AS June 2023 Paper 1 Q10
10 The plane P has normal vector \(2\mathbf{i} + a\mathbf{j} - \mathbf{k}\), where \(a\) is a positive constant, and the point \((3, -1, 1)\) lies in P. The plane \(x - z = 3\) makes an angle of \(45^\circ\) with P.
Find the cartesian equation of P. [7]
| Scheme | Marks | AO |
|---|---|---|
| Normal to \(x - z = 3\) is vector \(\mathbf{i} - \mathbf{k}\) | B1 | 3.1a |
| \(\cos 45^\circ = \dfrac{(\mathbf{i} - \mathbf{k}).(2\mathbf{i} + a\mathbf{j} - \mathbf{k})}{\sqrt{2}\sqrt{a^2 + 5}}\) | M1 | 1.1a |
| \(\Rightarrow \dfrac{1}{\sqrt{2}} = \dfrac{3}{\sqrt{2}\sqrt{a^2 + 5}}\) \(\Rightarrow \sqrt{a^2 + 5} = 3\) | A1 | 1.1 |
| \(\Rightarrow a = 2\) | A1 | 1.1 |
| Plane equation is \(2x + ay - z = k\) | M1 | 1.1 |
| \((3, -1, 1)\) lies in plane \(\Rightarrow 6 - a - 1 = k\) | M1 | 3.1a |
| \(\Rightarrow k = 3\) so plane equation is \(2x + 2y - z = 3\) | A1 | 3.2a |
| [7] |
Notes
B1: may be implied by \(\mathbf{i} - \mathbf{k}\) seen (or column vector)
M1: (1st) allow 1 slip
A1: (1st) must have \(1/\sqrt{2}\) for \(\cos 45^\circ\)
M1: (2nd) or with their \(a\)
M1: (3rd) substituting \((3, -1, 1)\) into plane equation (\(k = 5 -\) their \(a\))