2020 H2 Maths Paper 1 Q1 — Worked Solution

Part (i)

2 marks

We find a normal to $\pi_{1}$ using the cross product of the two vectors lying in the plane.

\begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix} \times \begin{pmatrix} 1 \\ - 5 \\ - 2 \end{pmatrix}

= \begin{pmatrix} (1)( - 2) - (0)( - 5) \\ (0)(1) - (1)( - 2) \\ (1)( - 5) - (1)(1) \end{pmatrix}

= \begin{pmatrix} - 2 \\ 2 \\ - 6 \end{pmatrix}

Dividing by $- 2$ to simplify,

\begin{pmatrix} - 2 \\ 2 \\ - 6 \end{pmatrix} = - 2\begin{pmatrix} 1 \\ - 1 \\ 3 \end{pmatrix}

To check your answer:

\begin{pmatrix} 1 \\ - 1 \\ 3 \end{pmatrix} \cdot \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix} = 1 - 1 + 0 = 0

and

\begin{pmatrix} 1 \\ - 1 \\ 3 \end{pmatrix} \cdot \begin{pmatrix} 1 \\ - 5 \\ - 2 \end{pmatrix} = 1 + 5 - 6 = 0

, confirming the vector is perpendicular to both vectors in the plane. ✓

$\therefore$ a normal to $\pi_{1}$ is $\begin{pmatrix} 1 \\ - 1 \\ 3 \end{pmatrix}$ .

Part (ii)

2 marks

Approach

The acute angle between two planes equals the acute angle between their normal vectors, found with the dot-product formula $\cos\theta = \frac{\left| \mathbf{n}_{1} \cdot \mathbf{n}_{2} \right|}{\left| \mathbf{n}_{1} \right|\left| \mathbf{n}_{2} \right|}$ .

A normal to $\pi_{2}$ is $\begin{pmatrix} 4 \\ 5 \\ - 6 \end{pmatrix}$ , read directly from its cartesian equation. Let $\theta$ be the acute angle between $\pi_{1}$ and $\pi_{2}$ .

\cos\theta = \frac{\left| \begin{pmatrix} 1 \\ - 1 \\ 3 \end{pmatrix} \cdot \begin{pmatrix} 4 \\ 5 \\ - 6 \end{pmatrix} \right|}{\left| \begin{pmatrix} 1 \\ - 1 \\ 3 \end{pmatrix} \right|\left| \begin{pmatrix} 4 \\ 5 \\ - 6 \end{pmatrix} \right|}

We compute the dot product and the two magnitudes.

\begin{pmatrix} 1 \\ - 1 \\ 3 \end{pmatrix} \cdot \begin{pmatrix} 4 \\ 5 \\ - 6 \end{pmatrix} = 4 - 5 - 18 = - 19

\left| \begin{pmatrix} 1 \\ - 1 \\ 3 \end{pmatrix} \right| = \sqrt{11},\ \ \left| \begin{pmatrix} 4 \\ 5 \\ - 6 \end{pmatrix} \right| = \sqrt{77}

\cos\theta = \frac{19}{\sqrt{11} \times \sqrt{77}}

= \frac{19}{\sqrt{847}}

= \frac{19\sqrt{7}}{77}

\approx 0.6528

\theta \approx {49.2}^{\circ}\ \left( 1\ \text{d.p.} \right)

To check your answer:

\theta \approx {49.2}^{\circ}

lies between

0^{\circ}

and

90^{\circ}

, confirming it is the acute angle between the planes. ✓

$\therefore$ the acute angle between $\pi_{1}$ and $\pi_{2}$ is ${49.2}^{\circ}$ .

2020 H2 Maths Paper 1, Question 1

Part (i)

Part (ii)

This is how we mark, and how we teach.