C4 June 2012 Q8
8. Relative to a fixed origin \(O\), the point \(A\) has position vector \((10\mathbf{i} + 2\mathbf{j} + 3\mathbf{k})\), and the point \(B\) has position vector \((8\mathbf{i} + 3\mathbf{j} + 4\mathbf{k})\).
The line \(l\) passes through the points \(A\) and \(B\).
(a) Find the vector \(\overrightarrow{AB}\). (2)
(b) Find a vector equation for the line \(l\). (2)
The point \(C\) has position vector \((3\mathbf{i} + 12\mathbf{j} + 3\mathbf{k})\).
The point \(P\) lies on \(l\). Given that the vector \(\overrightarrow{CP}\) is perpendicular to \(l\),
(c) find the position vector of the point \(P\). (6)
| Scheme | Marks |
|---|---|
| \(\overrightarrow{AB} = \begin{pmatrix} 8 \\ 3 \\ 4 \end{pmatrix} - \begin{pmatrix} 10 \\ 2 \\ 3 \end{pmatrix} = \begin{pmatrix} -2 \\ 1 \\ 1 \end{pmatrix}\) | M1 A1 |
| (2) |
| Scheme | Marks |
|---|---|
| \(\mathbf{r} = \begin{pmatrix} 10 \\ 2 \\ 3 \end{pmatrix} + t\begin{pmatrix} -2 \\ 1 \\ 1 \end{pmatrix}\) \(\mathbf{r} = \begin{pmatrix} 8 \\ 3 \\ 4 \end{pmatrix} + t\begin{pmatrix} -2 \\ 1 \\ 1 \end{pmatrix}\) | M1 A1ft |
| (2) |
| Scheme | Marks |
|---|---|
| \(\overrightarrow{CP} = \begin{pmatrix} 10 - 2t \\ 2 + t \\ 3 + t \end{pmatrix} - \begin{pmatrix} 3 \\ 12 \\ 3 \end{pmatrix} = \begin{pmatrix} 7 - 2t \\ t - 10 \\ t \end{pmatrix}\) | M1 A1 |
| \(\begin{pmatrix} 7 - 2t \\ t - 10 \\ t \end{pmatrix}.\begin{pmatrix} -2 \\ 1 \\ 1 \end{pmatrix} = -14 + 4t + t - 10 + t = 0\) | M1 |
| Leading to \(\qquad t = 4\) | A1 |
| Position vector of \(P\) is \(\begin{pmatrix} 10 - 8 \\ 2 + 4 \\ 3 + 4 \end{pmatrix} = \begin{pmatrix} 2 \\ 6 \\ 7 \end{pmatrix}\) | M1 A1 |
| (6) | |
| (10 marks) |
Notes
In the printed scheme a bracket joins these method marks: each later M mark is dependent on the M mark before it.
Alternative working for (c)
| Scheme | Marks |
|---|---|
| \(\overrightarrow{CP} = \begin{pmatrix} 8 - 2t \\ 3 + t \\ 4 + t \end{pmatrix} - \begin{pmatrix} 3 \\ 12 \\ 3 \end{pmatrix} = \begin{pmatrix} 5 - 2t \\ t - 9 \\ t + 1 \end{pmatrix}\) | M1 A1 |
| \(\begin{pmatrix} 5 - 2t \\ t - 9 \\ t + 1 \end{pmatrix}.\begin{pmatrix} -2 \\ 1 \\ 1 \end{pmatrix} = -10 + 4t + t - 9 + t + 1 = 0\) | M1 |
| Leading to \(\qquad t = 3\) | A1 |
| Position vector of \(P\) is \(\begin{pmatrix} 8 - 6 \\ 3 + 3 \\ 4 + 3 \end{pmatrix} = \begin{pmatrix} 2 \\ 6 \\ 7 \end{pmatrix}\) | M1 A1 |
| (6) |