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Date	May 2022	Marks available	4	Reference code	22M.1.AHL.TZ1.11
Level	Additional Higher Level	Paper	Paper 1 (without calculator)	Time zone	Time zone 1
Command term	Show that	Question number	11	Adapted from	N/A

Question

Consider the three planes

$\prod_{1} : 2 x - y + z = 4$

$\prod_{2} : x - 2 y + 3 z = 5$

$\prod_{3} : - 9 x + 3 y - 2 z = 32$

Show that the three planes do not intersect.

[4]

Verify that the point $P (1, - 2, 0)$ lies on both $\prod_{1}$ and $\prod_{2}$ .

[1]

b.i.

Find a vector equation of $L$ , the line of intersection of $\prod_{1}$ and $\prod_{2}$ .

[4]

b.ii.

Find the distance between $L$ and $\prod_{3}$ .

[6]

Markscheme

METHOD 1

attempt to eliminate a variable M1

obtain a pair of equations in two variables

EITHER

$- 3 x + z = - 3$ and A1

$- 3 x + z = 44$ A1

$- 5 x + y = - 7$ and A1

$- 5 x + y = 40$ A1

$3 x - z = 3$ and A1

$3 x - z = - \frac{79}{5}$ A1

THEN

the two lines are parallel ( $- 3 \neq 44$ or $- 7 \neq 40$ or $3 \neq - \frac{79}{5}$ ) R1

Note: There are other possible pairs of equations in two variables.
To obtain the final R1, at least the initial M1 must have been awarded.

hence the three planes do not intersect AG

METHOD 2

vector product of the two normals $= (\begin{matrix} - 1 \\ - 5 \\ - 3 \end{matrix})$ (or equivalent) A1

$r = (\begin{matrix} 1 \\ - 2 \\ 0 \end{matrix}) + λ (\begin{matrix} 1 \\ 5 \\ 3 \end{matrix})$ (or equivalent) A1

Note: Award A0 if “ $r =$ ” is missing. Subsequent marks may still be awarded.

Attempt to substitute $(1 + λ, - 2 + 5 λ, 3 λ)$ in $\prod_{3}$ M1

$- 9 (1 + λ) + 3 (- 2 + 5 λ) - 2 (3 λ) = 32$

$- 15 = 32$ , a contradiction R1

hence the three planes do not intersect AG

METHOD 3

attempt to eliminate a variable M1

$- 3 y + 5 z = 6$ A1

$- 3 y + 5 z = 100$ A1

$0 = 94$ , a contradiction R1

Note: Accept other equivalent alternatives. Accept other valid methods.
To obtain the final R1, at least the initial M1 must have been awarded.

hence the three planes do not intersect AG

[4 marks]

$\prod_{1} : 2 + 2 + 0 = 4$ and $\prod_{2} : 1 + 4 + 0 = 5$ A1

[1 mark]

b.i.

METHOD 1

attempt to find the vector product of the two normals M1

$(\begin{matrix} 2 \\ - 1 \\ 1 \end{matrix}) \times (\begin{matrix} 1 \\ - 2 \\ 3 \end{matrix})$

$= (\begin{matrix} - 1 \\ - 5 \\ - 3 \end{matrix})$ A1

$r = (\begin{matrix} 1 \\ - 2 \\ 0 \end{matrix}) + λ (\begin{matrix} 1 \\ 5 \\ 3 \end{matrix})$ A1A1

Note: Award A1A0 if “ $r =$ ” is missing.
Accept any multiple of the direction vector.
Working for (b)(ii) may be seen in part (a) Method 2. In this case penalize lack of “ $r =$ ” only once.

METHOD 2

attempt to eliminate a variable from $\prod_{1}$ and $\prod_{2}$ M1

$3 x - z = 3$ OR $3 y - 5 z = - 6$ OR $5 x - y = 7$

Let $x = t$

substituting $x = t$ in $3 x - z = 3$ to obtain

$z = - 3 + 3 t$ and $y = 5 t - 7$ (for all three variables in parametric form) A1

$r = (\begin{matrix} 0 \\ - 7 \\ - 3 \end{matrix}) + λ (\begin{matrix} 1 \\ 5 \\ 3 \end{matrix})$ A1A1

Note: Award A1A0 if “ $r =$ ” is missing.
Accept any multiple of the direction vector. Accept other position vectors which satisfy both the planes $\prod_{1}$ and $\prod_{2}$ .

[4 marks]

b.ii.

METHOD 1

the line connecting $L$ and $\prod_{3}$ is given by $L_{1}$

attempt to substitute position and direction vector to form $L_{1}$ (M1)

$s = (\begin{matrix} 1 \\ - 2 \\ 0 \end{matrix}) + t (\begin{matrix} - 9 \\ 3 \\ - 2 \end{matrix})$ A1

substitute $(1 - 9 t, - 2 + 3 t, - 2 t)$ in $\prod_{3}$ M1

$- 9 (1 - 9 t) + 3 (- 2 + 3 t) - 2 (- 2 t) = 32$

$94 t = 47 \Rightarrow t = \frac{1}{2}$ A1

attempt to find distance between $(1, - 2, 0)$ and their point $(- \frac{7}{2}, - \frac{1}{2}, - 1)$ (M1)

$= |(\begin{matrix} 1 \\ - 2 \\ 0 \end{matrix}) + \frac{1}{2} (\begin{matrix} - 9 \\ 3 \\ - 2 \end{matrix}) - (\begin{matrix} 1 \\ - 2 \\ 0 \end{matrix})| = \frac{1}{2} \sqrt{{(- 9)}^{2} + 3^{2} + {(- 2)}^{2}}$

$= \frac{\sqrt{94}}{2}$ A1

METHOD 2

unit normal vector equation of $\prod_{3}$ is given by $\frac{(\begin{matrix} - 9 \\ 3 \\ 2 \end{matrix}) \cdot (\begin{matrix} x \\ y \\ z \end{matrix})}{\sqrt{81 + 9 + 4}}$ (M1)

$= \frac{32}{\sqrt{94}}$ A1

let $\prod_{4}$ be the plane parallel to $\prod_{3}$ and passing through $P$ ,
then the normal vector equation of $\prod_{4}$ is given by

$(\begin{matrix} - 9 \\ 3 \\ 2 \end{matrix}) \cdot (\begin{matrix} x \\ y \\ z \end{matrix}) = (\begin{matrix} - 9 \\ 3 \\ 2 \end{matrix}) \cdot (\begin{matrix} 1 \\ - 2 \\ 0 \end{matrix}) = - 15$ M1

unit normal vector equation of $\prod_{4}$ is given by

$\frac{(\begin{matrix} - 9 \\ 3 \\ 2 \end{matrix}) \cdot (\begin{matrix} x \\ y \\ z \end{matrix})}{\sqrt{81 + 9 + 4}} = \frac{- 15}{\sqrt{94}}$ A1

distance between the planes is $\frac{32}{\sqrt{94}} - \frac{- 15}{\sqrt{94}}$ (M1)

$= \frac{47}{\sqrt{94}} (= \frac{\sqrt{94}}{2})$ A1

[6 marks]

Examiners report

Part (a) was well attempted using a variety of approaches. Most candidates were able to gain marks for part (a) through attempts to eliminate a variable with many subsequently making algebraic errors. Part (b)(i) was well done. For part (b)(ii) few successful attempts were noted, many candidates failed to use an appropriate notation "r =" while giving the vector equation of a line. Part (c) proved to be challenging for most candidates with very few correct answers seen. Many candidates did not attempt part (c).

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b.i.

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b.ii.

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