A2 June 2019 Paper 1 Q5
5 A plane has equation \(\mathbf{r}.\begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} = 7\)
A line has equation \(\mathbf{r} = \begin{bmatrix} 2 \\ 0 \\ 1 \end{bmatrix} + \mu\begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix}\)
Calculate the acute angle between the line and the plane.
Give your answer to the nearest \(0.1^\circ\) [3 marks]
| Scheme | Marks | AO |
|---|---|---|
| Finds scalar (or vector) product of the correct vectors PI by seeing AWRT \(35^\circ\) | M1 | 1.1a |
| Divides their scalar product (or magnitude of vector product) of their vectors by product of their magnitudes PI by seeing AWRT \(35^\circ\) | M1 | 1.1a |
| Deduces the correct angle, correct to at least 1dp | A1 | 2.2a |
| (3 marks) |
Typical solution
\[\begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix}\cdot\begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} = 2\]Moduli of vectors are \(\sqrt{2}\) and \(\sqrt{3}\)
Let \(\alpha\) be angle between normal & line
\[\cos\alpha = \frac{2}{\sqrt{6}}\]Angle between plane & line
\[= 90 - \alpha = 54.7^\circ\]