# Question 1:

## Part (a) — Main Method (Cross Product)

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\pm\overrightarrow{AB} = \pm\begin{pmatrix}4\\-4\\-1\end{pmatrix}$, $\pm\overrightarrow{BC} = \pm\begin{pmatrix}-1\\5\\2\end{pmatrix}$, $\pm\overrightarrow{AC} = \pm\begin{pmatrix}3\\1\\1\end{pmatrix}$ | M1 | Attempts any 2 of these vectors. Allow written as coordinates. |
| $\overrightarrow{AB}\times\overrightarrow{AC} = \begin{pmatrix}4\\-4\\-1\end{pmatrix}\times\begin{pmatrix}3\\1\\1\end{pmatrix} = \begin{pmatrix}-3\\-7\\16\end{pmatrix}$ | dM1 | Attempts vector product of 2 appropriate vectors. If no working shown, look for at least 2 correct elements. |
| Area $= \frac{1}{2}\sqrt{3^2+7^2+16^2} = \frac{1}{2}\sqrt{314}$ | A1 | Correct exact area. Allow recovery from sign errors in vector product e.g. $\pm 3\mathbf{i} \pm 7\mathbf{j} \pm 16\mathbf{k}$ |

> Note: A correct exact area of $\frac{1}{2}\sqrt{314}$ with no evidence of incorrect work scores full marks.

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## Part (a) — Alternative 1 (Cosine Rule)

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\pm\overrightarrow{AB} = \pm\begin{pmatrix}4\\-4\\-1\end{pmatrix}$, $\pm\overrightarrow{BC} = \pm\begin{pmatrix}-1\\5\\2\end{pmatrix}$, $\pm\overrightarrow{AC} = \pm\begin{pmatrix}3\\1\\1\end{pmatrix}$ | M1 | Attempts any 2 of these vectors. |
| $\|\pm\overrightarrow{AB}\| = \sqrt{4^2+4^2+1^2}$, $\|\pm\overrightarrow{BC}\| = \sqrt{1^2+5^2+2^2}$, $\|\pm\overrightarrow{AC}\| = \sqrt{3^2+1^2+1^2}$ $\cos A = \frac{33+11-30}{2\sqrt{33}\sqrt{11}} = \frac{7\sqrt{3}}{33}$ or $\cos B = \frac{30+33-11}{2\sqrt{30}\sqrt{33}} = \frac{13\sqrt{2}}{3\sqrt{55}}$ or $\cos C = \frac{30+11-33}{2\sqrt{30}\sqrt{11}} = \frac{\sqrt{8}}{\sqrt{165}}$ (For reference $A=68.44...°$, $B=34.27...°$, $C=77.27...°$) **or** $\cos A = \frac{\overrightarrow{AB}\cdot\overrightarrow{AC}}{\sqrt{33}\sqrt{11}} = \frac{12-4-1}{\sqrt{33}\sqrt{11}}$ | dM1 | Attempts magnitude of all 3 sides and cosine of one angle using correctly applied cosine rule, **or** finds magnitude of 2 sides and cosine of included angle using correctly applied scalar product. |
| Area $= \frac{1}{2}\sqrt{11}\sqrt{33}\sin A = \frac{1}{2}\sqrt{314}$ **or** Area $= \frac{1}{2}\sqrt{30}\sqrt{33}\sin B = \frac{1}{2}\sqrt{314}$ **or** Area $= \frac{1}{2}\sqrt{30}\sqrt{11}\sin C = \frac{1}{2}\sqrt{314}$ | A1 | Correct exact area. Allow recovery from sign errors that do not affect calculations e.g. $\pm\overrightarrow{AB}=\pm4\mathbf{i}\pm4\mathbf{j}\pm\mathbf{k}$, $\pm\overrightarrow{BC}=\pm\mathbf{i}\pm5\mathbf{j}\pm2\mathbf{k}$, $\pm\overrightarrow{AC}=\pm3\mathbf{i}\pm\mathbf{j}\pm\mathbf{k}$. Allow decimals as long as correct exact area found. |

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## Part (a) — Alternative 2 (Scalar Product / Altitude)

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\pm\overrightarrow{AB} = \pm\begin{pmatrix}4\\-4\\-1\end{pmatrix}$, $\pm\overrightarrow{BC} = \pm\begin{pmatrix}-1\\5\\2\end{pmatrix}$, $\pm\overrightarrow{AC} = \pm\begin{pmatrix}3\\1\\1\end{pmatrix}$ | M1 | Attempts any 2 of these vectors. |
| $A$ to $BC$ is $\sqrt{AB^2 - \left(\frac{\overrightarrow{AB}\cdot\overrightarrow{BC}}{BC}\right)^2} = \sqrt{\frac{157}{15}}$ **or** $B$ to $CA$ is $\sqrt{BC^2 - \left(\frac{\overrightarrow{BC}\cdot\overrightarrow{CA}}{CA}\right)^2} = \sqrt{\frac{314}{11}}$ **or** $C$ to $BA$ is $\sqrt{AC^2 - \left(\frac{\overrightarrow{AC}\cdot\overrightarrow{AB}}{AB}\right)^2} = \sqrt{\frac{314}{33}}$ | dM1 | Attempts one of the altitudes of triangle $ABC$ using a correct method. |
| Area $= \frac{1}{2}\sqrt{30}\sqrt{\frac{157}{15}} = \frac{1}{2}\sqrt{314}$ **or** Area $= \frac{1}{2}\sqrt{11}\sqrt{\frac{314}{11}} = \frac{1}{2}\sqrt{314}$ **or** Area $= \frac{1}{2}\sqrt{33}\sqrt{\frac{314}{33}} = \frac{1}{2}\sqrt{314}$ | A1 | Correct exact area. Allow work in decimals as long as correct exact area is found. |

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## Part (a) — Alternative 3 (Vector Products of position vectors)

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\mathbf{a}\times\mathbf{b} = \begin{pmatrix}0\\4\\-16\end{pmatrix}$, $\mathbf{b}\times\mathbf{c} = \begin{pmatrix}0\\-8\\20\end{pmatrix}$, $\mathbf{c}\times\mathbf{a} = \begin{pmatrix}-3\\-3\\12\end{pmatrix}$ | M1 | Attempts these vector products. |
| $\mathbf{a}\times\mathbf{b}+\mathbf{b}\times\mathbf{c}+\mathbf{c}\times\mathbf{a} = \begin{pmatrix}-3\\-7\\16\end{pmatrix}$ | dM1 | Adds the appropriate vector products. |
| Area $= \frac{1}{2}\sqrt{3^2+7^2+16^2} = \frac{1}{2}\sqrt{314}$ | A1 | Correct exact area. Allow work in decimals as long as correct exact area is found. |

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## Part (b)

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\pm\overrightarrow{AD} = \pm\begin{pmatrix}2\\-2\\k-1\end{pmatrix}$, $\pm\overrightarrow{BD} = \pm\begin{pmatrix}-2\\2\\k\end{pmatrix}$, $\pm\overrightarrow{CD} = \pm\begin{pmatrix}-1\\-3\\k-2\end{pmatrix}$ | M1 | Attempts one of these vectors. |
| $\overrightarrow{AB}\times\overrightarrow{AC}\cdot\overrightarrow{AD} = \begin{pmatrix}-3\\-7\\16\end{pmatrix}\cdot\begin{pmatrix}2\\-2\\k-1\end{pmatrix} = -6+14+16k-16$ **or** $\overrightarrow{AB}\times\overrightarrow{AC}\cdot\overrightarrow{BD} = \begin{pmatrix}-3\\-7\\16\end{pmatrix}\cdot\begin{pmatrix}-2\\2\\k\end{pmatrix} = 6-14+16k$ **or** $\overrightarrow{AB}\times\overrightarrow{AC}\cdot\overrightarrow{CD} = \begin{pmatrix}-3\\-7\\16\end{pmatrix}\cdot\begin{pmatrix}-1\\-3\\k-2\end{pmatrix} = 3+21+16k-32$ | dM1 | Attempts a suitable triple product to obtain a scalar quantity ($\frac{1}{6}$ not required). Must be forming the triple product correctly, not the magnitude of a vector. Must be an attempt at the tetrahedron $ABCD$. |
| Volume $= \frac{1}{3}\|8k-4\|$ | A1 | Must see modulus and must be 2 terms but allow equivalents e.g. $\frac{4}{3}\|2k-1\|$, $\frac{1}{6}\|16k-8\|$, $\frac{1}{6}\|8-16k\|$. Award once a correct answer is seen and apply isw if necessary. |

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