C4 January 2006 Q6
6. The line \(l_1\) has vector equation
\[\mathbf{r} = 8\mathbf{i} + 12\mathbf{j} + 14\mathbf{k} + \lambda(\mathbf{i} + \mathbf{j} - \mathbf{k}),\]
where \(\lambda\) is a parameter.
The point \(A\) has coordinates \((4, 8, a)\), where \(a\) is a constant. The point \(B\) has coordinates \((b, 13, 13)\), where \(b\) is a constant. Points \(A\) and \(B\) lie on the line \(l_1\).
(a) Find the values of \(a\) and \(b\). (3)
Given that the point \(O\) is the origin, and that the point \(P\) lies on \(l_1\) such that \(OP\) is perpendicular to \(l_1\),
(b) find the coordinates of \(P\). (5)
(c) Hence find the distance \(OP\), giving your answer as a simplified surd. (2)
| Scheme | Marks |
|---|---|
| \(\lambda = -4 \to a = 18, \qquad \mu = 1 \to b = 9\) | M1 A1, A1 |
| (3) |
| Scheme | Marks |
|---|---|
| \(\begin{pmatrix}8 + \lambda\\12 + \lambda\\14 - \lambda\end{pmatrix} \bullet \begin{pmatrix}1\\1\\-1\end{pmatrix} = 0\) | M1 |
| \(\therefore 8 + \lambda + 12 + \lambda - 14 + \lambda = 0\) | A1 |
| Solves to obtain \(\lambda\) \((\lambda = -2)\) | dM1 |
| Then substitutes value for \(\lambda\) to give \(P\) at the point \((6, 10, 16)\) (any form) | M1, A1 |
| (5) |
| Scheme | Marks |
|---|---|
| \(OP = \sqrt{36 + 100 + 256}\) | M1 |
| \(\left(= \sqrt{392}\right) = 14\sqrt{2}\) | A1 cao |
| (2) | |
| (10 marks) |