| Exam Board | Edexcel |
|---|---|
| Module | P4 (Pure Mathematics 4) |
| Year | 2023 |
| Session | January |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors: Lines & Planes |
| Type | Point on line satisfying condition |
| Difficulty | Standard +0.3 This is a standard Further Maths vectors question requiring routine techniques: finding a direction vector, writing a line equation, and using perpendicularity (dot product = 0) to find a point. While it involves multiple steps, each is algorithmic with no novel insight required, making it slightly easier than the average A-level question. |
| Spec | 1.10c Magnitude and direction: of vectors1.10g Problem solving with vectors: in geometry4.04a Line equations: 2D and 3D, cartesian and vector forms4.04c Scalar product: calculate and use for angles |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(\overrightarrow{AB} = (\pm)\left[(8\mathbf{i}+3\mathbf{j}-7\mathbf{k})-(2\mathbf{i}-3\mathbf{j}+5\mathbf{k})\right] = \ldots\) | M1 | Attempts to subtract vectors \(\overrightarrow{OA}\) and \(\overrightarrow{OB}\) either way around. May be implied by two correct components |
| \(\overrightarrow{AB} = 6\mathbf{i}+6\mathbf{j}-12\mathbf{k}\) | A1 | o.e. |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(\mathbf{r} = \begin{pmatrix}2\\-3\\5\end{pmatrix} + \lambda\begin{pmatrix}1\\1\\-2\end{pmatrix}\) o.e. | B1ft | Any correct equation for the line; may use correct or follow-through multiple of \(\overrightarrow{AB}\) for direction and any point on the line. Must start \(\mathbf{r} = \ldots\) or accept \(x\mathbf{i}+y\mathbf{j}+z\mathbf{k} = \ldots\) (\(l = \ldots\) is B0) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| Attempts \(\pm\overrightarrow{CP} = \pm\begin{pmatrix}2+\lambda-3\\-3+\lambda-5\\5-2\lambda-2\end{pmatrix}\) | M1 | Attempts \(\pm\overrightarrow{CP}\) using point \(C\) and a general point on \(l\) |
| \(\overrightarrow{CP} \bullet k\begin{pmatrix}1\\1\\-2\end{pmatrix} = 0 \Rightarrow 1(\lambda-1)+1(\lambda-8)-2(-2\lambda+3)=0\) | dM1 | Sets scalar product of their \(\overrightarrow{CP}\) and direction of \(l\) (or \(\overrightarrow{AB}\)) to 0; proceeds to equation in \(\lambda\). Condone sign slips if intention is clear. Alt: attempts to minimise distance \(CP\) by completing the square or differentiation |
| \(\Rightarrow \lambda = \frac{5}{2}\) | A1 | Use of \(\overrightarrow{AB}\) in \(\overrightarrow{CP}\) gives \(\lambda = \frac{5}{12}\); use of \(\overrightarrow{OB}\): \(\lambda = \frac{-7}{2}\) or \(\frac{-7}{12}\) |
| \(\overrightarrow{OP} = \begin{pmatrix}2\\-3\\5\end{pmatrix}+\frac{5}{2}\begin{pmatrix}1\\1\\-2\end{pmatrix} = \frac{9}{2}\mathbf{i}-\frac{1}{2}\mathbf{j}\) | ddM1, A1 | Substitutes \(\lambda\) from correct method into \(l\); Accept as coordinates and accept \(P = \ldots\) instead of \(\overrightarrow{OP}\) |
## Question 6:
### Part (a)(i):
| Working | Mark | Guidance |
|---------|------|----------|
| $\overrightarrow{AB} = (\pm)\left[(8\mathbf{i}+3\mathbf{j}-7\mathbf{k})-(2\mathbf{i}-3\mathbf{j}+5\mathbf{k})\right] = \ldots$ | M1 | Attempts to subtract vectors $\overrightarrow{OA}$ and $\overrightarrow{OB}$ either way around. May be implied by two correct components |
| $\overrightarrow{AB} = 6\mathbf{i}+6\mathbf{j}-12\mathbf{k}$ | A1 | o.e. |
### Part (a)(ii):
| Working | Mark | Guidance |
|---------|------|----------|
| $\mathbf{r} = \begin{pmatrix}2\\-3\\5\end{pmatrix} + \lambda\begin{pmatrix}1\\1\\-2\end{pmatrix}$ o.e. | B1ft | Any correct equation for the line; may use correct or follow-through multiple of $\overrightarrow{AB}$ for direction and any point on the line. Must start $\mathbf{r} = \ldots$ or accept $x\mathbf{i}+y\mathbf{j}+z\mathbf{k} = \ldots$ ($l = \ldots$ is B0) |
### Part (b):
| Working | Mark | Guidance |
|---------|------|----------|
| Attempts $\pm\overrightarrow{CP} = \pm\begin{pmatrix}2+\lambda-3\\-3+\lambda-5\\5-2\lambda-2\end{pmatrix}$ | M1 | Attempts $\pm\overrightarrow{CP}$ using point $C$ and a general point on $l$ |
| $\overrightarrow{CP} \bullet k\begin{pmatrix}1\\1\\-2\end{pmatrix} = 0 \Rightarrow 1(\lambda-1)+1(\lambda-8)-2(-2\lambda+3)=0$ | dM1 | Sets scalar product of their $\overrightarrow{CP}$ and direction of $l$ (or $\overrightarrow{AB}$) to 0; proceeds to equation in $\lambda$. Condone sign slips if intention is clear. Alt: attempts to minimise distance $CP$ by completing the square or differentiation |
| $\Rightarrow \lambda = \frac{5}{2}$ | A1 | Use of $\overrightarrow{AB}$ in $\overrightarrow{CP}$ gives $\lambda = \frac{5}{12}$; use of $\overrightarrow{OB}$: $\lambda = \frac{-7}{2}$ or $\frac{-7}{12}$ |
| $\overrightarrow{OP} = \begin{pmatrix}2\\-3\\5\end{pmatrix}+\frac{5}{2}\begin{pmatrix}1\\1\\-2\end{pmatrix} = \frac{9}{2}\mathbf{i}-\frac{1}{2}\mathbf{j}$ | ddM1, A1 | Substitutes $\lambda$ from correct method into $l$; Accept as coordinates and accept $P = \ldots$ instead of $\overrightarrow{OP}$ |
---\begin{enumerate}
\item Relative to a fixed origin $O$.
\end{enumerate}
\begin{itemize}
\item the point $A$ has position vector $2 \mathbf { i } - 3 \mathbf { j } + 5 \mathbf { k }$
\item the point $B$ has position vector $8 \mathbf { i } + 3 \mathbf { j } - 7 \mathbf { k }$
\end{itemize}
The line $l$ passes through $A$ and $B$.\\
(a) (i) Find $\overrightarrow { A B }$\\
(ii) Find a vector equation for the line $l$
The point $C$ has position vector $3 \mathbf { i } + 5 \mathbf { j } + 2 \mathbf { k }$\\
The point $P$ lies on $l$\\
Given that $\overrightarrow { C P }$ is perpendicular to $l$\\
(b) find the position vector of the point $P$
\hfill \mbox{\textit{Edexcel P4 2023 Q6 [8]}}