AS October 2021 Paper 1 Q3
3 Three planes have the following equations.
\[\begin{aligned} 2x - 3y + z &= -3, \\ x - 4y + 2z &= 1, \\ -3x - 2y + 3z &= 14. \end{aligned}\](a)
(i) Write the system of equations in matrix form. [1]
(ii) Hence find the point of intersection of the planes. [2]
(b) In this question you must show detailed reasoning.
Find the acute angle between the planes \(2x - 3y + z = -3\) and \(x - 4y + 2z = 1\). [4]
Find the acute angle between the planes \(2x - 3y + z = -3\) and \(x - 4y + 2z = 1\). [4]
| Scheme | Marks | AO |
|---|---|---|
| (i) \(\begin{pmatrix} 2 & -3 & 1 \\ 1 & -4 & 2 \\ -3 & -2 & 3 \end{pmatrix}\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} -3 \\ 1 \\ 14 \end{pmatrix}\) | B1 | 2.5 |
| [1] | ||
| (ii) \([\begin{pmatrix} x \\ y \\ z \end{pmatrix} =]\ \begin{pmatrix} 2 & -3 & 1 \\ 1 & -4 & 2 \\ -3 & -2 & 3 \end{pmatrix}^{-1}\begin{pmatrix} -3 \\ 1 \\ 14 \end{pmatrix}\) | M1 | 1.1 |
| \(\Rightarrow x = -1,\ y = 2,\ z = 5\) | A1 | 1.1 |
| [2] |
Notes
(a)(i)
B1: allow 1 slip
(a)(ii)
M1: or \(\mathbf{M}^{-1}\begin{pmatrix} -3 \\ 1 \\ 14 \end{pmatrix}\) soi
A1: BC allow unsupported answers
| Scheme | Marks | AO |
|---|---|---|
| DR Normal vectors are \(2\mathbf{i} - 3\mathbf{j} + \mathbf{k}\) and \(\mathbf{i} - 4\mathbf{j} + 2\mathbf{k}\) | M1 | 1.1a |
| \(\cos\theta = \dfrac{2 \times 1 + (-3) \times (-4) + 1 \times 2}{\sqrt{2^2 + (-3)^2 + 1^2} \times \sqrt{1^2 + (-4)^2 + 2^2}}\) | M1 | 1.1 |
| \(= \dfrac{16}{\sqrt{14}\sqrt{21}}\) | A1 | 1.1 |
| \(\theta = 21.1^\circ\) | A1 | 1.1 |
| [4] |
Notes
M1: (1st) soi
A1: (1st) \((0.9331\ldots)\)
A1: (2nd) or 0.368 rad, \(21^\circ\) or 0.37 rad or better