A2 June 2022 Paper 1 Q13
13 The points A and B have coordinates \((4, 0, -1)\) and \((10, 4, -3)\) respectively. The planes \(\Pi_1\) and \(\Pi_2\) have equations \(x - 2y = 5\) and \(2x + 3y - z = -4\) respectively.
(a) Find the acute angle between the line AB and the plane \(\Pi_1\). [4]
(b) Show that the line AB meets \(\Pi_1\) and \(\Pi_2\) at the same point, whose coordinates should be specified. [5]
(c)
(i) Find \((\mathbf{i} - 2\mathbf{j}) \times (2\mathbf{i} + 3\mathbf{j} - \mathbf{k})\). [1]
(ii) Hence find the acute angle between the planes \(\Pi_1\) and \(\Pi_2\). [3]
(iii) Find the shortest distance between the point A and the line of intersection of the planes \(\Pi_1\) and \(\Pi_2\). [4]
| Scheme | Marks | AO |
|---|---|---|
| \(\overrightarrow{\mathrm{AB}} = 6\mathbf{i} + 4\mathbf{j} - 2\mathbf{k}\) | B1 | 1.1 |
| angle between line and normal is \(\theta\) where \(\cos\theta = \dfrac{(3\mathbf{i} + 2\mathbf{j} - \mathbf{k}) \cdot (\mathbf{i} - 2\mathbf{j})}{\sqrt{14}\sqrt{5}} = \dfrac{-1}{\sqrt{14}\sqrt{5}}\) | M1 A1 | 3.1a 1.1 |
| \(\theta = 96.9^\circ\), so angle with plane is \(6.9^\circ\) | A1 | 1.1 |
| [4] |
Notes
B1: Any multiple soi
A1: \(7^\circ\) or better 0.1198 rad
| Scheme | Marks | AO |
|---|---|---|
| eqn of AB is \([\mathbf{r} =]\,4\mathbf{i} - \mathbf{k} + \lambda(3\mathbf{i} + 2\mathbf{j} - \mathbf{k})\) | B1ft | 2.1 |
| Substituting \(x = 4 + 3\lambda,\ y = 2\lambda,\ z = -1 - \lambda\): \(4 + 3\lambda - 4\lambda = 5 \Rightarrow \lambda = -1\) | M1 | 1.1 |
| meets \(\Pi_1\) at \((1, -2, 0)\) (say C) | A1 | 2.2a |
| substituting into \(\Pi_2\) | M1 | 3.1a |
| \(2 \times 1 + 3 \times (-2) - 0 = -4\), so C lies on \(\Pi_2\) | A1 | 2.2a |
| [5] |
Notes
B1ft: Soi Note may use another position vector as long as it is correct
M1: or solving with \(\Pi_2\)
Alternative method
| Scheme | Marks |
|---|---|
| eqn of AB is \([\mathbf{r} =]\,4\mathbf{i} - \mathbf{k} + \lambda(3\mathbf{i} + 2\mathbf{j} - \mathbf{k})\) | B1 |
| Substituting \(x = 4 + 3\lambda,\ y = 2\lambda,\ z = -1 - \lambda\): \(4 + 3\lambda - 4\lambda = 5 \Rightarrow \lambda = -1\) | M1 |
| Substituting \(x = 4 + 3\lambda,\ y = 2\lambda,\ z = -1 - \lambda\): \(2(4 + 3\lambda) + 3(2\lambda) - (-1 - \lambda) = -4 \Rightarrow \lambda = -1\) | M1 |
| Finding equal \(\lambda\) for both planes | A1 |
| \((1, -2, 0)\) | A1 |
| [5] |
B1: soi
| Scheme | Marks | AO |
|---|---|---|
| (i) \((\mathbf{i} - 2\mathbf{j}) \times (2\mathbf{i} + 3\mathbf{j} - \mathbf{k}) = 2\mathbf{i} + \mathbf{j} + 7\mathbf{k}\) | B1 | 1.1 |
| [1] |
| Scheme | Marks | AO |
|---|---|---|
| (ii) \(\sqrt{54}\) | B1 | 3.1a |
| \(\sqrt{5} \times \sqrt{14}\sin\theta = \sqrt{54}\) | M1 | 1.1 |
| \(\Rightarrow \theta = 61.4^\circ\) | A1 | 1.1 |
| [3] |
Notes
B1: soi
A1: 1.07 rad
| Scheme | Marks | AO |
|---|---|---|
| (iii) \(2\mathbf{i} + \mathbf{j} + 7\mathbf{k}\) is direction vector of line of intersection | B1ft | 3.1a |
| \(\begin{pmatrix} 3 \\ 2 \\ -1 \end{pmatrix} \times \begin{pmatrix} 2 \\ 1 \\ 7 \end{pmatrix} = \begin{pmatrix} 15 \\ -23 \\ -1 \end{pmatrix}\) | M1 | 1.1 |
| \(d = \dfrac{\sqrt{15^2 + (-23)^2 + (-1)^2}}{\sqrt{2^2 + 1^2 + 7^2}}\) | M1 | 1.1 |
| \(= 3.74\) | A1cao | 1.1 |
| [4] |
Notes
B1ft: Soi by using in both numerator and denominator in correct method for \(d\)