2018 H2 Maths Paper 2 Q3 — Worked Solution

Part (i)

1 mark

Since $\text{ABCD}$ is a parallelogram, the diagonals $AC$ and $BD$ bisect each other, so the midpoint of $AC$ equals the midpoint of $BD$ .

\frac{1}{2}\left( \overrightarrow{OA} + \overrightarrow{OC} \right) = \frac{1}{2}\left( \overrightarrow{OB} + \overrightarrow{OD} \right)

\Rightarrow \overrightarrow{OD} = \overrightarrow{OA} + \overrightarrow{OC} - \overrightarrow{OB}

= \begin{pmatrix} 5 \\ - 4 \\ 1 \end{pmatrix} + \begin{pmatrix} - 5 \\ 4 \\ 2 \end{pmatrix} - \begin{pmatrix} 5 \\ 4 \\ 0 \end{pmatrix}

= \begin{pmatrix} - 5 \\ - 4 \\ 3 \end{pmatrix}

$\therefore$ $D = ( - 5,\ - 4,\ 3)$ .

Part (ii)

3 marks

Approach

The plane $\text{BCE}$ is determined by the two points $B$ , $C$ and the apex $E$ . We find two direction vectors in the plane, take their cross product for a normal, then use one point to fix the constant.

\overrightarrow{BC} = \begin{pmatrix} - 5 \\ 4 \\ 2 \end{pmatrix} - \begin{pmatrix} 5 \\ 4 \\ 0 \end{pmatrix} = \begin{pmatrix} - 10 \\ 0 \\ 2 \end{pmatrix},\ \ \overrightarrow{BE} = \begin{pmatrix} 0 \\ 0 \\ 10 \end{pmatrix} - \begin{pmatrix} 5 \\ 4 \\ 0 \end{pmatrix} = \begin{pmatrix} - 5 \\ - 4 \\ 10 \end{pmatrix}

A normal to the plane is

\overrightarrow{n} = \overrightarrow{BC} \times \overrightarrow{BE}

= \begin{pmatrix} - 10 \\ 0 \\ 2 \end{pmatrix} \times \begin{pmatrix} - 5 \\ - 4 \\ 10 \end{pmatrix}

= \begin{pmatrix} (0)(10) - (2)( - 4) \\ (2)( - 5) - ( - 10)(10) \\ ( - 10)( - 4) - (0)( - 5) \end{pmatrix}

= \begin{pmatrix} 8 \\ 90 \\ 40 \end{pmatrix}

Dividing by the common factor $2$ , we may take $\overrightarrow{n} = \begin{pmatrix} 4 \\ 45 \\ 20 \end{pmatrix}$ . Using point $B$ ,

\overrightarrow{n} \cdot \overrightarrow{OB} = 4(5) + 45(4) + 20(0) = 200

$\therefore$ the cartesian equation of face $\text{BCE}$ is $4x + 45y + 20z = 200$ .

To check your answer: substituting

C( - 5,4,2)

:

4( - 5) + 45(4) + 20(2) = - 20 + 180 + 40 = 200

. ✓

Part (iii)

3 marks

Approach

The angle between the two planes equals the angle between their normal vectors (taking the acute angle). We already have the normal to face $\text{BCE}$ ; we now find a normal to the base $\text{ABCD}$ using $\overrightarrow{AB}$ and $\overrightarrow{AD}$ , then apply the formula for the angle between two planes.

\overrightarrow{AB} = \begin{pmatrix} 0 \\ 8 \\ - 1 \end{pmatrix},\ \ \overrightarrow{AD} = \begin{pmatrix} - 10 \\ 0 \\ 2 \end{pmatrix}

A normal to the base is

\overrightarrow{m} = \overrightarrow{AB} \times \overrightarrow{AD} = \begin{pmatrix} (8)(2) - ( - 1)(0) \\ ( - 1)( - 10) - (0)(2) \\ (0)(0) - (8)( - 10) \end{pmatrix} = \begin{pmatrix} 8 \\ 5 \\ 40 \end{pmatrix}

Let $\theta$ be the angle between the two planes. Using the normals $\overrightarrow{n} = \begin{pmatrix} 4 \\ 45 \\ 20 \end{pmatrix}$ and $\overrightarrow{m} = \begin{pmatrix} 8 \\ 5 \\ 40 \end{pmatrix}$ ,

\cos\theta = \frac{\left| \overrightarrow{n} \cdot \overrightarrow{m} \right|}{\left| \overrightarrow{n} \right|\left| \overrightarrow{m} \right|}

= \frac{\left| 4(8) + 45(5) + 20(40) \right|}{\sqrt{4^{2} + 45^{2} + 20^{2}}\sqrt{8^{2} + 5^{2} + 40^{2}}}

= \frac{1057}{\sqrt{2441}\sqrt{1689}}

By GC, $\theta = {58.6}^{\circ}$ (1 d.p.).

$\therefore$ the angle between face $\text{BCE}$ and the base is ${58.6}^{\circ}$ .

Part (iv)

5 marks

Approach

The midpoint of $AD$ is a point not on face $\text{BCE}$ ; its shortest distance to that plane is the perpendicular distance, found directly from the cartesian equation of the plane using the normal vector from part (ii).

Let $M$ be the midpoint of $AD$ .

M = \frac{1}{2}\left( \begin{pmatrix} 5 \\ - 4 \\ 1 \end{pmatrix} + \begin{pmatrix} - 5 \\ - 4 \\ 3 \end{pmatrix} \right) = \begin{pmatrix} 0 \\ - 4 \\ 2 \end{pmatrix}

Using the plane $4x + 45y + 20z = 200$ with normal $\overrightarrow{n} = \begin{pmatrix} 4 \\ 45 \\ 20 \end{pmatrix}$ , $\left| \overrightarrow{n} \right| = \sqrt{2441}$ ,

\text{Distance} = \frac{\left| 4(0) + 45( - 4) + 20(2) - 200 \right|}{\sqrt{2441}}

= \frac{| - 180 + 40 - 200|}{\sqrt{2441}}

= \frac{340}{\sqrt{2441}}

By GC, this is $6.88$ (3 s.f.).

$\therefore$ the shortest distance from the midpoint of $AD$ to face $\text{BCE}$ is $\frac{340}{\sqrt{2441}} = 6.88\ \text{units}$ .

2018 H2 Maths Paper 2, Question 3

Part (i)

Part (ii)

Part (iii)

Part (iv)

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