AS June 2019 Paper 1 Q4
4
(a) Find \(\mathbf{M}^{-1}\), where \(\mathbf{M} = \begin{pmatrix} 1 & 2 & 3 \\ -1 & 1 & 2 \\ -2 & 1 & 2 \end{pmatrix}\). [1]
(b) Hence find, in terms of the constant \(k\), the point of intersection of the planes\[\begin{aligned} x + 2y + 3z &= 19, \\ -x + y + 2z &= 4, \\ -2x + y + 2z &= k. \end{aligned}\][3]
(c) In this question you must show detailed reasoning.
Find the acute angle between the planes \(x + 2y + 3z = 19\) and \(-x + y + 2z = 4\). [4]
Find the acute angle between the planes \(x + 2y + 3z = 19\) and \(-x + y + 2z = 4\). [4]
| Scheme | Marks | AO |
|---|---|---|
| \(\mathbf{M}^{-1} = \begin{pmatrix} 0 & 1 & -1 \\ 2 & -8 & 5 \\ -1 & 5 & -3 \end{pmatrix}\) | B1 | 1.1 |
| [1] |
Notes
B1: BC
| Scheme | Marks | AO |
|---|---|---|
| \(\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 & 1 & -1 \\ 2 & -8 & 5 \\ -1 & 5 & -3 \end{pmatrix}\begin{pmatrix} 19 \\ 4 \\ k \end{pmatrix}\) | M1 | 1.1a |
| \(x = 4 - k,\ y = 6 + 5k,\ z = 1 - 3k\) | A2,1,0 | 1.1,1.1 |
| [3] |
Notes
A2,1,0: Allow SCB2 for correct answer found using elimination
condone \(\begin{pmatrix} 4 - k \\ 6 + 5k \\ 1 - 3k \end{pmatrix}\)
| Scheme | Marks | AO |
|---|---|---|
| DR normals are \(\mathbf{i} + 2\mathbf{j} + 3\mathbf{k}\) and \(-\mathbf{i} + \mathbf{j} + 2\mathbf{k}\) | B1 | 1.2 |
| \(\cos\theta = \dfrac{1 \times (-1) + 2 \times 1 + 3 \times 2}{\sqrt{1^2 + 2^2 + 3^2}\sqrt{(-1)^2 + 1^2 + 2^2}}\) | M1 | 3.1a |
| \(= \dfrac{7}{\sqrt{14}\sqrt{6}}\) | A1 | 1.1 |
| \(\Rightarrow \theta = 40.2^\circ\) or 0.702 rads | A1 | 1.1 |
| [4] |
Notes
B1: soi
A1: (1st) correct expression
oe eg \(\dfrac{\sqrt{21}}{6}\)
A1: (2nd) \(40^\circ\) or 0.70 or better
mark final answer