| Exam Board | Edexcel |
|---|---|
| Module | P4 (Pure Mathematics 4) |
| Year | 2023 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors: Lines & Planes |
| Type | Reflection in plane |
| Difficulty | Standard +0.8 This is a multi-part Further Maths vectors question requiring: (a) standard line equation from two points, (b) perpendicular projection using dot product, (c) reflection using the foot of perpendicular as midpoint, (d) distance calculation. While systematic, it requires careful coordinate work across multiple steps and understanding of geometric relationships in 3D, placing it moderately above average difficulty. |
| Spec | 1.10b Vectors in 3D: i,j,k notation1.10c Magnitude and direction: of vectors1.10f Distance between points: using position 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 |
| Attempts direction vector by subtracting \((5\mathbf{i}+6\mathbf{j}+3\mathbf{k})\) and \((4\mathbf{i}+8\mathbf{j}+\mathbf{k})\) either way around | M1 | If no method shown, look for two correct components of \((\pm\mathbf{i}\pm2\mathbf{j}\pm2\mathbf{k})\) or multiple |
| E.g. \(\mathbf{r} = 4\mathbf{i}+8\mathbf{j}+\mathbf{k}+\lambda(\mathbf{i}-2\mathbf{j}+2\mathbf{k})\), \(\mathbf{r}=5\mathbf{i}+6\mathbf{j}+3\mathbf{k}+\mu(\mathbf{i}-2\mathbf{j}+2\mathbf{k})\) | A1 | Any correct equation including lhs of '\(\mathbf{r}=\)'; allow column vector form |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\overrightarrow{PC} = \begin{pmatrix}4+\lambda\\8-2\lambda\\1+2\lambda\end{pmatrix} - \begin{pmatrix}2\\-2\\1\end{pmatrix} = \begin{pmatrix}2+\lambda\\10-2\lambda\\2\lambda\end{pmatrix}\) | M1 | Attempts general vector from \(P\) to \(l\) forming \(\overrightarrow{PC}\) where \(C\) is general point on \(l\) |
| Uses \(\overrightarrow{PC}\cdot(\mathbf{i}-2\mathbf{j}+2\mathbf{k})=0\); e.g. \(1(2+\lambda)-2(10-2\lambda)+2\times2\lambda=0 \Rightarrow \lambda = \ldots\) | dM1 | Dependent on previous M; scalar product with gradient must be attempted |
| \(\lambda=2 \Rightarrow \mathbf{c}=(4+2)\mathbf{i}+(8-4)\mathbf{j}+(1+4)\mathbf{k}\) | ddM1 | Uses \(\lambda\) to find \(C\) |
| \((6,\ 4,\ 5)\) | A1 | Correct coordinates or vector |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\overrightarrow{OP'} = \mathbf{p}+2\overrightarrow{PC} = 2\mathbf{i}-2\mathbf{j}+\mathbf{k}+2(4\mathbf{i}+6\mathbf{j}+4\mathbf{k})\) | M1 | Correct strategy; can be implied by two correct coordinates using their \(P\) and \(C\) |
| \((10,\ 10,\ 9)\) | A1 | Correct coordinates or vector |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \( | PP' | = 2 |
| \(4\sqrt{17}\) | A1 | CAO |
# Question 4:
**Part (a):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| Attempts direction vector by subtracting $(5\mathbf{i}+6\mathbf{j}+3\mathbf{k})$ and $(4\mathbf{i}+8\mathbf{j}+\mathbf{k})$ either way around | M1 | If no method shown, look for two correct components of $(\pm\mathbf{i}\pm2\mathbf{j}\pm2\mathbf{k})$ or multiple |
| E.g. $\mathbf{r} = 4\mathbf{i}+8\mathbf{j}+\mathbf{k}+\lambda(\mathbf{i}-2\mathbf{j}+2\mathbf{k})$, $\mathbf{r}=5\mathbf{i}+6\mathbf{j}+3\mathbf{k}+\mu(\mathbf{i}-2\mathbf{j}+2\mathbf{k})$ | A1 | Any correct equation including lhs of '$\mathbf{r}=$'; allow column vector form |
**Part (b):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\overrightarrow{PC} = \begin{pmatrix}4+\lambda\\8-2\lambda\\1+2\lambda\end{pmatrix} - \begin{pmatrix}2\\-2\\1\end{pmatrix} = \begin{pmatrix}2+\lambda\\10-2\lambda\\2\lambda\end{pmatrix}$ | M1 | Attempts general vector from $P$ to $l$ forming $\overrightarrow{PC}$ where $C$ is general point on $l$ |
| Uses $\overrightarrow{PC}\cdot(\mathbf{i}-2\mathbf{j}+2\mathbf{k})=0$; e.g. $1(2+\lambda)-2(10-2\lambda)+2\times2\lambda=0 \Rightarrow \lambda = \ldots$ | dM1 | Dependent on previous M; scalar product with gradient must be attempted |
| $\lambda=2 \Rightarrow \mathbf{c}=(4+2)\mathbf{i}+(8-4)\mathbf{j}+(1+4)\mathbf{k}$ | ddM1 | Uses $\lambda$ to find $C$ |
| $(6,\ 4,\ 5)$ | A1 | Correct coordinates or vector |
**Part (c):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\overrightarrow{OP'} = \mathbf{p}+2\overrightarrow{PC} = 2\mathbf{i}-2\mathbf{j}+\mathbf{k}+2(4\mathbf{i}+6\mathbf{j}+4\mathbf{k})$ | M1 | Correct strategy; can be implied by two correct coordinates using their $P$ and $C$ |
| $(10,\ 10,\ 9)$ | A1 | Correct coordinates or vector |
**Part (d):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $|PP'| = 2|\overrightarrow{PC}| = 2\sqrt{4^2+6^2+4^2} = \ldots$ | M1 | Correct method using their values; must be complete method, not distance squared |
| $4\sqrt{17}$ | A1 | CAO |
---\begin{enumerate}
\item Relative to a fixed origin $O$,
\end{enumerate}
\begin{itemize}
\item the point $A$ has position vector $4 \mathbf { i } + 8 \mathbf { j } + \mathbf { k }$
\item the point $B$ has position vector $5 \mathbf { i } + 6 \mathbf { j } + 3 \mathbf { k }$
\item the point $P$ has position vector $2 \mathbf { i } - 2 \mathbf { j } + \mathbf { k }$
\end{itemize}
The straight line $l$ passes through $A$ and $B$.\\
(a) Find a vector equation for $l$.
The point $C$ lies on $l$ so that $P C$ is perpendicular to $l$.\\
(b) Find the coordinates of $C$.
The point $P ^ { \prime }$ is the reflection of $P$ in the line $l$.\\
(c) Find the coordinates of $P ^ { \prime }$\\
(d) Hence find $\left| \overrightarrow { P P ^ { \prime } } \right|$, giving your answer as a simplified surd.
\hfill \mbox{\textit{Edexcel P4 2023 Q4 [10]}}