## Question 11(a):

$\begin{pmatrix}7\\4\\9\end{pmatrix} + \lambda\begin{pmatrix}1\\1\\4\end{pmatrix} = \begin{pmatrix}-6\\-7\\3\end{pmatrix} + \mu\begin{pmatrix}5\\4\\b\end{pmatrix}$ giving $7+\lambda = -6+5\mu$, $4+\lambda = -7+4\mu$, $9+4\lambda = 3+b\mu$ | M1 | Writes down any two equations for coordinates of point of intersection. Must attempt to set coordinates equal but condone slips.

Full method to find both $\lambda$ and $\mu$ from equations 1 and 2 and use in equation 3 to find $b$ | dM1 |

$(1)-(2) \Rightarrow 3 = 1+\mu \Rightarrow \mu = 2$; sub $\mu=2$ into (1) $\Rightarrow \lambda = -3$; $9-12 = 3+2b \Rightarrow b = -3$ | A1 | Completely correct work including $\lambda=-3$, $\mu=2$ and substitution into both sides of third equation to give $b=-3$

Position vector of intersection substituting $\lambda=-3$ into $l_1$ or $\mu=2$ into $l_2$ | dM1 | Substitutes their $\lambda$ into $l_1$ or $\mu$ into $l_2$. May be implied by at least 2 correct coordinates for $X$.

$X = (4, 1, -3)$ | A1 | Correct coordinates or vector. Correct coordinates implies M1A1.

---

## Question 11(b) Way 1:

$\pm\overrightarrow{XA} = \pm\begin{pmatrix}2\\2\\8\end{pmatrix}$, $\quad \pm\overrightarrow{XB} = \pm\begin{pmatrix}10\\8\\-6\end{pmatrix}$ | M1 | Attempts difference between coordinates $X$ and $A$, $X$ and $B$. Could be implied by calculation of lengths $AX$ and $BX$. Allow slips but must be subtracting.

$\pm\overrightarrow{XA}\cdot\pm\overrightarrow{XB} = |XA||XB|\cos\theta \Rightarrow 20+16-48 = \sqrt{72}\sqrt{200}\cos\theta$ | dM1A1 | M1: Attempt scalar product of $\overrightarrow{XA}$ and $\overrightarrow{XB}$ or $\overrightarrow{AX}$ and $\overrightarrow{BX}$ etc. A1: Correct unsimplified expression $20+16-48 = \sqrt{72}\sqrt{200}\cos\theta$

$\cos\theta = \frac{-12}{\sqrt{72}\times\sqrt{200}} \Rightarrow \theta = \arccos\left(-\frac{1}{10}\right)$ | A1* | Given answer. Must be an intermediate line with $\cos\theta = \ldots$ or $\theta = \ldots$

# Question (b) Way 2:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\mathbf{d}_1 = \begin{pmatrix}1\\1\\4\end{pmatrix}$, $\mathbf{d}_2 = \begin{pmatrix}5\\4\\-3\end{pmatrix}$ | M1 | Uses $b = -3$ and the direction vectors **or multiples of the direction vectors** |
| $\mathbf{d}_1 \cdot \mathbf{d}_2 = \|\mathbf{d}_1\|\|\mathbf{d}_2\|\cos\theta \Rightarrow 5+4-12 = \sqrt{18}\sqrt{50}\cos\theta$ | dM1A1 | M1: Attempt scalar product of direction vectors. A1: correct un-simplified expression $5+4-12=\sqrt{18}\sqrt{50}\cos\theta$ |
| $\cos\theta = \frac{-3}{\sqrt{18}\times\sqrt{50}} \Rightarrow \theta = \arccos\left(-\frac{1}{10}\right)$* | A1* | Given answer. Must be intermediate line with $\cos\theta =..$ or $\theta = \ldots$ |

# Question (b) Way 3:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\pm\overrightarrow{XA} = \pm\begin{pmatrix}2\\2\\8\end{pmatrix}$, $\pm\overrightarrow{XB} = \pm\begin{pmatrix}10\\8\\-6\end{pmatrix}$ | M1 | Attempts difference between coordinates $X$ and $A$, $X$ and $B$. Could be implied by calculation of lengths $AX$ and $BX$. Allow slips but must be subtracting. |
| $\|AB\|^2 = \|XA\|^2 + \|XB\|^2 - 2\|XA\|\|XB\|\cos\theta \Rightarrow 8^2+6^2+14^2 = 72+200-2\sqrt{72}\sqrt{200}\cos\theta$ | dM1A1 | M1: Uses $\overrightarrow{AB}$ with correct attempt at cosine rule. A1: correct un-simplified expression $8^2+6^2+14^2=72+200-2\sqrt{72}\sqrt{200}\cos\theta$ |
| $\cos\theta = \frac{-24}{2\sqrt{72}\times\sqrt{200}} \Rightarrow \theta = \arccos\left(-\frac{1}{10}\right)$* | A1* | Given answer. Must be intermediate line with $\cos\theta =..$ or $\theta = \ldots$ |

# Question (c):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\cos\theta = -\frac{1}{10} \Rightarrow \sin\theta = \frac{\sqrt{99}}{10}$ | B1 | e.g. $\sqrt{\frac{99}{100}}$, $\frac{3\sqrt{11}}{10}$. May be implied by correct exact area. |
| Area $= \frac{1}{2} \times XA \times XB \times \sin\theta = \frac{1}{2}\times 6\sqrt{2}\times 10\sqrt{2}\times\frac{3\sqrt{11}}{10}$ | M1 | Uses Area $= \frac{1}{2}XA \times XB \times \sin\theta$. Must use angle given by $\cos^{-1}\left(-\frac{1}{10}\right)$ |
| $A = 18\sqrt{11}$ | A1 | Accept e.g. $A = 9\sqrt{44}$, $\sqrt{3564}$ |
| **Note:** $A = \frac{1}{2}\times 6\sqrt{2}\times 10\sqrt{2}\times\sin(95.7391...) = 18\sqrt{11}$ **scores all 3 marks** | | **(3) (12 marks total)** |

---