| Exam Board | Edexcel |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Session | Specimen |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors: Cross Product & Distances |
| Type | Area of triangle using cross product |
| Difficulty | Standard +0.3 This is a standard Further Maths FP3 vectors question testing routine application of cross product for area, scalar triple product for volume, and using the volume formula to find perpendicular distance. All three parts follow textbook methods with no novel insight required, though the multi-step nature and Further Maths content places it slightly above average A-level difficulty. |
| Spec | 1.10c Magnitude and direction: of vectors1.10d Vector operations: addition and scalar multiplication4.04g Vector product: a x b perpendicular vector |
| Answer | Marks | Guidance |
|---|---|---|
| \(\overrightarrow{AB} = (-1, 3, -1); \quad \overrightarrow{AC} = (-1, 3, 1)\) | M1A1 | |
| \(\overrightarrow{AB} \times \overrightarrow{AC} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -1 & 3 & -1 \\ -1 & 3 & 1 \end{vmatrix}\) | ||
| \(= \mathbf{i}(3+3) + \mathbf{j}(1+1) + \mathbf{k}(-3+3)\) | M1, A1, A1 | |
| \(= 6\mathbf{i} + 2\mathbf{j}\) | ||
| Area of \(\triangle ABC = \frac{1}{2} | \overrightarrow{AB} \times \overrightarrow{AC} | = \frac{1}{2}\sqrt{36+4} = \sqrt{10}\) square units |
| Answer | Marks | Guidance |
|---|---|---|
| Volume of tetrahedron \(= \frac{1}{6}\left | \overrightarrow{AD} \cdot (\overrightarrow{AB} \times \overrightarrow{AC})\right | \) |
| \(= \frac{1}{6} | -12+8 | = \frac{2}{3}\) cubic units |
| Answer | Marks | Guidance |
|---|---|---|
| Unit vector in direction \(\overrightarrow{AB} \times \overrightarrow{AC}\) is: \(\mathbf{n} = \frac{1}{\sqrt{40}}(6\mathbf{i} + 2\mathbf{j}) = \frac{1}{\sqrt{10}}(3\mathbf{i} + \mathbf{j})\) | M1 | |
| \(p = | \mathbf{n} \cdot \overrightarrow{AD} | = \frac{1}{\sqrt{10}} |
## (a)
$\overrightarrow{AB} = (-1, 3, -1); \quad \overrightarrow{AC} = (-1, 3, 1)$ | M1A1 |
$\overrightarrow{AB} \times \overrightarrow{AC} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -1 & 3 & -1 \\ -1 & 3 & 1 \end{vmatrix}$ | |
$= \mathbf{i}(3+3) + \mathbf{j}(1+1) + \mathbf{k}(-3+3)$ | M1, A1, A1 |
$= 6\mathbf{i} + 2\mathbf{j}$ | |
Area of $\triangle ABC = \frac{1}{2}|\overrightarrow{AB} \times \overrightarrow{AC}| = \frac{1}{2}\sqrt{36+4} = \sqrt{10}$ square units | M1A1ft | (7)
## (b)
Volume of tetrahedron $= \frac{1}{6}\left|\overrightarrow{AD} \cdot (\overrightarrow{AB} \times \overrightarrow{AC})\right|$ | |
$= \frac{1}{6}|-12+8| = \frac{2}{3}$ cubic units | M1A1 | (2)
## (c)
Unit vector in direction $\overrightarrow{AB} \times \overrightarrow{AC}$ is: $\mathbf{n} = \frac{1}{\sqrt{40}}(6\mathbf{i} + 2\mathbf{j}) = \frac{1}{\sqrt{10}}(3\mathbf{i} + \mathbf{j})$ | M1 |
$p = |\mathbf{n} \cdot \overrightarrow{AD}| = \frac{1}{\sqrt{10}}|(3\mathbf{i} + \mathbf{j}) \cdot (-2\mathbf{i} + 4\mathbf{j})| = \frac{1}{\sqrt{10}}|-6+4| = \frac{2}{\sqrt{10}}$ units | M1A1 | (3)
---The points $A$, $B$, $C$, and $D$ have position vectors
$$\mathbf{a} = 2\mathbf{i} + \mathbf{k}, \quad \mathbf{b} = \mathbf{i} + 3\mathbf{j}, \quad \mathbf{c} = \mathbf{i} + 3\mathbf{j} + 2\mathbf{k}, \quad \mathbf{d} = 4\mathbf{j} + \mathbf{k}$$
respectively.
\begin{enumerate}[label=(\alph*)]
\item Find $\overrightarrow{AB} \times \overrightarrow{AC}$ and hence find the area of triangle $ABC$. [7]
\item Find the volume of the tetrahedron $ABCD$. [2]
\item Find the perpendicular distance of $D$ from the plane containing $A$, $B$ and $C$. [3]
\end{enumerate}
(Total 12 marks)
\hfill \mbox{\textit{Edexcel FP3 Q8 [12]}}