A2 June 2019 Paper 1 Q8
8 The equation of a plane is \(4x + 2y + z = 7\).
The point \(A\) has coordinates \((9, 6, 1)\) and the point \(B\) is the reflection of \(A\) in the plane.
Find the coordinates of the point \(B\). [6]
| Scheme | Marks | AO |
|---|---|---|
| \(AB\) has direction \(\begin{pmatrix} 4 \\ 2 \\ 1 \end{pmatrix}\) | B1 | 1.1 |
| and any point on it is \((9 + 4\lambda,\ 6 + 2\lambda,\ 1 + \lambda)\) | M1 | 3.1a |
| If this point lies on plane then \(4(9 + 4\lambda) + 2(6 + 2\lambda) + (1 + \lambda) = 7\) \(\Rightarrow 49 + 21\lambda = 7 \Rightarrow 21\lambda = -42 \Rightarrow \lambda = -2\) | M1 | 3.1a |
| So \(B\) is where \(\lambda = -4\) | M1 A1 | 1.1a 1.1 |
| \(\Rightarrow B\) has coordinates \((-7, -2, -3)\) | A1 | 3.2a |
| [6] |
Notes
B1: Direction. Alternative methods possible
M1: Attempt to find point on line
M1: Attempt to find \(\lambda\)
M1: Double \(\lambda\)
A1: \(\lambda\) soi
A1: Must be coordinates
Alternative method
| Scheme | Marks |
|---|---|
| Distance between point and plane \(= \dfrac{42}{\sqrt{21}} = 2\sqrt{21}\) Distance between point and reflected point \(= 4\sqrt{21}\) | M1 |
| Reflected point is \((x, y, z) \Rightarrow (x - 9)^2 + (y - 6)^2 + (z - 1)^2\) Any point on normal line is \((9 + 4\lambda, 6 + 2\lambda, 1 + \lambda)\) | B1 |
| \(\Rightarrow 16\lambda^2 + 4\lambda^2 + \lambda^2 = 336 \Rightarrow 21\lambda^2 = 336 \Rightarrow \lambda^2 = 16\) | M1 |
| \(\Rightarrow \lambda = \pm 4\) | A1 |
| \((25, 14, 5)\) is the same side \(\Rightarrow (-7, -2, -3)\) | A1 |
| [6] |