AS June 2022 Paper 1 Q2
2
(a) Show that the vector \(\mathbf{i} + 4\mathbf{j} + 2\mathbf{k}\) is parallel to the plane \(2x + y - 3z = 10\). [3]
(b) Determine the acute angle between the planes \(2x + y - 3z = 10\) and \(x - y - 3z = 3\). [4]
| Scheme | Marks | AO |
|---|---|---|
| \((\mathbf{i} + 4\mathbf{j} + 2\mathbf{k}).(2\mathbf{i} + \mathbf{j} - 3\mathbf{k})\) | M1 | 3.1a |
| \(= 1 \times 2 + 4 \times 1 + 2 \times (-3) = 0\) | A1 | 1.1 |
| so \(\mathbf{i} + 4\mathbf{j} + 2\mathbf{k}\) is perpendicular to normal to the plane, hence parallel to the plane | A1 | 3.2a |
| [3] |
Notes
M1: scalar product of \((\mathbf{i} + 4\mathbf{j} + 2\mathbf{k})\) and \((2\mathbf{i} + \mathbf{j} - 3\mathbf{k})\)
A1: (1st) \(= 0\)
A1: (2nd) justification given
(corrected from the printed mark scheme: the last line printed “so \(2\mathbf{i} + \mathbf{j} - 3\mathbf{k}\) is perpendicular to normal to the plane”; \(2\mathbf{i} + \mathbf{j} - 3\mathbf{k}\) is the normal itself, so this should be \(\mathbf{i} + 4\mathbf{j} + 2\mathbf{k}\))
| Scheme | Marks | AO |
|---|---|---|
| \(\cos\theta = \dfrac{(2\mathbf{i} + \mathbf{j} - 3\mathbf{k}).(\mathbf{i} - \mathbf{j} - 3\mathbf{k})}{\sqrt{2^2 + 1^2 + (-3)^2}\sqrt{1^2 + (-1)^2 + (-3)^2}}\) | M1 M1 | 1.1a 1.1 |
| \(= \dfrac{2 - 1 + 9}{\sqrt{14}\sqrt{11}}\ \left[= \dfrac{10}{\sqrt{14}\sqrt{11}}\right]\) | A1 | 1.1 |
| \(\theta = 36.3^\circ\) | A1 | 1.1 |
| [4] |
Notes
M1: (1st) finding angle between normals
M1: (2nd) formula correct
A1: (2nd) \(36^\circ\) or 0.63 rads or better