IB Questionbank

User interface language: English | Español

Date	November Example questions	Marks available	1	Reference code	EXN.2.AHL.TZ0.11
Level	Additional Higher Level	Paper	Paper 2	Time zone	Time zone 0
Command term	Find	Question number	11	Adapted from	N/A

Question

The points $A (5, - 2, 5)$ $A (5, - 2, 5)$ , $B (5, 4, - 1)$ $B (5, 4, - 1)$ , $C (- 1, - 2, - 1)$ $C (- 1, - 2, - 1)$ and $D (7, - 4, - 3)$ $D (7, - 4, - 3)$ are the vertices of a right-pyramid.

The line $L$ $L$ passes through the point $D$ $D$ and is perpendicular to $Π$ $Π$ .

Find the vectors $- - \to AB$ $\vec{AB}$ and $- - \to AC$ $\vec{AC}$ .

[2]

Use a vector method to show that $B ˆ A C = 60 °$ $B \hat{A} C = 60 °$ .

[3]

Show that the Cartesian equation of the plane $Π$ $Π$ that contains the triangle $ABC$ $ABC$ is $- x + y + z = - 2$ $- x + y + z = - 2$ .

[3]

Find a vector equation of the line $L$ $L$ .

[1]

d.i.

Hence determine the minimum distance, $d_{min}$ $d_{min}$ , from $D$ $D$ to $Π$ $Π$ .

[4]

d.ii.

Find the volume of right-pyramid $ABCD$ $ABCD$ .

[4]

Markscheme

* 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.

$- - \to AB = ⎛ ⎜ ⎝ \begin{matrix} 06 - 6 \end{matrix} ⎞ ⎟ ⎠ ⎛ ⎜ ⎝ = 6 ⎛ ⎜ ⎝ \begin{matrix} 01 - 1 \end{matrix} ⎞ ⎟ ⎠ ⎞ ⎟ ⎠$ $\vec{AB} = (\begin{matrix} 0 \\ 6 \\ - 6 \end{matrix}) (= 6 (\begin{matrix} 0 \\ 1 \\ - 1 \end{matrix}))$ A1

$- - \to AC = ⎛ ⎜ ⎝ \begin{matrix} - 6 0 - 6 \end{matrix} ⎞ ⎟ ⎠ ⎛ ⎜ ⎝ = 6 ⎛ ⎜ ⎝ \begin{matrix} - 1 0 - 1 \end{matrix} ⎞ ⎟ ⎠ ⎞ ⎟ ⎠$ $\vec{AC} = (\begin{matrix} - 6 \\ 0 \\ - 6 \end{matrix}) (= 6 (\begin{matrix} - 1 \\ 0 \\ - 1 \end{matrix}))$ A1

[2 marks]

attempts to use $cos B ˆ A C = \frac{- - \to AB \cdot - - \to AC}{∣ ∣ ∣ - - \to AB ∣ ∣ ∣ ∣ ∣ ∣ - - \to AC ∣ ∣ ∣}$ $\cos B \hat{A} C = \frac{\vec{AB} \cdot \vec{AC}}{|\vec{AB}| |\vec{AC}|}$ (M1)

$= \frac{⎛ ⎜ ⎝ \begin{matrix} 06 - 6 \end{matrix} ⎞ ⎟ ⎠ \cdot ⎛ ⎜ ⎝ \begin{matrix} - 6 0 - 6 \end{matrix} ⎞ ⎟ ⎠}{\sqrt{72} \times \sqrt{72}}$ $= \frac{(\begin{matrix} 0 \\ 6 \\ - 6 \end{matrix}) \cdot (\begin{matrix} - 6 \\ 0 \\ - 6 \end{matrix})}{\sqrt{72} \times \sqrt{72}}$ A1

$= \frac{1}{2}$ $= \frac{1}{2}$ A1

so $B ˆ A C = 60 °$ $B \hat{A} C = 60 °$ AG

[3 marks]

attempts to find a vector normal to $Π$ $Π$ M1

for example, $- - \to AB \times - - \to AC = ⎛ ⎜ ⎝ \begin{matrix} - 36 3636 \end{matrix} ⎞ ⎟ ⎠ ⎛ ⎜ ⎝ = 36 ⎛ ⎜ ⎝ \begin{matrix} - 1 11 \end{matrix} ⎞ ⎟ ⎠ ⎞ ⎟ ⎠$ $\vec{AB} \times \vec{AC} = (\begin{matrix} - 36 \\ 36 \\ 36 \end{matrix}) (= 36 (\begin{matrix} - 1 \\ 1 \\ 1 \end{matrix}))$ leading to A1

a vector normal to $Π$ $Π$ is $n = (\begin{matrix} - 1 \\ 1 \\ 1 \end{matrix})$

EITHER

substitutes $(5, - 2, - 5)$ (or $(5, 4, - 1)$ or $(- 1, - 2, - 1)$ ) into $- x + y + z = d$ and attempts to find the value of $d$

for example, $d = - 5 - 2 + 5 (= - 2)$ M1

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

for example, $(\begin{matrix} x \\ y \\ z \end{matrix}) \cdot (\begin{matrix} - 1 \\ 1 \\ 1 \end{matrix}) = (\begin{matrix} 5 \\ - 2 \\ 5 \end{matrix}) \cdot (\begin{matrix} - 1 \\ 1 \\ 1 \end{matrix})$