A2 June 2019 Paper 1 Q12
12 Three intersecting lines \(L_1\), \(L_2\) and \(L_3\) have equations
\[L_1\!: \frac{x}{2} = \frac{y}{3} = \frac{z}{1}, \quad L_2\!: \frac{x}{1} = \frac{y}{2} = \frac{z}{-4} \quad \text{and} \quad L_3\!: \frac{x - 1}{1} = \frac{y - 2}{1} = \frac{z + 4}{5}.\]Find the area of the triangle enclosed by these lines. [9]
| Scheme | Marks | AO |
|---|---|---|
| \(L_1\) and \(L_3\): \(\lambda = -4 + 5\nu,\ -8 + 10\nu = 1 + \nu\) | M1 | 1.1b |
| \(\Rightarrow \nu = 1,\ \lambda = 1\), so meet at \((2, 3, 1)\) | B1 | 2.2a |
| \(L_2\) and \(L_3\) meet at \((1, 2, -4)\) | B1 | 2.2a |
| \(L_1\) and \(L_2\) meet at origin | B1 | 2.2a |
| (4) | ||
| \(\cos\theta = \dfrac{(2\mathbf{i} + 3\mathbf{j} + \mathbf{k}).(\mathbf{i} + 2\mathbf{j} - 4\mathbf{k})}{\sqrt{2^2 + 3^2 + 1^2}\sqrt{1^2 + 2^2 + (-4)^2}}\) | M1 | 3.1a |
| \(= \dfrac{4}{\sqrt{14}\sqrt{21}}\) | A1 | 1.1b |
| \(\theta = 76.5^\circ\) (1.335) | A1 | 1.1b |
| (3) | ||
| Area \(= \frac{1}{2}\sqrt{14}\sqrt{21}\sin 76.5^\circ\) | M1 | 1.1a |
| \(= 8.34\) [units\(^2\)] | A1 | 3.2a |
| (2) | ||
| [9] |
Notes
M1: attempt to solve (any pair)
B1, B1, B1: by solving or inspection; soi
M1: or cosine rule: sides \(\sqrt{21}\), \(\sqrt{14}\), \(\sqrt{27}\)
A1: or \(\cos^{-1}\dfrac{10}{\sqrt{14}\sqrt{27}}\), \(\cos^{-1}\dfrac{17}{\sqrt{21}\sqrt{27}}\)
A1: \(59.0^\circ\) (1.030), \(44.4^\circ\) (0.775); any correct angle
M1: or \(\frac{1}{2}\sqrt{14}\sqrt{27}\sin 59.0^\circ\) or \(\frac{1}{2}\sqrt{21}\sqrt{27}\sin 44.4^\circ\)
A1: art 8.3 or \(\sqrt{278}/2\)
Alternative solution (last 5 marks)
| Scheme | Marks |
|---|---|
| \((2\mathbf{i} + 3\mathbf{j} + \mathbf{k}) \times (\mathbf{i} + 2\mathbf{j} - 4\mathbf{k})\) \(= -14\mathbf{i} + 9\mathbf{j} + \mathbf{k}\) | M1 A1 |
| (2) | |
| Area \(= \frac{1}{2} \times \sqrt{14^2 + 9^2 + 1^2}\) \(= \frac{1}{2}\sqrt{278} = 8.34\) [units\(^2\)] | M1 A2 |
| (3) |
M1: using cross product: \((-2\mathbf{i} - 3\mathbf{j} - \mathbf{k}) \times (-\mathbf{i} - \mathbf{j} - 5\mathbf{k}) = 14\mathbf{i} - 9\mathbf{j} - \mathbf{k}\), \((-\mathbf{i} - 2\mathbf{j} + 4\mathbf{k}) \times (\mathbf{i} + \mathbf{j} + 5\mathbf{k}) = -14\mathbf{i} + 9\mathbf{j} + \mathbf{k}\)
M1: or \(\frac{1}{2}\) base \(\times\) height: \(\frac{1}{2} \times \sqrt{14} \times \sqrt{287}/\sqrt{14}\), etc
A2: art 8.34