| Exam Board | Edexcel |
|---|---|
| Module | P4 (Pure Mathematics 4) |
| Year | 2023 |
| Session | October |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors: Lines & Planes |
| Type | Triangle and parallelogram areas |
| Difficulty | Standard +0.8 This is a multi-part Further Maths vectors question requiring: (a) simple observation of common point, (b) standard dot product for angle, (c) area calculation using cross product with isosceles constraint, (d) solving for position using distance equation. Parts (c) and (d) require careful algebraic manipulation and the isosceles constraint adds problem-solving depth beyond routine exercises, placing it moderately above average difficulty. |
| Spec | 1.10c Magnitude and direction: of vectors1.10d Vector operations: addition and scalar multiplication1.10h Vectors in kinematics: uniform acceleration in vector form4.04a Line equations: 2D and 3D, cartesian and vector forms4.04c Scalar product: calculate and use for angles |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \((2, 3, -7)\) | B1 | Correct coordinates or position vector e.g. \(x=2, y=3, z=-7\) or \(2\mathbf{i}+3\mathbf{j}-7\mathbf{k}\). Condone \((2\mathbf{i}, 3\mathbf{j}, -7\mathbf{k})\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\begin{pmatrix}1\\2\\2\end{pmatrix}\bullet\begin{pmatrix}4\\-1\\8\end{pmatrix} = 1\times4+2\times-1+2\times8=(18)\) | M1 | Any non-zero multiples of these vectors can be used. If method not explicit, implied by 2 correct components |
| \(18=\sqrt{1^2+2^2+2^2}\times\sqrt{4^2+(-1)^2+8^2}\cos\theta\) | dM1 | Full attempt at \(\mathbf{a}\cdot\mathbf{b}=\ |
| \(\cos\theta=\dfrac{2}{3}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\pm\left(\begin{pmatrix}4\\-1\\8\end{pmatrix}-\begin{pmatrix}1\\2\\2\end{pmatrix}\right)=\pm\begin{pmatrix}3\\-3\\6\end{pmatrix}\) | M1 | Attempts \(\pm\left(\alpha\begin{pmatrix}4\\-1\\8\end{pmatrix}-\beta\begin{pmatrix}1\\2\\2\end{pmatrix}\right)\) |
| \(3^2+3^2+6^2=1^2+2^2+2^2+4^2+(-1)^2+8^2-2\sqrt{1^2+2^2+2^2}\times\sqrt{4^2+(-1)^2+8^2}\cos\theta\) | dM1 | Full attempt at cosine rule using appropriate lengths. Depends on first M mark |
| \(\cos\theta=\dfrac{2}{3}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\overrightarrow{PQ}=6\mathbf{i}+12\mathbf{j}+12\mathbf{k}\Rightarrow PQ=\sqrt{6^2+12^2+12^2}=(18)\) or \(PQ=6\times\sqrt{1^2+2^2+2^2}=(18)\) | M1 | Uses \(\lambda=6\) to find length \(PQ\). If method not explicit, implied by 2 correct components |
| Area \(QPR=\dfrac{1}{2}\times18^2\times\sqrt{1-\left(\dfrac{2}{3}\right)^2}=54\sqrt{5}\) | M1 A1 | Full attempt at \(\frac{1}{2}ab\sin C\) where \(a=b=\ |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \((4\mu)^2+\mu^2+(8\mu)^2=6^2+12^2+12^2 \Rightarrow \mu=\dfrac{"18"}{\sqrt{4^2+(-1)^2+8^2}}=(\pm)2\) | M1 | Attempts correct method for at least one value of \(\mu\). Pythagoras used correctly, both sides consistent. Must use length of \(PQ\) not \(OQ\) |
| \(\mathbf{r}=\begin{pmatrix}2\\3\\-7\end{pmatrix}\pm2\begin{pmatrix}4\\-1\\8\end{pmatrix}\) | dM1 | Attempts one correct position. Depends on first M mark |
| \((10, 1, 9)\) and \((-6, 5, -23)\) | A1 | Gives both possible coordinates. Allow as coordinates, vectors, or as \(x=\ldots, y=\ldots, z=\ldots\) |
# Question 6:
## Part (a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $(2, 3, -7)$ | B1 | Correct coordinates or position vector e.g. $x=2, y=3, z=-7$ or $2\mathbf{i}+3\mathbf{j}-7\mathbf{k}$. Condone $(2\mathbf{i}, 3\mathbf{j}, -7\mathbf{k})$ |
## Part (b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\begin{pmatrix}1\\2\\2\end{pmatrix}\bullet\begin{pmatrix}4\\-1\\8\end{pmatrix} = 1\times4+2\times-1+2\times8=(18)$ | M1 | Any non-zero multiples of these vectors can be used. If method not explicit, implied by 2 correct components |
| $18=\sqrt{1^2+2^2+2^2}\times\sqrt{4^2+(-1)^2+8^2}\cos\theta$ | dM1 | Full attempt at $\mathbf{a}\cdot\mathbf{b}=\|\mathbf{a}\|\|\mathbf{b}\|\cos\theta$. Depends on first M mark |
| $\cos\theta=\dfrac{2}{3}$ | A1 | |
## Part (b) Alternative (cosine rule):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\pm\left(\begin{pmatrix}4\\-1\\8\end{pmatrix}-\begin{pmatrix}1\\2\\2\end{pmatrix}\right)=\pm\begin{pmatrix}3\\-3\\6\end{pmatrix}$ | M1 | Attempts $\pm\left(\alpha\begin{pmatrix}4\\-1\\8\end{pmatrix}-\beta\begin{pmatrix}1\\2\\2\end{pmatrix}\right)$ |
| $3^2+3^2+6^2=1^2+2^2+2^2+4^2+(-1)^2+8^2-2\sqrt{1^2+2^2+2^2}\times\sqrt{4^2+(-1)^2+8^2}\cos\theta$ | dM1 | Full attempt at cosine rule using appropriate lengths. Depends on first M mark |
| $\cos\theta=\dfrac{2}{3}$ | A1 | |
## Part (c):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\overrightarrow{PQ}=6\mathbf{i}+12\mathbf{j}+12\mathbf{k}\Rightarrow PQ=\sqrt{6^2+12^2+12^2}=(18)$ or $PQ=6\times\sqrt{1^2+2^2+2^2}=(18)$ | M1 | Uses $\lambda=6$ to find length $PQ$. If method not explicit, implied by 2 correct components |
| Area $QPR=\dfrac{1}{2}\times18^2\times\sqrt{1-\left(\dfrac{2}{3}\right)^2}=54\sqrt{5}$ | M1 A1 | Full attempt at $\frac{1}{2}ab\sin C$ where $a=b=\|\overrightarrow{PQ}\|$ and $C$ is their $\theta$. A1: $54\sqrt{5}$ or exact equivalent e.g. $\dfrac{162\sqrt{5}}{3}$ |
## Part (d):
| Answer | Mark | Guidance |
|--------|------|----------|
| $(4\mu)^2+\mu^2+(8\mu)^2=6^2+12^2+12^2 \Rightarrow \mu=\dfrac{"18"}{\sqrt{4^2+(-1)^2+8^2}}=(\pm)2$ | M1 | Attempts correct method for at least one value of $\mu$. Pythagoras used correctly, both sides consistent. Must use length of $PQ$ not $OQ$ |
| $\mathbf{r}=\begin{pmatrix}2\\3\\-7\end{pmatrix}\pm2\begin{pmatrix}4\\-1\\8\end{pmatrix}$ | dM1 | Attempts one correct position. Depends on first M mark |
| $(10, 1, 9)$ and $(-6, 5, -23)$ | A1 | Gives both possible coordinates. Allow as coordinates, vectors, or as $x=\ldots, y=\ldots, z=\ldots$ |
---\begin{enumerate}
\item The line $l _ { 1 }$ has equation $\mathbf { r } = \left( \begin{array} { r } 2 \\ 3 \\ - 7 \end{array} \right) + \lambda \left( \begin{array} { l } 1 \\ 2 \\ 2 \end{array} \right)$ where $\lambda$ is a scalar parameter.
\end{enumerate}
The line $l _ { 2 }$ has equation $\mathbf { r } = \left( \begin{array} { r } 2 \\ 3 \\ - 7 \end{array} \right) + \mu \left( \begin{array} { r } 4 \\ - 1 \\ 8 \end{array} \right)$ where $\mu$ is a scalar parameter.\\
Given that $l _ { 1 }$ and $l _ { 2 }$ meet at the point $P$\\
(a) state the coordinates of $P$
Given that the angle between lines $l _ { 1 }$ and $l _ { 2 }$ is $\theta$\\
(b) find the value of $\cos \theta$, giving the answer as a fully simplified fraction.
The point $Q$ lies on $l _ { 1 }$ where $\lambda = 6$\\
Given that point $R$ lies on $l _ { 2 }$ such that triangle $Q P R$ is an isosceles triangle with $P Q = P R$\\
(c) find the exact area of triangle $Q P R$\\
(d) find the coordinates of the possible positions of point $R$
\hfill \mbox{\textit{Edexcel P4 2023 Q6 [10]}}