FP3 June 2009 Q7
7. The lines \(l_1\) and \(l_2\) have equations \[\mathbf{r} = \begin{pmatrix} 1 \\ -1 \\ 2 \end{pmatrix} + \lambda\begin{pmatrix} -1 \\ 3 \\ 4 \end{pmatrix} \quad \text{and} \quad \mathbf{r} = \begin{pmatrix} \alpha \\ -4 \\ 0 \end{pmatrix} + \mu\begin{pmatrix} 0 \\ 3 \\ 2 \end{pmatrix}.\]
If the lines \(l_1\) and \(l_2\) intersect, find
(a) the value of \(\alpha\), (4)
(b) an equation for the plane containing the lines \(l_1\) and \(l_2\), giving your answer in the form \(ax + by + cz + d = 0\), where \(a\), \(b\), \(c\) and \(d\) are constants. (4)
For other values of \(\alpha\), the lines \(l_1\) and \(l_2\) do not intersect and are skew lines.
Given that \(\alpha = 2\),
(c) find the shortest distance between the lines \(l_1\) and \(l_2\). (3)
| Scheme | Marks |
|---|---|
| If the lines meet, \(-1 + 3\lambda = -4 + 3\mu\) and \(2 + 4\lambda = 2\mu\) | M1 |
| Solve to give \(\lambda = 0\) \((\mu = 1\) but this need not be seen\()\). | M1 A1 |
| Also \(1 - \lambda = \alpha\) and so \(\alpha = 1\). | B1 |
| (4) |
| Scheme | Marks |
|---|---|
| \(\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -1 & 3 & 4 \\ 0 & 3 & 2 \end{vmatrix} = -6\mathbf{i} + 2\mathbf{j} - 3\mathbf{k}\) is perpendicular to both lines and hence to the plane | M1 A1 |
| The plane has equation \(\mathbf{r} \cdot \mathbf{n} = \mathbf{a} \cdot \mathbf{n}\), which is \(-6x + 2y - 3z = -14\), | M1 |
| i.e. \(-6x + 2y - 3z + 14 = 0\). | A1 o.a.e. |
| (4) |
Notes
OR (b) Alternative scheme
| Scheme | Marks |
|---|---|
| Use \((1, -1, 2)\) and \((\alpha, -4, 0)\) in equation \(ax + by + cz + d = 0\) | M1 |
| And third point so three equations, and attempt to solve | M1 |
| Obtain \(6x - 2y + 3z =\) | A1 |
| \((6x - 2y + 3z) - 14 = 0\) | A1 o.a.e. |
| (4) |
| Scheme | Marks |
|---|---|
| \((\mathbf{a}_1 - \mathbf{a}_2) = \mathbf{i} - 3\mathbf{j} - 2\mathbf{k}\) | M1 |
| Use formula \(\dfrac{(\mathbf{a}_1 - \mathbf{a}_2) \cdot \mathbf{n}}{|\mathbf{n}|} = \dfrac{(\mathbf{i} - 3\mathbf{j} - 2\mathbf{k}) \cdot (-6\mathbf{i} + 2\mathbf{j} - 3\mathbf{k})}{\sqrt{(36 + 4 + 9)}} = \left(\dfrac{-6}{7}\right)\) | M1 |
| Distance is \(\dfrac{6}{7}\) | A1 |
| (3) | |
| (11 marks) |