A2 June 2019 Paper 1 Q10
10 The points \(A(5, -4, 6)\) and \(B(6, -6, 8)\) lie on the line \(L\). The point \(C\) is \((15, -5, 9)\).
(a) \(D\) is the point on \(L\) that is closest to \(C\).
Find the coordinates of \(D\). [6 marks]
(b) Hence find, in exact form, the shortest distance from \(C\) to \(L\). [2 marks]
| Scheme | Marks | AO |
|---|---|---|
| Obtains an equation of \(L\). Condone one error in their direction vector. Condone lack of “\(\mathbf{r} =\)”, PI by correct \(\mathbf{v}\) | M1 | 1.1a |
| Obtains a correct equation of \(L\). Condone lack of “\(\mathbf{r} =\)”, PI by correct \(\mathbf{v}\) | A1 | 1.1b |
| Obtains their correct general vector from line to \(C\) | B1F | 3.1a |
| Finds scalar product of their \(\mathbf{v}\) and their \(\overrightarrow{AB}\) | M1 | 3.1a |
| Solves to find the correct \(\mu\) for their equation. | A1F | 1.1b |
| Finds correct \(D\) | A1 | 3.2a |
Typical solution
\[\mathbf{r} = \begin{bmatrix} 5 \\ -4 \\ 6 \end{bmatrix} + \mu\begin{bmatrix} 1 \\ -2 \\ 2 \end{bmatrix}\]\[\mathbf{v} = \begin{bmatrix} -10 + \mu \\ 1 - 2\mu \\ -3 + 2\mu \end{bmatrix}\]\[0 = \begin{bmatrix} 1 \\ -2 \\ 2 \end{bmatrix}\cdot\begin{bmatrix} -10 + \mu \\ 1 - 2\mu \\ -3 + 2\mu \end{bmatrix}\]\[-10 + \mu - 2 + 4\mu - 6 + 4\mu = 0\]\[\mu = 2\]\[D = (7,\ -8,\ 10)\]| Scheme | Marks | AO |
|---|---|---|
| Obtains their components of \(\overrightarrow{CD}\), must have their correct magnitude, but ignore sign. Allow one error. | M1 | 1.1a |
| Obtains their correct \(CD\), in exact form. | A1F | 1.1b |
| (8 marks) |