| Exam Board | Edexcel |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Session | Specimen |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors: Cross Product & Distances |
| Type | Volume of tetrahedron using scalar triple product |
| Difficulty | Challenging +1.2 This is a multi-part Further Maths question requiring systematic application of standard vector techniques: finding plane equations via cross products, computing volume using scalar triple product, using parametric lines with plane equations, and applying the volume scaling formula. While it involves multiple steps and Further Maths content (making it harder than typical A-level), each part follows well-established procedures without requiring novel geometric insight or particularly complex algebraic manipulation. |
| Spec | 1.10g Problem solving with vectors: in geometry4.04b Plane equations: cartesian and vector forms4.04g Vector product: a x b perpendicular vector |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\mathbf{AB} \times \mathbf{AC} = \begin{pmatrix}-1\\-1\\-1\end{pmatrix} \times \begin{pmatrix}1\\-1\\2\end{pmatrix} = \begin{pmatrix}-2-1\\-1+2\\1+1\end{pmatrix} = \begin{pmatrix}-3\\1\\2\end{pmatrix}\) | M1 | 1.1b — Attempting suitable cross product |
| \(\mathbf{r} \cdot \begin{pmatrix}-3\\1\\2\end{pmatrix} = \begin{pmatrix}0\\1\\0\end{pmatrix} \cdot \begin{pmatrix}-3\\1\\2\end{pmatrix} = 1\) | M1 | 1.1b — Complete method for Cartesian equation |
| Hence \(-3x + y + 2z = 1\) | A1 | 1.1b — Accept any equivalent form |
| Total: (3) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Volume of Tetrahedron \(= \frac{1}{6} | \mathbf{n} \cdot (\mathbf{AD}) | \) |
| \(= \frac{1}{6}\left | \begin{pmatrix}-3\\1\\2\end{pmatrix} \cdot \left(\begin{pmatrix}10\\5\\5\end{pmatrix} - \begin{pmatrix}1\\2\\1\end{pmatrix}\right)\right | \) |
| \(= \frac{1}{6} | {-27 + 3 + 8} | = \frac{8}{3}\) |
| Total: (3) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\mathbf{AE} = k\mathbf{AC}\) so \(E\) is \((1+k,\ 2-k,\ 1+2k)\) | M1 | 3.1a — Uses \(E\) on \(AC\) to find expression for \(E\) |
| \(E\) lies on plane so \(2(1+k) - 3(2-k) + 3 = 0\), leading to \(k = \ldots\) | M1 | 3.1a — Uses \(E\) in plane \(\Pi\) to form and solve expression in \(k\) |
| Hence \(k = \frac{1}{5}\) | A1 | 1.1b |
| Total: (3) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Volume \(ABEF = \frac{1}{6}(\mathbf{AB} \times \mathbf{AE}) \cdot \mathbf{AF} = \frac{1}{6}\left(\mathbf{AB} \times \frac{1}{5}\mathbf{AC}\right) \cdot \frac{1}{9}\mathbf{AD}\) | M1 | 3.1a — Uses formula for volume of tetrahedron, substitutes for \(\mathbf{AE}\) and \(\mathbf{AF}\) |
| \(= \frac{1}{45}\left(\frac{1}{6}(\mathbf{AB} \times \mathbf{AC}) \cdot \mathbf{AD}\right)\) and hence result | A1* | 2.2a — Deduces result; use of \(\frac{1}{6}(\mathbf{AB}\times\mathbf{AC})\cdot\mathbf{AD}\) required, no errors seen |
| Total: (2) | (11 marks) |
# Question 6:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\mathbf{AB} \times \mathbf{AC} = \begin{pmatrix}-1\\-1\\-1\end{pmatrix} \times \begin{pmatrix}1\\-1\\2\end{pmatrix} = \begin{pmatrix}-2-1\\-1+2\\1+1\end{pmatrix} = \begin{pmatrix}-3\\1\\2\end{pmatrix}$ | M1 | 1.1b — Attempting suitable cross product |
| $\mathbf{r} \cdot \begin{pmatrix}-3\\1\\2\end{pmatrix} = \begin{pmatrix}0\\1\\0\end{pmatrix} \cdot \begin{pmatrix}-3\\1\\2\end{pmatrix} = 1$ | M1 | 1.1b — Complete method for Cartesian equation |
| Hence $-3x + y + 2z = 1$ | A1 | 1.1b — Accept any equivalent form |
| **Total: (3)** | | |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Volume of Tetrahedron $= \frac{1}{6}|\mathbf{n} \cdot (\mathbf{AD})|$ | M1 | 3.1a — Identifies suitable vectors, attempts correct formula |
| $= \frac{1}{6}\left|\begin{pmatrix}-3\\1\\2\end{pmatrix} \cdot \left(\begin{pmatrix}10\\5\\5\end{pmatrix} - \begin{pmatrix}1\\2\\1\end{pmatrix}\right)\right|$ | M1 | 1.1b — Correct scalar triple product form using $\mathbf{n}$ from (a) |
| $= \frac{1}{6}|{-27 + 3 + 8}| = \frac{8}{3}$ | A1 | 1.1b |
| **Total: (3)** | | |
## Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\mathbf{AE} = k\mathbf{AC}$ so $E$ is $(1+k,\ 2-k,\ 1+2k)$ | M1 | 3.1a — Uses $E$ on $AC$ to find expression for $E$ |
| $E$ lies on plane so $2(1+k) - 3(2-k) + 3 = 0$, leading to $k = \ldots$ | M1 | 3.1a — Uses $E$ in plane $\Pi$ to form and solve expression in $k$ |
| Hence $k = \frac{1}{5}$ | A1 | 1.1b |
| **Total: (3)** | | |
## Part (d):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Volume $ABEF = \frac{1}{6}(\mathbf{AB} \times \mathbf{AE}) \cdot \mathbf{AF} = \frac{1}{6}\left(\mathbf{AB} \times \frac{1}{5}\mathbf{AC}\right) \cdot \frac{1}{9}\mathbf{AD}$ | M1 | 3.1a — Uses formula for volume of tetrahedron, substitutes for $\mathbf{AE}$ and $\mathbf{AF}$ |
| $= \frac{1}{45}\left(\frac{1}{6}(\mathbf{AB} \times \mathbf{AC}) \cdot \mathbf{AD}\right)$ and hence result | A1* | 2.2a — Deduces result; use of $\frac{1}{6}(\mathbf{AB}\times\mathbf{AC})\cdot\mathbf{AD}$ required, no errors seen |
| **Total: (2)** | | **(11 marks)** |
---\begin{enumerate}
\item A tetrahedron has vertices $A ( 1,2,1 ) , B ( 0,1,0 ) , C ( 2,1,3 )$ and $D ( 10,5,5 )$.
\end{enumerate}
Find\\
(a) a Cartesian equation of the plane $A B C$.\\
(b) the volume of the tetrahedron $A B C D$.
The plane $\Pi$ has equation $2 x - 3 y + 3 = 0$\\
The point $E$ lies on the line $A C$ and the point $F$ lies on the line $A D$.\\
Given that $\Pi$ contains the point $B$, the point $E$ and the point $F$,\\
(c) find the value of $k$ such that $\overrightarrow { A E } = k \overrightarrow { A C }$.
Given that $\overrightarrow { A F } = \frac { 1 } { 9 } \overrightarrow { A D }$\\
(d) show that the volume of the tetrahedron $A B C D$ is 45 times the volume of the tetrahedron $A B E F$.
\hfill \mbox{\textit{Edexcel FP1 Q6 [11]}}