AS June 2024 Paper 1 Q9
9 In this question you must show detailed reasoning.
Find a vector \(\mathbf{v}\) which has the following properties.
- It is a unit vector.
- It is parallel to the plane \(2x + 2y + z = 10\).
- It makes an angle of \(45^\circ\) with the normal to the plane \(x + z = 5\). [8]
| Scheme | Marks | AO |
|---|---|---|
| DR Let \(\mathbf{v} = a\mathbf{i} + b\mathbf{j} + c\mathbf{k}\) | M1 | 3.1a |
| \(\sqrt{a^2 + b^2 + c^2} = 1\) | B1 | 1.1 |
| \((a\mathbf{i} + b\mathbf{j} + c\mathbf{k}).(2\mathbf{i} + 2\mathbf{j} + \mathbf{k}) = 0\) | M1 | 3.1a |
| \(\Rightarrow 2a + 2b + c = 0\) | A1 | 1.1 |
| \(\dfrac{(a\mathbf{i} + b\mathbf{j} + c\mathbf{k}).(\mathbf{i} + \mathbf{k})}{\sqrt{a^2 + b^2 + c^2}.\sqrt{2}} = (\pm)\cos 45^\circ\) | M1 | 3.1a |
| \(\Rightarrow a + c = \sqrt{a^2 + b^2 + c^2}\) | A1 | 1.1 |
| \(\Rightarrow a + c = 1\) \(c = 1 - a,\ b = -\tfrac{1}{2}(a + 1) \Rightarrow a^2 + \tfrac{1}{4}(a + 1)^2 + (1 - a)^2 = 1\) \(\Rightarrow 9a^2 - 6a + 1 = 0 \Rightarrow (3a - 1)^2 = 0\) | M1 | 2.1 |
| \(a = \tfrac{1}{3}, b = -\tfrac{2}{3}, c = \tfrac{2}{3} \Rightarrow \mathbf{v} = \tfrac{1}{3}\mathbf{i} - \tfrac{2}{3}\mathbf{j} + \tfrac{2}{3}\mathbf{k}\) | A1 | 3.2a |
| [8] |
Notes
M1: (1st) define vector \(\mathbf{v}\)
B1: oe
M1: (2nd) scalar product with normal to plane = 0
M1: (3rd) use of formula for angle between vector and normal to \(x + z = 5\)
M1: (4th) Complete method for obtaining an equation in one unknown dep both previous M1 marks and B1 obtained
A1: (final) (\(\mathbf{v} = -\tfrac{1}{3}\mathbf{i} + \tfrac{2}{3}\mathbf{j} - \tfrac{2}{3}\mathbf{k}\) is also a valid solution, following use of the negative sign from the optional \(\pm\) above)