| Exam Board | Edexcel |
|---|---|
| Module | P4 (Pure Mathematics 4) |
| Year | 2021 |
| Session | October |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors: Cross Product & Distances |
| Type | Find reflection of point in line or plane |
| Difficulty | Standard +0.8 This is a multi-part Further Maths question requiring systematic application of vector methods: finding the nearest point on a line using dot product perpendicularity, calculating distance, and using geometric reasoning to find a reflection. While methodical rather than requiring deep insight, it demands careful algebraic manipulation across multiple steps and understanding of 3D vector geometry, placing it moderately above average difficulty. |
| Spec | 4.04c Scalar product: calculate and use for angles4.04h Shortest distances: between parallel lines and between skew lines |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| General point on \(l\): \(\begin{pmatrix}4-4\lambda\\2-3\lambda\\-3+5\lambda\end{pmatrix}\); \(\overrightarrow{AX} = \begin{pmatrix}-5-4\lambda\\5-3\lambda\\-5+5\lambda\end{pmatrix}\) | M1 | States or uses a general point on \(l\) and attempts \(\overrightarrow{AX}\). Condone slips either way around |
| \(\overrightarrow{AX}\cdot\begin{pmatrix}-4\\-3\\5\end{pmatrix} = 0 \Rightarrow 20+16\lambda-15+9\lambda-25+25\lambda = 0 \Rightarrow \lambda = \frac{2}{5}\) | dM1 A1 | dM1: Uses a correct method to find \(\lambda\). Via scalar product \(\overrightarrow{AX}\cdot\begin{pmatrix}-4\\-3\\5\end{pmatrix}=0\), or via differentiation of minimum distance \((-5-4\lambda)^2+(5-3\lambda)^2+(-5+5\lambda)^2\) |
| (i) Substitutes \(\lambda\) into \(\begin{pmatrix}4-4\lambda\\2-3\lambda\\-3+5\lambda\end{pmatrix} \Rightarrow X = \left(\frac{12}{5}, \frac{4}{5}, -1\right)\) | dM1 A1 | dM1: Uses their \(\lambda\) to find coordinates of \(X\). Condone use of position vector |
| (ii) \(\overrightarrow{AX} = -\frac{33}{5}\mathbf{i} + \frac{19}{5}\mathbf{j} - 3\mathbf{k}\); Shortest distance \(= \sqrt{\left(-\frac{33}{5}\right)^2+\left(\frac{19}{5}\right)^2+(-3)^2} = \sqrt{67}\) | M1 A1 | M1: Uses correct method to find distance \(AX\) using \(A\) and their \(X\). Award if correct method seen for two of three coordinates. A1: \(\sqrt{67}\) (note \(d=\sqrt{67}\) so \(d^2=67\) is common and fine) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Uses \(\overrightarrow{AX} = \overrightarrow{XB}\) or similar correct method; point \(B\) has position vector \(-\frac{21}{5}\mathbf{i} + \frac{23}{5}\mathbf{j} - 4\mathbf{k}\) | M1 A1 | M1: Uses a correct method to find \(B\). Method can be implied by two correct coordinates. A1: Correct position vector for \(B\). Condone use of coordinates |
## Question 7:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| General point on $l$: $\begin{pmatrix}4-4\lambda\\2-3\lambda\\-3+5\lambda\end{pmatrix}$; $\overrightarrow{AX} = \begin{pmatrix}-5-4\lambda\\5-3\lambda\\-5+5\lambda\end{pmatrix}$ | M1 | States or uses a general point on $l$ and attempts $\overrightarrow{AX}$. Condone slips either way around |
| $\overrightarrow{AX}\cdot\begin{pmatrix}-4\\-3\\5\end{pmatrix} = 0 \Rightarrow 20+16\lambda-15+9\lambda-25+25\lambda = 0 \Rightarrow \lambda = \frac{2}{5}$ | dM1 A1 | dM1: Uses a correct method to find $\lambda$. Via scalar product $\overrightarrow{AX}\cdot\begin{pmatrix}-4\\-3\\5\end{pmatrix}=0$, or via differentiation of minimum distance $(-5-4\lambda)^2+(5-3\lambda)^2+(-5+5\lambda)^2$ |
| (i) Substitutes $\lambda$ into $\begin{pmatrix}4-4\lambda\\2-3\lambda\\-3+5\lambda\end{pmatrix} \Rightarrow X = \left(\frac{12}{5}, \frac{4}{5}, -1\right)$ | dM1 A1 | dM1: Uses their $\lambda$ to find coordinates of $X$. Condone use of position vector |
| (ii) $\overrightarrow{AX} = -\frac{33}{5}\mathbf{i} + \frac{19}{5}\mathbf{j} - 3\mathbf{k}$; Shortest distance $= \sqrt{\left(-\frac{33}{5}\right)^2+\left(\frac{19}{5}\right)^2+(-3)^2} = \sqrt{67}$ | M1 A1 | M1: Uses correct method to find distance $AX$ using $A$ and their $X$. Award if correct method seen for two of three coordinates. A1: $\sqrt{67}$ (note $d=\sqrt{67}$ so $d^2=67$ is common and fine) |
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Uses $\overrightarrow{AX} = \overrightarrow{XB}$ or similar correct method; point $B$ has position vector $-\frac{21}{5}\mathbf{i} + \frac{23}{5}\mathbf{j} - 4\mathbf{k}$ | M1 A1 | M1: Uses a correct method to find $B$. Method can be implied by two correct coordinates. A1: Correct position vector for $B$. Condone use of coordinates |7. With respect to a fixed origin $O$,
\begin{itemize}
\item the line $l$ has equation $\mathbf { r } = \left( \begin{array} { r } 4 \\ 2 \\ - 3 \end{array} \right) + \lambda \left( \begin{array} { r } - 4 \\ - 3 \\ 5 \end{array} \right)$ where $\lambda$ is a scalar constant
\item the point $A$ has position vector $9 \mathbf { i } - 3 \mathbf { j } + 2 \mathbf { k }$
\end{itemize}
Given that $X$ is the point on $l$ nearest to $A$,
\begin{enumerate}[label=(\alph*)]
\item find
\begin{enumerate}[label=(\roman*)]
\item the coordinates of $X$
\item the shortest distance from $A$ to $l$.
Give your answer in the form $\sqrt { d }$, where $d$ is an integer.
The point $B$ is the image of $A$ after reflection in $l$.
\end{enumerate}\item Find the position vector of $B$.
Solutions relying on calculator technology are not acceptable.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{08756c4b-6619-42da-ac8a-2bf065c01de8-26_668_661_408_644}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{center}
\end{figure}
\end{enumerate}
\hfill \mbox{\textit{Edexcel P4 2021 Q7 [9]}}