AS June 2025 Paper 1 Q2
2 In this question you must show detailed reasoning.
Find the acute angle between the vector \(3\mathbf{i} + 2\mathbf{j} - \mathbf{k}\) and the normal vector to the plane \(2x + 3y + z = 6\). [4]
| Scheme | Marks | AO |
|---|---|---|
| DR Normal to plane is \(2\mathbf{i} + 3\mathbf{j} + \mathbf{k}\) | M1 | 3.1a |
| \(\cos\theta = \dfrac{(3\mathbf{i} + 2\mathbf{j} - \mathbf{k}).(2\mathbf{i} + 3\mathbf{j} + \mathbf{k})}{\sqrt{3^2 + 2^2 + 1^2}\sqrt{2^2 + 3^2 + (-1)^2}}\) | M1 | 1.1 |
| \(= \dfrac{11}{14}\) | A1 | 1.1 |
| \(\theta = 38.2^\circ\) | A1 | 1.1 |
| [4] |
Notes
M1: soi
M1: angle between vectors and modulus formulae used
A1: 38.2 or better or 0.667 rads or better