| Exam Board | Edexcel |
|---|---|
| Module | P4 (Pure Mathematics 4) |
| Year | 2022 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors: Lines & Planes |
| Type | Triangle and parallelogram areas |
| Difficulty | Standard +0.3 This is a straightforward multi-part vectors question requiring standard techniques: finding a line equation from two points, using perpendicularity condition (dot product = 0), and calculating triangle area using ½|AB × AC|. All steps are routine A-level procedures with no novel insight required, making it slightly easier than average. |
| Spec | 1.10b Vectors in 3D: i,j,k notation1.10c Magnitude and direction: of vectors1.10g Problem solving with vectors: in geometry4.04a Line equations: 2D and 3D, cartesian and vector forms |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\overrightarrow{AB} = \begin{pmatrix}4\\7\\-5\end{pmatrix} = 4\mathbf{i}+7\mathbf{j}-5\mathbf{k}\) | M1 | Attempts \(\overrightarrow{AB}\); implied by 2 out of 3 correct coordinates |
| \(\mathbf{r} = \mathbf{i}-4\mathbf{j}+3\mathbf{k}+\lambda(4\mathbf{i}+7\mathbf{j}-5\mathbf{k})\) or \(\mathbf{r} = 5\mathbf{i}+3\mathbf{j}-2\mathbf{k}+\lambda(4\mathbf{i}+7\mathbf{j}-5\mathbf{k})\) | M1A1 | M1: \(\overrightarrow{OA}+\lambda\times\overrightarrow{AB}\) or \(\overrightarrow{OB}+\lambda\times\overrightarrow{AB}\); A1: any correct equation with \(\mathbf{r}=...\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\overrightarrow{AC} = \begin{pmatrix}2\\p+4\\-4\end{pmatrix} = 2\mathbf{i}+(p+4)\mathbf{j}-4\mathbf{k}\) | M1 | Attempts \(\overrightarrow{AC}\); implied by 2 out of 3 correct coordinates |
| \(\begin{pmatrix}2\\p+4\\-4\end{pmatrix}\cdot\begin{pmatrix}4\\7\\-5\end{pmatrix} = 8+7p+28+20=0 \Rightarrow p=-8\) | M1A1 | M1: takes scalar product of \(\overrightarrow{AB}\) and \(\overrightarrow{AC}\) to form/solve linear equation in \(p\); A1: \(p=-8\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \( | AB | = \sqrt{4^2+7^2+(-5)^2} = \sqrt{90}\) or \( |
| Area \(= \frac{1}{2}\times\sqrt{90}\times 6 = 9\sqrt{10}\) | dM1A1 | dM1: attempts exact area of triangle \(ABC\); dependent on previous M mark; A1: \(9\sqrt{10}\) |
# Question 6:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\overrightarrow{AB} = \begin{pmatrix}4\\7\\-5\end{pmatrix} = 4\mathbf{i}+7\mathbf{j}-5\mathbf{k}$ | M1 | Attempts $\overrightarrow{AB}$; implied by 2 out of 3 correct coordinates |
| $\mathbf{r} = \mathbf{i}-4\mathbf{j}+3\mathbf{k}+\lambda(4\mathbf{i}+7\mathbf{j}-5\mathbf{k})$ or $\mathbf{r} = 5\mathbf{i}+3\mathbf{j}-2\mathbf{k}+\lambda(4\mathbf{i}+7\mathbf{j}-5\mathbf{k})$ | M1A1 | M1: $\overrightarrow{OA}+\lambda\times\overrightarrow{AB}$ or $\overrightarrow{OB}+\lambda\times\overrightarrow{AB}$; A1: any correct equation with $\mathbf{r}=...$ |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\overrightarrow{AC} = \begin{pmatrix}2\\p+4\\-4\end{pmatrix} = 2\mathbf{i}+(p+4)\mathbf{j}-4\mathbf{k}$ | M1 | Attempts $\overrightarrow{AC}$; implied by 2 out of 3 correct coordinates |
| $\begin{pmatrix}2\\p+4\\-4\end{pmatrix}\cdot\begin{pmatrix}4\\7\\-5\end{pmatrix} = 8+7p+28+20=0 \Rightarrow p=-8$ | M1A1 | M1: takes scalar product of $\overrightarrow{AB}$ and $\overrightarrow{AC}$ to form/solve linear equation in $p$; A1: $p=-8$ |
## Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $|AB| = \sqrt{4^2+7^2+(-5)^2} = \sqrt{90}$ or $|AC| = \sqrt{2^2+(-4)^2+(-4)^2} = 6$ | M1 | Attempts magnitude of either $\overrightarrow{AB}$ or $\overrightarrow{AC}$ using their $p$ |
| Area $= \frac{1}{2}\times\sqrt{90}\times 6 = 9\sqrt{10}$ | dM1A1 | dM1: attempts exact area of triangle $ABC$; dependent on previous M mark; A1: $9\sqrt{10}$ |
---\begin{enumerate}
\item Relative to a fixed origin $O$,
\end{enumerate}
\begin{itemize}
\item the point $A$ has position vector $\quad \mathbf { i } - 4 \mathbf { j } + 3 \mathbf { k }$
\item the point $B$ has position vector $5 \mathbf { i } + 3 \mathbf { j } - 2 \mathbf { k }$
\item the point $C$ has position vector $3 \mathbf { i } + p \mathbf { j } - \mathbf { k }$\\
where $p$ is a constant.\\
The line $l$ passes through $A$ and $B$.\\
(a) Find a vector equation for the line $l$
\end{itemize}
Given that $\overrightarrow { A C }$ is perpendicular to $l$\\
(b) find the value of $p$\\
(c) Hence find the area of triangle $A B C$, giving your answer as a surd in simplest form.
\hfill \mbox{\textit{Edexcel P4 2022 Q6 [9]}}