## Question 5(a):

| Working/Answer | Mark | Guidance |
|---|---|---|
| $\pm\overrightarrow{DE}=\pm\begin{pmatrix}-4\\3\\1\end{pmatrix}$, $\pm\overrightarrow{DF}=\pm\begin{pmatrix}-1\\5\\2\end{pmatrix}$, $\pm\overrightarrow{EF}=\pm\begin{pmatrix}3\\2\\1\end{pmatrix}$ | M1 | Correct vectors |
| $\text{Area}=\frac{1}{2}\left|\overrightarrow{DE}\times\overrightarrow{DF}\right|=\frac{1}{2}\begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\-4&3&1\\-1&5&2\end{vmatrix}=\frac{1}{2}\sqrt{1^2+7^2+17^2}$ | M1 | Correct cross product method |
| $=\frac{1}{2}\sqrt{339}\text{ (cm}^2\text{)}$ | A1* | Correct area |

**Alternative using trigonometry:**

| Working/Answer | Mark | Guidance |
|---|---|---|
| $DE=\sqrt{26}$, $DF=\sqrt{30}$, $EF=\sqrt{14}$ | M1 | Correct side lengths |
| $\cos DEF=\frac{26+14-30}{2\sqrt{26}\sqrt{14}}\Rightarrow DEF=\cos^{-1}\frac{5}{\sqrt{26}\sqrt{14}}$; Area $DEF=\frac{1}{2}\times\sqrt{26}\sqrt{14}\sqrt{1-\frac{25}{364}}=\ldots$ | M1 | Cosine rule and area formula |
| $=\frac{1}{2}\times\sqrt{26}\sqrt{14}\cdot\frac{\sqrt{339}}{\sqrt{26}\sqrt{14}}=\frac{1}{2}\sqrt{339}$ | A1 | Correct area |

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## Question 5(b):

| Working/Answer | Mark | Guidance |
|---|---|---|
| Attempt to find $T$, the 4th vertex: $\begin{pmatrix}2\\-1\\8\end{pmatrix}+\lambda\overrightarrow{AD}=\begin{pmatrix}1\\4\\10\end{pmatrix}+\mu\overrightarrow{CF}\Rightarrow\lambda=\ldots$ or $\mu=\ldots$; or e.g. $DF=\sqrt{30}$, $AC=\sqrt{120}$, Linear SF $=2$, $AT=2AD$ | M1 | Correct strategy for finding $T$ |
| $T(1,1,15)$ | A1 | Correct coordinates |
| $\overrightarrow{AT}=\begin{pmatrix}-2\\4\\14\end{pmatrix}$, $\overrightarrow{BT}=\begin{pmatrix}6\\-2\\12\end{pmatrix}$, $\overrightarrow{CT}=\begin{pmatrix}0\\-6\\10\end{pmatrix}$; $\overrightarrow{AT}\times\overrightarrow{BT}\cdot\overrightarrow{CT}=\begin{vmatrix}-2&4&14\\6&-2&12\\0&-6&10\end{vmatrix}=\ldots$ | M1 | Scalar triple product for volume |
| $V=\frac{1}{6}\left|-2(-20+72)-4(60)+14(-36)\right|=\frac{424}{3}$ | A1 | Correct volume of tetrahedron $ABCT$ |
| $\frac{1}{6}\overrightarrow{DT}\times\overrightarrow{ET}\cdot\overrightarrow{FT}=\frac{1}{6}\begin{vmatrix}-1&2&7\\3&-1&6\\0&-3&5\end{vmatrix}=\frac{1}{6}\left|-1(13)-2(15)+7(-9)\right|=\frac{53}{3}$ | M1 | Scalar triple product for second tetrahedron volume |

## Question (Part a):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Attempts to find 2 edges of triangle $DEF$ by subtracting components | M1 | Must be subtracting components |
| Uses correct process of vector product to attempt area including use of Pythagoras | M1 | |
| Deduces correct area with no errors | A1* | Correct area with no errors |

**Alternative for part (a):**

| Answer/Working | Mark | Guidance |
|---|---|---|
| Attempts to find 3 edges of triangle $DEF$ by subtracting components | M1 | Must be subtracting components |
| A complete method for the area | M1 | Allow decimals for this mark; must work in exact terms for A mark |
| Deduces correct area with no errors and no decimal work | A1* | |

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## Question (Part b) – Main Method:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Adopts correct strategy to find fourth vertex of tetrahedron e.g. finding where two edges intersect or uses linear scale factor | M1 | |
| Correct coordinates for the other vertex | A1 | |
| Uses scalar triple product between appropriate vectors to find volume of smaller or larger tetrahedron | M1 | |
| Correct volume for either tetrahedron | A1 | |
| Makes further progress by finding volume of other tetrahedron or calculates volume scale factor using appropriate method | M1 | e.g. using ratios or finding area of triangle $DEF$ and comparing with triangle $ABC$ |
| $\text{Volume} = \frac{424}{3} - \frac{53}{3} = \ldots$ or $\text{Volume} = \frac{7}{8} \times \frac{424}{3} = \ldots$ or $\text{Volume} = 7 \times \frac{53}{3} = \ldots$ | dM1 | Completes the problem by finding required volume of frustum. Depends on all previous method marks |
| $= \dfrac{371}{3} \text{ cm}^3$ | A1 | **(7 marks total)** |

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## Question (Part b) – Alternative (splits into 4 tetrahedra, $M$ = midpoint of $AC$):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $Vol_{ABMD} = \frac{1}{6}\overrightarrow{AB} \times \overrightarrow{AD} \bullet \overrightarrow{AM} = \ldots$ | M1 | Adopts correct strategy to find volume of one tetrahedron |
| $= \dfrac{106}{3}$ | A1 | Any correct volume |
| $Vol_{MBFC} = \frac{1}{6}\overrightarrow{BF} \times \overrightarrow{BC} \bullet \overrightarrow{BM} = \ldots\left(\dfrac{106}{3}\right)$ | M1 | Adopts correct strategy for at least 2 other tetrahedra |
| $Vol_{DMFB} = \frac{1}{6}\overrightarrow{DF} \times \overrightarrow{DM} \bullet \overrightarrow{DB} = \ldots\left(\dfrac{106}{3}\right)$ | A1 | Correct volumes |
| $Vol_{EDFB} = \frac{1}{6}\overrightarrow{ED} \times \overrightarrow{EF} \bullet \overrightarrow{EB} = \ldots\left(\dfrac{53}{3}\right)$ | M1 | Makes further progress finding all relevant tetrahedra |
| $ABMD + MBFC + DMFB + EDFB = 3 \times \dfrac{106}{3} + \dfrac{53}{3}$ | dM1 | Completes the problem by finding required volume of frustum. Depends on all previous method marks |
| $= \dfrac{371}{3} \text{ cm}^3$ | A1 | |