## Question 11(a)(i):

| Working/Answer | Marks | Guidance |
|---|---|---|
| Direction vectors are $\begin{pmatrix}-2\\1\\-3\end{pmatrix}$ and $\begin{pmatrix}2\\-1\\3\end{pmatrix}$; these are multiples of each other ($l_1$ direction $= -1 \times$ direction of $l_2$), so lines are parallel | E1 (2.4) | Must reference direction vectors |

## Question 11(a)(ii):

| Working/Answer | Marks | Guidance |
|---|---|---|
| Forms vector from point on $l_1$ to point on $l_2$: $A(1,5,-1)$, $B(-3,2,7)$; $\overrightarrow{AB} = \begin{pmatrix}-4\\-3\\8\end{pmatrix}$ | M1 (3.1a) | PI by $\sqrt{89}$ |
| Obtains correct $\overrightarrow{AB}$, PI by $\sqrt{89}$ | A1 (1.1b) | Or correct parametrised $\overrightarrow{PQ}$ |
| Forms $\overrightarrow{AB} \times \hat{l} = \frac{1}{\sqrt{14}}\begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\-4&-3&8\\2&-1&3\end{vmatrix} = \frac{1}{\sqrt{14}}\begin{pmatrix}1\\28\\10\end{pmatrix}$ | M1 (3.1a) | Or forms scalar product of $\overrightarrow{PQ}$ with direction vector |
| Uses distance $= k|\overrightarrow{AB} \times \begin{pmatrix}2\\-1\\3\end{pmatrix}|$ or $\overrightarrow{PQ}\cdot\begin{pmatrix}2\\-1\\3\end{pmatrix} = 0$ | M1 (1.1a) | |
| Distance $= \frac{1}{\sqrt{14}}\sqrt{1^2+28^2+10^2} = 7.95$ (3 s.f.) | R1 (2.1) | Rigorous argument required |

## Question 11(b):

| Working/Answer | Marks | Guidance |
|---|---|---|
| Expresses both lines in parametric form with different parameters | M1 (3.1a) | Condone same parameter for both lines |
| $\mathbf{r} = \begin{pmatrix}1\\5\\-1\end{pmatrix} + \mu\begin{pmatrix}-2\\1\\-3\end{pmatrix}$ and $\mathbf{r} = \begin{pmatrix}-5\\12\\-4\end{pmatrix} + \lambda\begin{pmatrix}4\\0\\9\end{pmatrix}$; equates with two different parameters | M1 (1.1a) | |
| Solves two simultaneous equations from components: $y: 5+\mu = 12 \Rightarrow \mu = 7$; sub. into $x$: $\lambda = -2$ | M1 (1.1b) | |
| Verifies third components equal: $\mu = 7$, $\lambda = -2$ satisfy $z$ equation | A1 (2.2a) | |
| Lines meet at $(-13, 12, -22)$ | R1 (2.1) | Completes rigorous argument |