Solved: The points $A(5, -2, 5)$, $B(5, 4, -1)$, $C(-1, -2, -1)$ and $D(7, -4, -3)$ are... [IB Mathematics Analysis and Approaches (AA)]

Question

HLPaper 2

The points $A(5, -2, 5)$ , $B(5, 4, -1)$ , $C(-1, -2, -1)$ and $D(7, -4, -3)$ are the vertices of a right-pyramid.The line $L$ passes through the point $D$ and is perpendicular to plane $\Phi$

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

[2]

Verified

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.

$\overrightarrow{AB} = (0, 6, -6) = 6(0, 1, -1)$ A1 $\overrightarrow{AC} = (-6, 0, -6) = 6(-1, 0, -1)$ A1

[2 marks]

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

[3]

Verified

Solution

attempts to find a vector normal to ${\Phi}$ M1 for example, ${\overrightarrow{AB}} \times {\overrightarrow{AC}} = \begin{pmatrix}-36\\ 36\\ 36\end{pmatrix} = 36\begin{pmatrix}-1\\ 1\\ 1\end{pmatrix}$ leading to A1 a vector normal to ${\Phi}$ is ${\mathbf{n}} = \begin{pmatrix}-1\\ 1\\ 1\end{pmatrix}$

EITHER substitutes ${\begin{pmatrix}5\\ -2\\ -5\end{pmatrix}}$ (or ${\begin{pmatrix}5\\ 4\\ -1\end{pmatrix}}$ or ${\begin{pmatrix}-1\\ -2\\ -1\end{pmatrix}}$ ) into ${-x+y+z=d}$ and attempts to find the value of ${d}$ for example, ${d=-5-2+5 = -2}$ M1

OR attempts to use ${\mathbf{r} \cdot \mathbf{n} = \mathbf{a} \cdot \mathbf{n}}$ M1 for example, ${\begin{pmatrix}x\\ y\\ z\end{pmatrix}} \cdot \begin{pmatrix}-1\\ 1\\ 1\end{pmatrix} = \begin{pmatrix}5\\ -2\\ 5\end{pmatrix} \cdot \begin{pmatrix}-1\\ 1\\ 1\end{pmatrix}$

THEN leading to the Cartesian equation of ${\Phi}$ as ${-x+y+z=-2}$ AG

[3 marks]

Find a vector equation of the line $L$ .

[1]

Verified

Solution

$r = \left( \begin{array}{ccc} 7 \\ -4 \\ -3 \end{array} \right) + \lambda \left( \begin{array}{ccc} -1 \\ 1 \\ 1 \end{array} \right)$ , $\lambda \in \mathbb{R}$

[1 mark]

Hence determine the minimum distance, $d_{\text{min}}$ , from $D$ to $\Phi$ .

[4]

Verified

Solution

The normal to the plane is

$\mathbf{n}=\begin{pmatrix} -1 \\ 1 \\ 1 \end{pmatrix}$

substitutes ${x=7-\lambda,y=-4+\lambda,z=-3+\lambda}$ into ${-x+y+z=-2}$ (M1) $-(7-\lambda)+(-4+\lambda)+(-3+\lambda)=-2$ $3\lambda=12$ A1 shows a correct calculation for finding ${\left|4\begin{pmatrix}-1\\1\\1\end{pmatrix}\right|}$ M1 $d_{\text{min}}=4\sqrt{3}$ A1

[4 marks]

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

[3]

Verified

Solution

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

$= \frac{{|-6 \quad 6 \quad -6| \cdot |-6 \quad 0 \quad -6|}}{{\sqrt{72} \times \sqrt{72}}}$

(A1) $= \frac{1}{2}$ (A1) so $B\hat A C = 60°$ AG

[3 marks]

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

[4]

Verified

Solution

let the area of triangle $ABC$ be $A$

EITHER attempts to find $A = \frac{1}{2}\left|\vec{AB} \times \vec{AC}\right|$ , for example M1 $A = \frac{1}{2} \left|\begin{pmatrix} -36 \\ 36 \\ 36 \end{pmatrix}\right|$

OR attempts to find $\frac{1}{2}\left|\vec{AB}\right|\left|\vec{AC}\right| \sin \theta$ , for example M1 $A = \frac{1}{2} \times 6 \sqrt{2} \times 6 \sqrt{2} \times \frac{\sqrt{3}}{2}$ (where $\sin \frac{\pi}{3} = \frac{\sqrt{3}}{2}$ )

THEN $A = 18 \sqrt{3}$ A1 uses $V = \frac{1}{3}A h$ where $A$ is the area of triangle $ABC$ and $h = d_{\text{min}}$ M1 $V = \frac{1}{3} \times 18 \sqrt{3} \times 4 \sqrt{3}$ $= 72$ A1

[4 marks]

The points $A(5, -2, 5)$, $B(5, 4, -1)$, $C(-1, -2, -1)$ and $D(7, -4, -3)$ are the vertices of a right-pyramid.The line $L$ passes through the point $D$ and is perpendicular to plane $\Phi$

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