = 3$ . The line $L$ and the plane $Π_{3}$ intersect at the point $P$ .

The point $B (0, - 2, 0)$ lies on $L$ .

Find the vector $\vec{AB}$ and the vector $\vec{AC}$ .

[2]

a.i.

Hence find the equation of $Π_{1}$ , expressing your answer in the form $a x + b y + c z = d$ , where $a, b, c, d \in ℤ$ .

[5]

a.ii.

The line $L$ is the intersection of $Π_{1}$ and $Π_{2}$ . Verify that the vector equation of $L$ can be written as $r = (\begin{matrix} 0 \\ - 2 \\ 0 \end{matrix}) + λ (\begin{matrix} 1 \\ 1 \\ - 1 \end{matrix})$ .

[2]

Show that at the point $P, λ = \frac{3}{4}$ .

[2]

c.i.

Hence find the coordinates of $P$ .

[1]

c.ii.

Find the reflection of the point $B$ in the plane $Π_{3}$ .

[7]

d.i.

Hence find the vector equation of the line formed when $L$ is reflected in the plane $Π_{3}$ .

[2]

d.ii.

Markscheme

attempts to find either $\vec{AB}$ or $\vec{AC}$ (M1)

$\vec{AB} = (\begin{matrix} - 3 \\ - 2 \\ 0 \end{matrix})$ and $\vec{AC} = (\begin{matrix} - 2 \\ 1 \\ - 7 \end{matrix})$ A1

[2 marks]

a.i.

METHOD 1

attempts to find $\vec{AB} \times \vec{AC}$ (M1)

$\vec{AB} \times \vec{AC} = (\begin{matrix} 14 \\ - 21 \\ - 7 \end{matrix})$ A1

EITHER

equation of plane is of the form $14 x - 21 y - 7 z = d (2 x - 3 y - z = d)$ (A1)

substitutes a valid point e.g $(3, 0, 0)$ to obtain a value of $d$ M1

$d = 42 (d = 6)$

attempts to use $r \cdot n = a \cdot n$ (M1)

$r \cdot (\begin{matrix} 14 \\ - 21 \\ - 7 \end{matrix}) = (\begin{matrix} 3 \\ 0 \\ 0 \end{matrix}) \cdot (\begin{matrix} 14 \\ - 21 \\ - 7 \end{matrix}) (r \cdot (\begin{matrix} 14 \\ - 21 \\ - 7 \end{matrix}) = 42)$ A1

$r \cdot (\begin{matrix} 2 \\ - 3 \\ - 1 \end{matrix}) = (\begin{matrix} 3 \\ 0 \\ 0 \end{matrix}) \cdot (\begin{matrix} 2 \\ - 3 \\ - 1 \end{matrix}) (r \cdot (\begin{matrix} 2 \\ - 3 \\ - 1 \end{matrix}) = 6)$

THEN

$14 x - 21 y - 7 z = 42 (2 x - 3 y - z = 6)$ A1

METHOD 2

equation of plane is of the form $(\begin{matrix} x \\ y \\ z \end{matrix}) = (\begin{matrix} 3 \\ 0 \\ 0 \end{matrix}) + s (\begin{matrix} - 3 \\ - 2 \\ 0 \end{matrix}) + t (\begin{matrix} - 2 \\ 1 \\ - 7 \end{matrix})$ A1

attempts to form equations for $x, y, z$ in terms of their parameters (M1)

$x = 3 - 3 s - 2 t, y = - 2 s + t, z = - 7 t$ A1

eliminates at least one of their parameters (M1)

for example, $2 x - 3 y = 6 - 7 t (\Rightarrow 2 x - 3 y = 6 + z)$

$2 x - 3 y - z = 6$ A1

[5 marks]

a.ii.

METHOD 1

substitutes $r = (\begin{matrix} 0 \\ - 2 \\ 0 \end{matrix}) + λ (\begin{matrix} 1 \\ 1 \\ - 1 \end{matrix})$ into their $Π_{1}$ and $Π_{2}$ (given) (M1)

$Π_{1} : 2 λ - 3 (- 2 + λ) - (- λ) = 6$ and $Π_{2} : 3 λ - 3 (- 2 + λ) + 2 (- λ) = 2$ A1

Note: Award (M1)A0 for correct verification using a specific value of $λ$ .

so the vector equation of $L$ can be written as $r = (\begin{matrix} 0 \\ - 2 \\ 0 \end{matrix}) + λ (\begin{matrix} 1 \\ 1 \\ - 1 \end{matrix})$ AG

METHOD 2

EITHER

attempts to find $(\begin{matrix} 2 \\ - 3 \\ - 1 \end{matrix}) \times (\begin{matrix} 3 \\ - 1 \\ 2 \end{matrix})$ M1

$= (\begin{matrix} - 7 \\ - 7 \\ 7 \end{matrix})$

$(\begin{matrix} 2 \\ - 3 \\ - 1 \end{matrix}) \cdot (\begin{matrix} 1 \\ 1 \\ - 1 \end{matrix}) = (2 - 3 + 1) = 0$ and $(\begin{matrix} 3 \\ - 1 \\ 2 \end{matrix}) \cdot (\begin{matrix} 1 \\ 1 \\ - 1 \end{matrix}) = (3 - 1 - 2) = 0$ M1

THEN

substitutes $(0, - 2, 0)$ into $Π_{1}$ and $Π_{2}$

$Π_{1} : 2 (0) - 3 (- 2) - (0) = 6$ and $Π_{2} : 3 (0) - (- 2) + 2 (0) = 2$ A1

so the vector equation of $L$ can be written as $r = (\begin{matrix} 0 \\ - 2 \\ 0 \end{matrix}) + λ (\begin{matrix} 1 \\ 1 \\ - 1 \end{matrix})$ AG

METHOD 3

attempts to solve $2 x - 3 y - z = 6$ and $3 x - y + 2 z = 2$ (M1)

for example, $x = - λ, y = - 2 - λ, z = λ$ A1

Note: Award A1 for substituting $x = 0$ (or $y = - 2$ or $z = 0$ ) into $Π_{1}$ and $Π_{2}$ and solving simultaneously. For example, solving $- 3 y - z = 6$ and $- y + 2 z = 2$ to obtain $y = - 2$ and $z = 0$ .

so the vector equation of $L$ can be written as $r = (\begin{matrix} 0 \\ - 2 \\ 0 \end{matrix}) + λ (\begin{matrix} 1 \\ 1 \\ - 1 \end{matrix})$ AG

[2 marks]

substitutes the equation of $L$ into the equation of $Π_{3}$ (M1)

$2 λ + 2 λ = 3 \Rightarrow 4 λ = 3$ A1

$λ = \frac{3}{4}$ AG

[2 marks]

c.i.

$P$ has coordinates $(\frac{3}{4}, - \frac{5}{4}, - \frac{3}{4})$ A1

[1 mark]

c.ii.

normal to $Π_{3}$ is $n = (\begin{matrix} 2 \\ 0 \\ - 2 \end{matrix})$ (A1)

Note: May be seen or used anywhere.

considers the line normal to $Π_{3}$ passing through $B (0, - 2, 0)$ (M1)
$r = (\begin{matrix} 0 \\ - 2 \\ 0 \end{matrix}) + μ (\begin{matrix} 2 \\ 0 \\ - 2 \end{matrix})$ A1

EITHER

finding the point on the normal line that intersects $Π_{3}$
attempts to solve simultaneously with plane $2 x - 2 z = 3$ (M1)

$4 μ + 4 μ = 3$

$μ = \frac{3}{8}$ A1

point is $(\frac{3}{4}, - 2, - \frac{3}{4})$

$((\begin{matrix} 2 μ \\ - 2 \\ - 2 μ \end{matrix}) - (\begin{matrix} \frac{3}{4} \\ - \frac{5}{4} \\ - \frac{3}{4} \end{matrix})) \cdot (\begin{matrix} 2 \\ 0 \\ - 2 \end{matrix}) = 0$ (M1)

$4 μ - \frac{3}{2} + 4 μ - \frac{3}{2} = 0$

$μ = \frac{3}{8}$ A1

attempts to find the equation of the plane parallel to $Π_{3}$ containing $B' (x - z = 3)$ and solve simultaneously with $L$ (M1)

$2 μ' + 2 μ' = 3$

$μ' = \frac{3}{4}$ A1

THEN

so, another point on the reflected line is given by

$r = (\begin{matrix} 0 \\ - 2 \\ 0 \end{matrix}) + \frac{3}{4} (\begin{matrix} 2 \\ 0 \\ - 2 \end{matrix})$ (A1)

$\Rightarrow B' (\frac{3}{2}, - 2, - \frac{3}{2})$ A1

[7 marks]

d.i.

EITHER

attempts to find the direction vector of the reflected line using their $P$ and $B'$ (M1)

$\vec{PB'} = (\begin{matrix} \frac{3}{4} \\ - \frac{3}{4} \\ - \frac{3}{4} \end{matrix})$

attempts to find their direction vector of the reflected line using a vector approach (M1)

$\vec{PB'} = \vec{PB} + \vec{BB'} = - \frac{3}{4} (\begin{matrix} 1 \\ 1 \\ - 1 \end{matrix}) + \frac{3}{2} (\begin{matrix} 1 \\ 0 \\ - 1 \end{matrix})$

THEN

$r = (\begin{matrix} \frac{3}{2} \\ - 2 \\ - \frac{3}{2} \end{matrix}) + λ (\begin{matrix} \frac{3}{4} \\ - \frac{3}{4} \\ - \frac{3}{4} \end{matrix})$ (or equivalent) A1