A2 June 2021 Paper 1 Q11
11 The line \(L_1\) has equation \(\mathbf{r} = \begin{bmatrix}2 \\ 2 \\ 3\end{bmatrix} + \lambda\begin{bmatrix}2 \\ 3 \\ -1\end{bmatrix}\)
The line \(L_2\) has equation \(\mathbf{r} = \begin{bmatrix}6 \\ 4 \\ 1\end{bmatrix} + \mu\begin{bmatrix}-2 \\ 1 \\ 1\end{bmatrix}\)
(a) Find the acute angle between the lines \(L_1\) and \(L_2\), giving your answer to the nearest \(0.1^\circ\) [3 marks]
(b) The lines \(L_1\) and \(L_2\) lie in the plane \(\Pi_1\)
(i) Find the equation of \(\Pi_1\), giving your answer in the form \(\mathbf{r} \cdot \mathbf{n} = d\) [4 marks]
(ii) Hence find the shortest distance of the plane \(\Pi_1\) from the origin. [1 mark]
(c) The points \(A(4, -1, -1)\), \(B(1, 5, -7)\) and \(C(3, 4, -8)\) lie in the plane \(\Pi_2\)
Find the angle between the planes \(\Pi_1\) and \(\Pi_2\), giving your answer to the nearest \(0.1^\circ\) [4 marks]
| Scheme | Marks | AO |
|---|---|---|
| Obtains scalar (or vector) product of the direction vectors PI by seeing AWRT \(103^\circ\) | M1 | 1.1a |
| Divides their scalar product (or their magnitude of vector product) by product of the magnitudes PI by AWRT \(103^\circ\) | M1 | 1.1a |
| Deduces the correct angle, correct to at least 1dp | A1 | 2.2a |
| (3) |
Typical solution
\[\begin{bmatrix}2 \\ 3 \\ -1\end{bmatrix} \cdot \begin{bmatrix}-2 \\ 1 \\ 1\end{bmatrix} = -2\]Moduli of vectors are \(\sqrt{14}\) and \(\sqrt{6}\)
Let \(\alpha\) be angle between lines
\[\cos\alpha = \frac{-2}{\sqrt{14}\sqrt{6}} = \frac{-1}{\sqrt{21}}\]Angle between lines
\[= 180 - \alpha = 77.4^\circ\]| Scheme | Marks | AO |
|---|---|---|
| (i) Finds vector product of direction vectors | M1 | 3.1a |
| Obtains correct result (or multiple of it) | A1 | 1.1b |
| Takes scalar product of their normal vector \((\mathbf{n}_1)\) and a point in the plane | M1 | 2.2a |
| Obtains correct result in the correct form (or multiple of it) | A1 | 1.1b |
| (4) | ||
| (ii) Obtains the correct distance for their vector equation ACF | B1F | 2.2a |
| (1) |
Typical solution
(i)
\[\mathbf{n}_1 = \begin{bmatrix}2 \\ 3 \\ -1\end{bmatrix} \times \begin{bmatrix}-2 \\ 1 \\ 1\end{bmatrix} = 4\begin{bmatrix}1 \\ 0 \\ 2\end{bmatrix}\]\[d = \begin{bmatrix}2 \\ 2 \\ 3\end{bmatrix} \cdot \begin{bmatrix}1 \\ 0 \\ 2\end{bmatrix} = 8\]\[\mathbf{r} \cdot \begin{bmatrix}1 \\ 0 \\ 2\end{bmatrix} = 8\](ii)
\[\text{Distance to origin} = \frac{8}{\sqrt{5}}\]| Scheme | Marks | AO |
|---|---|---|
| Forms two direction vectors from the three points and identifies the need to take the cross product of them | M1 | 3.1a |
| Obtains correct result (or multiple of it) | A1 | 1.1b |
| Finds scalar (or vector) product of their normal vectors | M1 | 1.1a |
| Obtains correct angle (accept \(129.2^\circ\)) | A1 | 1.1b |
| (4) | ||
| (12 marks) |
Typical solution
\[\overrightarrow{AB} = \begin{bmatrix}-3 \\ 6 \\ -6\end{bmatrix} = -3\begin{bmatrix}1 \\ -2 \\ 2\end{bmatrix}, \quad \overrightarrow{AC} = \begin{bmatrix}-1 \\ 5 \\ -7\end{bmatrix}\]\[\mathbf{n}_2 = \begin{bmatrix}1 \\ -2 \\ 2\end{bmatrix} \times \begin{bmatrix}-1 \\ 5 \\ -7\end{bmatrix} = \begin{bmatrix}4 \\ 5 \\ 3\end{bmatrix}\]Let \(\beta\) be angle between planes
\[\cos\beta = \frac{\begin{bmatrix}1 \\ 0 \\ 2\end{bmatrix} \cdot \begin{bmatrix}4 \\ 5 \\ 3\end{bmatrix}}{\sqrt{5}\sqrt{50}} = \frac{2}{\sqrt{10}}\]\[\beta = 50.8^\circ\]