FP3 June 2013 (R) Q2
2. Two skew lines \(l_1\) and \(l_2\) have equations \[\begin{aligned} l_1&: \mathbf{r} = (\mathbf{i} - \mathbf{j} + \mathbf{k}) + \lambda(4\mathbf{i} + 3\mathbf{j} + 2\mathbf{k}) \\ l_2&: \mathbf{r} = (3\mathbf{i} + 7\mathbf{j} + 2\mathbf{k}) + \mu(-4\mathbf{i} + 6\mathbf{j} + \mathbf{k}) \end{aligned}\] respectively, where \(\lambda\) and \(\mu\) are real parameters.
(a) Find a vector in the direction of the common perpendicular to \(l_1\) and \(l_2\) (2)
(b) Find the shortest distance between these two lines. (5)
| Scheme | Marks |
|---|---|
| \(l_1: (\mathbf{i} - \mathbf{j} + \mathbf{k}) + \lambda(4\mathbf{i} + 3\mathbf{j} + 2\mathbf{k}) \qquad l_2: (3\mathbf{i} + 7\mathbf{j} + 2\mathbf{k}) + \lambda(-4\mathbf{i} + 6\mathbf{j} + \mathbf{k})\) | |
| \(\mathbf{n} = \begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k} \\ 4 & 3 & 2 \\ -4 & 6 & 1\end{vmatrix} = -9\mathbf{i} - 12\mathbf{j} + 36\mathbf{k}\) M1: Correct attempt at a vector product between \(4\mathbf{i} + 3\mathbf{j} + 2\mathbf{k}\) and \(-4\mathbf{i} + 6\mathbf{j} + \mathbf{k}\) (if the method is unclear then 2 components must be correct) allowing for the sign error in the \(y\) component. A1: Any multiple of \((3\mathbf{i} + 4\mathbf{j} - 12\mathbf{k})\) | M1A1 |
| (2) |
| Scheme | Marks |
|---|---|
| \(\mathbf{a}_1 - \mathbf{a}_2 = \pm(2\mathbf{i} + 8\mathbf{j} + \mathbf{k})\) M1: Attempt to subtract position vectors A1: Correct vector \(\pm(2\mathbf{i} + 8\mathbf{j} + \mathbf{k})\) (Allow as coordinates) | M1 A1 |
| So \(p = \dfrac{\begin{pmatrix}2 \\ 8 \\ 1\end{pmatrix} \bullet \begin{pmatrix}-9 \\ -12 \\ 36\end{pmatrix}}{\sqrt{9^2 + 12^2 + 36^2}}\) Correct formula for the distance using their vectors: \(\dfrac{\text{"}\pm(2\mathbf{i} + 8\mathbf{j} + \mathbf{k})\text{"} \bullet \text{"}\mathbf{n}\text{"}}{|\text{"}\mathbf{n}\text{"}|}\) | M1 |
| \(p = \dfrac{\pm 78}{\sqrt{1521}} = \dfrac{\pm 78}{39} = 2\) M1: Correctly forms a scalar product in the numerator and Pythagoras in the denominator. (Dependent on the previous method mark) A1: 2 (not \(-2\)) | dM1 A1 |
| (5) | |
| (7 marks) |
Notes
(b) Way 2
| Scheme | Marks |
|---|---|
| \((\mathbf{i} - \mathbf{j} + \mathbf{k}) \bullet (3\mathbf{i} + 4\mathbf{j} - 12\mathbf{k}) = -13\ (d_1)\) \((3\mathbf{i} + 7\mathbf{j} + 2\mathbf{k}) \bullet (3\mathbf{i} + 4\mathbf{j} - 12\mathbf{k}) = 13\ (d_2)\) M1: Attempt scalar product between their \(\mathbf{n}\) and either position vector A1: Both scalar products correct | M1A1 |
| \(\dfrac{\pm 13}{\sqrt{3^2 + 4^2 + 12^2}}\ (= 1)\) Divides either of their scalar products by the magnitude of their normal vector. \(\dfrac{d_1 \text{ or } d_2}{|\text{"}\mathbf{n}\text{"}|}\) | M1 |
| \(p = \dfrac{d_1}{|\text{"}\mathbf{n}\text{"}|} - \dfrac{d_2}{|\text{"}\mathbf{n}\text{"}|}\) or \(2 \times \dfrac{d_1}{|\text{"}\mathbf{n}\text{"}|}\) M1: Correct attempt to find the required distance i.e. subtracts their \(\dfrac{d_1}{|\text{"}\mathbf{n}\text{"}|}\) and \(\dfrac{d_2}{|\text{"}\mathbf{n}\text{"}|}\) or doubles their \(\dfrac{d_1}{|\text{"}\mathbf{n}\text{"}|}\) if \(|d_1| = |d_2|\). (Dependent on the previous method mark) A1: 2 (not \(-2\)) | dM1 A1 |
| (5) |