| Exam Board | Edexcel |
|---|---|
| Module | FP1 AS (Further Pure 1 AS) |
| Year | 2023 |
| Session | June |
| Marks | 9 |
| 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 straightforward application of cross product formulas from FP1. Part (a) requires computing the cross product symbolically, then using the parallel condition to set up two equations in p and q. Part (b) is direct recall that area = ½|AB × AC|. While it involves algebraic manipulation, it follows a standard template with no conceptual surprises. |
| Spec | 4.04g Vector product: a x b perpendicular vector |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Complete method to find \(p\) and \(q\): attempt \(\overrightarrow{AB} \times \overrightarrow{AC}\) set equal to multiple of \(2\mathbf{i}+3\mathbf{j}+4\mathbf{k}\), form and solve two simultaneous equations | M1 | 3.1a |
| \(\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ p & 4 & 6 \\ q & 4 & 5 \end{vmatrix} = (20-24)\mathbf{i} - (5p-6q)\mathbf{j} + (4p-4q)\mathbf{k}\) | M1 | 1.1b |
| Correct cross product | A1 | 1.1b |
| \(\overrightarrow{AB} \times \overrightarrow{AC} = -2(2\mathbf{i}+3\mathbf{j}+4\mathbf{k})\) or \(\begin{pmatrix}2\\3\\4\end{pmatrix} \times (\overrightarrow{AB} \times \overrightarrow{AC}) = \mathbf{0}\) | B1 | 2.2a |
| \(-5p-6q = -2\times3\) and \(4p-4q = -2\times4\) | M1 | 3.1a |
| Solves \(5p-6q=6\) and \(4p-4q=-8\) to find values for \(p\) and \(q\) | dM1 | 1.1b |
| \(p=-18,\ q=-16\) | A1 | 1.1b |
| (7 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Scalar product of both \(\overrightarrow{AB}\) and \(\overrightarrow{AC}\) with \(2\mathbf{i}+3\mathbf{j}+4\mathbf{k}\) set equal to 0 | M1 | 3.1a |
| \(\begin{pmatrix}p\\4\\6\end{pmatrix}\cdot\begin{pmatrix}2\\3\\4\end{pmatrix} = 2p+12+24\) or \(\begin{pmatrix}q\\4\\5\end{pmatrix}\cdot\begin{pmatrix}2\\3\\4\end{pmatrix} = 2q+12+20\) | M1, A1 | 1.1b |
| \(2p+36=0\) or \(2q+32=0\) | B1 | 2.2a |
| \(\Rightarrow p=\ldots\) or \(\Rightarrow q=\ldots\) | M1 | 1.1b |
| \(\Rightarrow p=\ldots\) and \(\Rightarrow q=\ldots\) | dM1 | 3.1a |
| \(p=-18,\ q=-16\) | A1 | 1.1b |
| (7 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Area \(= \frac{1}{2}\begin{vmatrix}-4\\-6\\-8\end{vmatrix} = \frac{1}{2}\left[\sqrt{(-4)^2+(-6)^2+(-8)^2}\right] = \ldots\) | M1 | 1.1b |
| Area \(= \sqrt{29}\) or \(\frac{1}{2}\sqrt{116}\) | A1 | 1.1b |
| (2 marks) |
## Question 5(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Complete method to find $p$ and $q$: attempt $\overrightarrow{AB} \times \overrightarrow{AC}$ set equal to multiple of $2\mathbf{i}+3\mathbf{j}+4\mathbf{k}$, form and solve two simultaneous equations | M1 | 3.1a |
| $\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ p & 4 & 6 \\ q & 4 & 5 \end{vmatrix} = (20-24)\mathbf{i} - (5p-6q)\mathbf{j} + (4p-4q)\mathbf{k}$ | M1 | 1.1b |
| Correct cross product | A1 | 1.1b |
| $\overrightarrow{AB} \times \overrightarrow{AC} = -2(2\mathbf{i}+3\mathbf{j}+4\mathbf{k})$ or $\begin{pmatrix}2\\3\\4\end{pmatrix} \times (\overrightarrow{AB} \times \overrightarrow{AC}) = \mathbf{0}$ | B1 | 2.2a |
| $-5p-6q = -2\times3$ and $4p-4q = -2\times4$ | M1 | 3.1a |
| Solves $5p-6q=6$ and $4p-4q=-8$ to find values for $p$ and $q$ | dM1 | 1.1b |
| $p=-18,\ q=-16$ | A1 | 1.1b |
| **(7 marks)** | | |
**Alt method:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| Scalar product of both $\overrightarrow{AB}$ and $\overrightarrow{AC}$ with $2\mathbf{i}+3\mathbf{j}+4\mathbf{k}$ set equal to 0 | M1 | 3.1a |
| $\begin{pmatrix}p\\4\\6\end{pmatrix}\cdot\begin{pmatrix}2\\3\\4\end{pmatrix} = 2p+12+24$ or $\begin{pmatrix}q\\4\\5\end{pmatrix}\cdot\begin{pmatrix}2\\3\\4\end{pmatrix} = 2q+12+20$ | M1, A1 | 1.1b |
| $2p+36=0$ or $2q+32=0$ | B1 | 2.2a |
| $\Rightarrow p=\ldots$ or $\Rightarrow q=\ldots$ | M1 | 1.1b |
| $\Rightarrow p=\ldots$ and $\Rightarrow q=\ldots$ | dM1 | 3.1a |
| $p=-18,\ q=-16$ | A1 | 1.1b |
| **(7 marks)** | | |
---
## Question 5(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Area $= \frac{1}{2}\begin{vmatrix}-4\\-6\\-8\end{vmatrix} = \frac{1}{2}\left[\sqrt{(-4)^2+(-6)^2+(-8)^2}\right] = \ldots$ | M1 | 1.1b |
| Area $= \sqrt{29}$ or $\frac{1}{2}\sqrt{116}$ | A1 | 1.1b |
| **(2 marks)** | | |
---\begin{enumerate}
\item The points $A , B$ and $C$ are the vertices of a triangle.
\end{enumerate}
Given that
\begin{itemize}
\item $\overrightarrow { A B } = \left( \begin{array} { l } p \\ 4 \\ 6 \end{array} \right)$ and $\overrightarrow { A C } = \left( \begin{array} { l } q \\ 4 \\ 5 \end{array} \right)$ where $p$ and $q$ are constants
\item $\overrightarrow { A B } \times \overrightarrow { A C }$ is parallel to $2 \mathbf { i } + 3 \mathbf { j } + 4 \mathbf { k }$\\
(a) determine the value of $p$ and the value of $q$\\
(b) Hence, determine the exact area of triangle $A B C$
\end{itemize}
\hfill \mbox{\textit{Edexcel FP1 AS 2023 Q5 [9]}}