# Question 1:

## Part (i)

| Answer | Marks | Guidance |
|--------|-------|----------|
| $\overrightarrow{AB} \times \overrightarrow{AC} = \begin{pmatrix}1\\-6\\16\end{pmatrix} \times \begin{pmatrix}9\\-12\\-3\end{pmatrix} = \begin{pmatrix}210\\147\\42\end{pmatrix} \left[= 21\begin{pmatrix}10\\7\\2\end{pmatrix}\right]$ | M1 | Evaluation of vector product; one correct element (FT) |
| | A2 | Give A1 for one correct element; Give A1 for a non-zero multiple |
| Equation of $P$ is $10x + 7y + 2z = d$ | M1 | |
| $10x + 7y + 2z = 40$ | A1 **[5]** | Accept $210x + 147y + 42z - 840 = 0$ etc |

## Part (ii)

| Answer | Marks | Guidance |
|--------|-------|----------|
| $AD = \left\|\dfrac{6(-3) + 3(12) + 2(-7) - 32}{\sqrt{6^2 + 3^2 + 2^2}}\right\|$ | M1 | For numerator; M0 if constant term omitted |
| | M1 | For denominator |
| $= \dfrac{28}{7} = 4$ | A1 **[3]** | |
| **OR:** $6(-3+6\lambda) + 3(12+3\lambda) + 2(-7+2\lambda) = 32$ | M1 | Equation for $\lambda$ |
| $\lambda = \dfrac{4}{7}$, $AD = \left\|\lambda\begin{pmatrix}6\\3\\2\end{pmatrix}\right\| = \dfrac{4}{7}\sqrt{6^2+3^2+2^2}$ | M1 | Using $\lambda$ to find distance AD; Independent of previous M1; But M0 if $\lambda = \pm1$ or $\lambda = 0$ |
| $= 4$ | A1 | |

## Part (iii)

| Answer | Marks | Guidance |
|--------|-------|----------|
| When $x=0$: $\begin{cases}7y+2z=40\\3y+2z=32\end{cases}$ giving $y=2$, $z=13$ | M1 | Finding a point on $L$ |
| | A1 | One correct point e.g. $(1, 1, 11.5)$; e.g. $(2,0,10)$, $\left(\frac{26}{3},-\frac{20}{3},0\right)$ |
| $\begin{pmatrix}10\\7\\2\end{pmatrix} \times \begin{pmatrix}6\\3\\2\end{pmatrix} = \begin{pmatrix}8\\-8\\-12\end{pmatrix}$ | M1 | Vector product of direction vectors |
| | A1 | Direction of $L$ correct; OR finding a second point on $L$ and using 2 points to find direction |
| Equation of $L$ is $\mathbf{r} = \begin{pmatrix}0\\2\\13\end{pmatrix} + \lambda\begin{pmatrix}2\\-2\\-3\end{pmatrix}$ | A1 FT **[5]** | Any correct form; Dependent on M1M1; Condone omission of '$\mathbf{r} =$' |
| **OR:** Eliminating $z$: $4x + 4y = 8$ | M1 | Eliminating one variable; Or $6y-4z=-40$ or $12x+8z=104$ |
| | A1 | |
| $x = \lambda$, $y = 2-\lambda$, $z = 13 - \frac{3}{2}\lambda$ | M1 A1A1 | Finding (e.g.) $y$ and $z$ in terms of $x$; Or A1 FT dependent on M1M1 |

## Part (iv)

| Answer | Marks | Guidance |
|--------|-------|----------|
| $\left[\begin{pmatrix}-3\\12\\-7\end{pmatrix} - \begin{pmatrix}0\\2\\13\end{pmatrix}\right] \times \begin{pmatrix}2\\-2\\-3\end{pmatrix} = \begin{pmatrix}-3\\10\\-20\end{pmatrix} \times \begin{pmatrix}2\\-2\\-3\end{pmatrix} = \begin{pmatrix}-70\\-49\\-14\end{pmatrix}$ | M1 | Appropriate vector product |
| | A2 FT | Give A1 if one error |
| Shortest distance is $\dfrac{\sqrt{70^2+49^2+14^2}}{\sqrt{2^2+2^2+3^2}} = \sqrt{\dfrac{7497}{17}}$ | M1 | Finding magnitude of vector product |
| | M1 | Complete method for finding distance; Dependent on previous M1M1 |
| Shortest distance is $21$ | A1 **[6]** | A0 for 21 resulting from wrong v.p. |
| **OR:** $\begin{pmatrix}2\lambda\\2-2\lambda\\13-3\lambda\end{pmatrix} - \begin{pmatrix}-3\\12\\-7\end{pmatrix} \cdot \begin{pmatrix}2\\-2\\-3\end{pmatrix} = 0$ | M1 | Allow one error |
| | A1 FT | |
| $\lambda = 2$ | M1 | Obtaining a value of $\lambda$; Dependent on previous M1 |
| | A1 FT | |
| Shortest distance is $\sqrt{(7)^2+(-14)^2+(14)^2}$ | M1 | Dependent on previous M1M1 |
| Shortest distance is $21$ | A1 | |

## Part (v)

| Answer | Marks | Guidance |
|--------|-------|----------|
| $\overrightarrow{AD} = \left(\pm\right)\dfrac{4}{7}\begin{pmatrix}6\\3\\2\end{pmatrix}$ | M1 | $\overrightarrow{AD}$ is a multiple of $\begin{pmatrix}6\\3\\2\end{pmatrix}$; M1 for $\overrightarrow{AD} = \begin{pmatrix}6\\3\\2\end{pmatrix}$ |
| | A1 FT | FT from (ii) |
| Volume is $\frac{1}{6}(\overrightarrow{AB} \times \overrightarrow{AC}) \cdot \overrightarrow{AD}$ | M1 | Appropriate scalar triple product; Just stated, $\frac{1}{6}$ not needed |
| $= \frac{1}{6} \times 21\begin{pmatrix}10\\7\\2\end{pmatrix} \cdot \frac{4}{7}\begin{pmatrix}6\\3\\2\end{pmatrix} = 2(60+21+4)$ | M1 | Evaluation of scalar triple product; Independent of previous M's, but must be numerical |
| $= 170$ | A1 **[5]** | |

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