## Question 4(a):

**Plane equations from normal vectors:**

$\mathbf{r} \cdot \mathbf{n} = \mathbf{a} \cdot \mathbf{n}$ applied to obtain plane equations | M1 | Uses $\mathbf{r}.\mathbf{n} = \mathbf{a}.\mathbf{n}$ at least once; accept $\mathbf{r}.\mathbf{n} = p$ form

$-x + 3y + 3z = -5$ and $2x - 5z = 16$ | A1 | Both correct equations

**Finding line of intersection:**

$x = \frac{16 + 5z}{2}$ (or equivalent parameter expression) | M1 | Obtains one variable/parameter in terms of one other variable

$z = \frac{2x-16}{5} \Rightarrow x = 5 + 3y + 3\!\left(\frac{2x-16}{5}\right) \Rightarrow x = -15y + 23$ | M1 | Obtains third variable in terms of other variables

$\left\{x = -15y + 23 = \frac{16+5z}{2}\right\}$ | A1 | Both correct equations (M1 on epen)

**Vector equation of line:**

$\frac{x-0}{1} = \frac{y - \frac{23}{15}}{-\frac{1}{15}} = \frac{z + \frac{16}{5}}{\frac{2}{5}} \Rightarrow \mathbf{r} = \begin{pmatrix}0\\\frac{23}{15}\\-\frac{16}{5}\end{pmatrix} + \lambda\begin{pmatrix}1\\-\frac{1}{15}\\\frac{2}{5}\end{pmatrix}$ | dM1 | Attempts vector equation; requires all previous M marks; $\mathbf{r}=$ may be missing

Any correct equation including $\mathbf{r} =$ e.g. $\mathbf{r} = \begin{pmatrix}23\\0\\6\end{pmatrix} + \lambda\begin{pmatrix}-15\\1\\-6\end{pmatrix}$ | A1 | Any correct equation including "$\mathbf{r}=$"

**Total: 7 marks**

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## Question 4(a) Alt (finds point and vector product of normals):

$-x+3y+3z=-5$ and $2x-5z=16$ | M1 A1 | As above

$x=0 \Rightarrow z = -\frac{16}{5}$ | M1 | Sets one variable equal to a value and finds another variable

$3y = -5 - 3\!\left(-\frac{16}{5}\right) \Rightarrow y = \frac{23}{15}$ giving point $\left(0, \frac{23}{15}, -\frac{16}{5}\right)$ | M1 A1 | Proceeds to find remaining variable; A1: correct values (M1 on epen)

$\begin{pmatrix}-1\\3\\3\end{pmatrix} \times \begin{pmatrix}2\\0\\-5\end{pmatrix} = \begin{pmatrix}-15\\1\\-6\end{pmatrix}$ then forms $\mathbf{r} = \begin{pmatrix}0\\\frac{23}{15}\\-\frac{16}{5}\end{pmatrix} + \lambda\begin{pmatrix}-15\\1\\-6\end{pmatrix}$ | dM1 | Attempts vector product of normals (two correct components if method unclear); forms vector equation with point and direction; requires all previous M marks

Any correct equation including "$\mathbf{r}=$" | A1 |

**Total: 7 marks**

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## Question 4(b) Way 1 (STP including $\overrightarrow{CD}$):

$\left\|\begin{pmatrix}-15\\1\\-6\end{pmatrix}\right\| = \sqrt{262} \Rightarrow \overrightarrow{CD} = \frac{5}{\sqrt{262}}\begin{pmatrix}-15\\1\\-6\end{pmatrix}$ | M1 | Attempts magnitude of direction vector and scales correctly to length 5

Let $C = (8,1,0)$: $\overrightarrow{CA} = \begin{pmatrix}-6\\3\\-5\end{pmatrix}$ and $\overrightarrow{CB} = \begin{pmatrix}-5\\5\\-2\end{pmatrix}$ | M1 | Finds vectors for any two edges other than $CD$; not scored if either vector in terms of a parameter unless assigned/eliminated

$\overrightarrow{CD}.\!\left(\overrightarrow{CA} \times \overrightarrow{CB}\right) = \frac{5}{\sqrt{262}}\begin{pmatrix}-15\\1\\-6\end{pmatrix}.\begin{pmatrix}-6\\3\\-5\end{pmatrix}\times\begin{pmatrix}-5\\5\\-2\end{pmatrix} = \ldots = \left\{-\frac{910}{\sqrt{262}}\right\}$ | M1 | Uses appropriate scalar triple product; must not include position vectors; M0 if clear evidence of inappropriate method

$V = \frac{1}{6}\left|\overrightarrow{CD}.\!\left(\overrightarrow{CA}\times\overrightarrow{CB}\right)\right| = \frac{455}{3\sqrt{262}}$ or $\frac{455\sqrt{262}}{786}$ | dM1 A1 | Divides STP by 6, obtains positive value; requires previous M mark; A1: correct exact value

**Total: 5 marks**

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## Question 4(b) Way 2 (STP not including $\overrightarrow{CD}$):

$\overrightarrow{CD} = \frac{5}{\sqrt{262}}\begin{pmatrix}-15\\1\\-6\end{pmatrix}$ | M1 | As Way 1

$\overrightarrow{AC} = \begin{pmatrix}6\\-3\\5\end{pmatrix}$ and $\overrightarrow{AB} = \begin{pmatrix}1\\2\\3\end{pmatrix}$ | M1 | Finds vectors for any two edges other than $CD$

Finds $\overrightarrow{OD}$ then $\overrightarrow{AD}$, computes $\overrightarrow{AD}.\!\left(\overrightarrow{AB}\times\overrightarrow{AC}\right) = \ldots = \left\{-\frac{910}{\sqrt{262}}\right\}$ | M1 | Uses appropriate scalar triple product; must not include position vectors

$V = \frac{1}{6}\left|\overrightarrow{AD}.\!\left(\overrightarrow{AB}\times\overrightarrow{AC}\right)\right| = \frac{455}{3\sqrt{262}}$ or $\frac{455\sqrt{262}}{786}$ | dM1 A1 | Divides by 6, positive value; A1: correct exact value

**Total: 5 marks**

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## Question 4(b) Way 3 (Triangle area + perpendicular distance × volume of pyramid):

$\overrightarrow{CD} = \frac{5}{\sqrt{262}}\begin{pmatrix}-15\\1\\-6\end{pmatrix}$ | M1 | Attempts magnitude of direction vector, scales to length 5

$\text{Area } \triangle ACD = \frac{1}{2}\left|\overrightarrow{CD}\times\overrightarrow{CA}\right| = \frac{1}{2}\left|\frac{5}{\sqrt{262}}\begin{pmatrix}-15\\1\\-6\end{pmatrix}\times\begin{pmatrix}-6\\3\\-5\end{pmatrix}\right| = \ldots = \left\{\frac{65\sqrt{19}}{2\sqrt{262}}\right\}$ | M1 | Uses formula for area of one face; must be full method (vector product and modulus); condone missing $\frac{1}{2}$

Shortest distance of $B(3,6,-2)$ to $-x+3y+3z=-5$: $\frac{|-1\times3+3\times6+3\times(-2)+5|}{\sqrt{(-1)^2+3^2+3^2}} = \ldots = \left\{\frac{14}{\sqrt{19}}\right\}$ | M1 | Obtains value for perpendicular height via formula or any credible method

$V = \frac{1}{3} \times \frac{65\sqrt{19}}{2\sqrt{262}} \times \frac{14}{\sqrt{19}} = \frac{455}{3\sqrt{262}}$ or $\frac{455\sqrt{262}}{786}$ | dM1 A1 | Uses $\frac{1}{3}\times\text{area}\,\Delta\times\text{perp. height}$, obtains positive value; $\frac{1}{2}$ must have been used for triangle area earlier unless using $\frac{1}{6}\times$; requires previous M mark; A1: either correct exact value

**Total: 5 marks**

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