| Exam Board | Edexcel |
|---|---|
| Module | P4 (Pure Mathematics 4) |
| Year | 2018 |
| Session | Specimen |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors: Cross Product & Distances |
| Type | Find parameter value for geometric condition |
| Difficulty | Standard +0.3 This is a structured multi-part vectors question with straightforward applications of standard techniques: substituting into a line equation, writing parallel line equations, finding distances using magnitude, using dot product for angles, and applying geometric relationships. While it has many parts, each step is routine for Further Maths students with no novel problem-solving required. Slightly easier than average due to its guided nature. |
| Spec | 1.10c Magnitude and direction: of vectors1.10d Vector operations: addition and scalar multiplication4.04a Line equations: 2D and 3D, cartesian and vector forms4.04c Scalar product: calculate and use for angles |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(A(3, 5, 0)\) | B1 | Allow \(A(3,5,0)\) or \(3\mathbf{i}+5\mathbf{j}\) or \(3\mathbf{i}+5\mathbf{j}+0\mathbf{k}\) or \(\begin{pmatrix}3\\5\\0\end{pmatrix}\) |
| Total: (1) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\{l_2:\} \mathbf{r} = \begin{pmatrix}1\\5\\2\end{pmatrix} + \lambda\begin{pmatrix}-5\\4\\3\end{pmatrix}\) with \(\mathbf{a}+\lambda\mathbf{d}\), \(\mathbf{a}\neq 0\), \(\mathbf{d}\neq 0\); with either \(\mathbf{a}=\mathbf{i}+5\mathbf{j}+2\mathbf{k}\) or \(\mathbf{d}=-5\mathbf{i}+4\mathbf{j}+3\mathbf{k}\), or a multiple of \(-5\mathbf{i}+4\mathbf{j}+3\mathbf{k}\) | M1 | |
| Correct vector equation using \(\mathbf{r}=\) or \(l=\) or \(l_2=\) | A1 | Allow parameters \(\mu\) or \(t\) instead of \(\lambda\) |
| Total: (2) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\overrightarrow{AP} = \overrightarrow{OP} - \overrightarrow{OA} = \begin{pmatrix}1\\5\\2\end{pmatrix} - \begin{pmatrix}3\\5\\0\end{pmatrix} = \begin{pmatrix}-2\\0\\2\end{pmatrix}\) | ||
| \(AP = \sqrt{(-2)^2+(0)^2+(2)^2} = \sqrt{8} = 2\sqrt{2}\) | M1 | Full method for finding \(AP\). Allow M1A1 for \(\begin{pmatrix}2\\0\\2\end{pmatrix}\) leading to \(AP=2\sqrt{2}\) |
| \(2\sqrt{2}\) | A1 | |
| Total: (2) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\overrightarrow{AP}=\begin{pmatrix}-2\\0\\2\end{pmatrix}\) and \(\mathbf{d}_2=\begin{pmatrix}-5\\4\\3\end{pmatrix}\) | M1 | Realisation that dot product is required between \(\overrightarrow{AP}\) or \(\overrightarrow{PA}\) and \(\pm K\mathbf{d}_2\) or \(\pm K\mathbf{d}_1\) |
| \(\{\cos\theta=\} \dfrac{\overrightarrow{AP}\cdot\mathbf{d}_2}{ | \overrightarrow{AP} | |
| \(\{\cos\theta\} = \dfrac{\pm(10+0+6)}{\sqrt{8}\cdot\sqrt{50}} = \dfrac{4}{5}\) | A1 cso | \(\cos\theta = \dfrac{4}{5}\) or exact equivalent |
| Total: (3) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\{\text{Area } APE =\} \dfrac{1}{2}(\text{their } 2\sqrt{2})^2\sin\theta\) | M1 | For \(\frac{1}{2}(2\sqrt{2})^2\sin(36.869...°)\) or \(\frac{1}{2}(2\sqrt{2})^2\sin(180°-36.869...°)\); candidates must use \(\theta\) from part (d) or apply correct method using \(\sin\theta=\frac{3}{5}\) from \(\cos\theta=\frac{4}{5}\) |
| \(= 2.4\) | A1 | awrt 2.40 |
| Total: (2) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\overrightarrow{PE} = (-5\lambda)\mathbf{i}+(4\lambda)\mathbf{j}+(3\lambda)\mathbf{k}\) and \(PE = \text{their } 2\sqrt{2}\) from part (c) | ||
| \(\{PE^2=\}\ (-5\lambda)^2+(4\lambda)^2+(3\lambda)^2 = (\text{their } 2\sqrt{2})^2\) | M1 | This mark can be implied. Allow special case 1st M1 for \(\lambda=2.5\) from comparing lengths |
| \(\Rightarrow 50\lambda^2=8 \Rightarrow \lambda^2=\dfrac{4}{25} \Rightarrow \lambda = \pm\dfrac{2}{5}\) | A1 | Either \(\lambda=\dfrac{2}{5}\) or \(\lambda=-\dfrac{2}{5}\) |
| \(l_2: \mathbf{r} = \begin{pmatrix}1\\5\\2\end{pmatrix} \pm \dfrac{2}{5}\begin{pmatrix}-5\\4\\3\end{pmatrix}\) | dM1 | Substitutes at least one value of \(\lambda\) into \(l_2\); dependent on previous M mark. Can be implied by at least 2 (out of 6) correct \(x,y,z\) ordinates |
| \(\{\overrightarrow{OE}\} = \begin{pmatrix}3\\\frac{17}{5}\\\frac{4}{5}\end{pmatrix}\) or \(\begin{pmatrix}3\\3.4\\0.8\end{pmatrix}\), \(\{\overrightarrow{OE}\} = \begin{pmatrix}-1\\\frac{33}{5}\\\frac{16}{5}\end{pmatrix}\) or \(\begin{pmatrix}-1\\6.6\\3.2\end{pmatrix}\) | A1 | At least one set of coordinates correct |
| Both sets of coordinates correct | A1 | |
| Total: (5) | ||
| Question Total: (15 marks) |
# Question 9:
## Part (a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $A(3, 5, 0)$ | B1 | Allow $A(3,5,0)$ or $3\mathbf{i}+5\mathbf{j}$ or $3\mathbf{i}+5\mathbf{j}+0\mathbf{k}$ or $\begin{pmatrix}3\\5\\0\end{pmatrix}$ |
| **Total: (1)** | | |
## Part (b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\{l_2:\} \mathbf{r} = \begin{pmatrix}1\\5\\2\end{pmatrix} + \lambda\begin{pmatrix}-5\\4\\3\end{pmatrix}$ with $\mathbf{a}+\lambda\mathbf{d}$, $\mathbf{a}\neq 0$, $\mathbf{d}\neq 0$; with either $\mathbf{a}=\mathbf{i}+5\mathbf{j}+2\mathbf{k}$ or $\mathbf{d}=-5\mathbf{i}+4\mathbf{j}+3\mathbf{k}$, or a multiple of $-5\mathbf{i}+4\mathbf{j}+3\mathbf{k}$ | M1 | |
| Correct vector equation using $\mathbf{r}=$ or $l=$ or $l_2=$ | A1 | Allow parameters $\mu$ or $t$ instead of $\lambda$ |
| **Total: (2)** | | |
## Part (c):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\overrightarrow{AP} = \overrightarrow{OP} - \overrightarrow{OA} = \begin{pmatrix}1\\5\\2\end{pmatrix} - \begin{pmatrix}3\\5\\0\end{pmatrix} = \begin{pmatrix}-2\\0\\2\end{pmatrix}$ | | |
| $AP = \sqrt{(-2)^2+(0)^2+(2)^2} = \sqrt{8} = 2\sqrt{2}$ | M1 | Full method for finding $AP$. Allow M1A1 for $\begin{pmatrix}2\\0\\2\end{pmatrix}$ leading to $AP=2\sqrt{2}$ |
| $2\sqrt{2}$ | A1 | |
| **Total: (2)** | | |
## Part (d):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\overrightarrow{AP}=\begin{pmatrix}-2\\0\\2\end{pmatrix}$ and $\mathbf{d}_2=\begin{pmatrix}-5\\4\\3\end{pmatrix}$ | M1 | Realisation that dot product is required between $\overrightarrow{AP}$ or $\overrightarrow{PA}$ and $\pm K\mathbf{d}_2$ or $\pm K\mathbf{d}_1$ |
| $\{\cos\theta=\} \dfrac{\overrightarrow{AP}\cdot\mathbf{d}_2}{|\overrightarrow{AP}||\mathbf{d}_2|} = \dfrac{\pm\begin{pmatrix}-2\\0\\2\end{pmatrix}\cdot\begin{pmatrix}-5\\4\\3\end{pmatrix}}{\sqrt{(-2)^2+(0)^2+(2)^2}\cdot\sqrt{(-5)^2+(4)^2+(3)^2}}$ | dM1 | Full method for finding $\cos\theta$, dependent on previous M |
| $\{\cos\theta\} = \dfrac{\pm(10+0+6)}{\sqrt{8}\cdot\sqrt{50}} = \dfrac{4}{5}$ | A1 cso | $\cos\theta = \dfrac{4}{5}$ or exact equivalent |
| **Total: (3)** | | |
## Part (e):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\{\text{Area } APE =\} \dfrac{1}{2}(\text{their } 2\sqrt{2})^2\sin\theta$ | M1 | For $\frac{1}{2}(2\sqrt{2})^2\sin(36.869...°)$ or $\frac{1}{2}(2\sqrt{2})^2\sin(180°-36.869...°)$; candidates must use $\theta$ from part (d) or apply correct method using $\sin\theta=\frac{3}{5}$ from $\cos\theta=\frac{4}{5}$ |
| $= 2.4$ | A1 | awrt 2.40 |
| **Total: (2)** | | |
## Part (f):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\overrightarrow{PE} = (-5\lambda)\mathbf{i}+(4\lambda)\mathbf{j}+(3\lambda)\mathbf{k}$ and $PE = \text{their } 2\sqrt{2}$ from part (c) | | |
| $\{PE^2=\}\ (-5\lambda)^2+(4\lambda)^2+(3\lambda)^2 = (\text{their } 2\sqrt{2})^2$ | M1 | This mark can be implied. Allow special case 1st M1 for $\lambda=2.5$ from comparing lengths |
| $\Rightarrow 50\lambda^2=8 \Rightarrow \lambda^2=\dfrac{4}{25} \Rightarrow \lambda = \pm\dfrac{2}{5}$ | A1 | Either $\lambda=\dfrac{2}{5}$ or $\lambda=-\dfrac{2}{5}$ |
| $l_2: \mathbf{r} = \begin{pmatrix}1\\5\\2\end{pmatrix} \pm \dfrac{2}{5}\begin{pmatrix}-5\\4\\3\end{pmatrix}$ | dM1 | Substitutes at least one value of $\lambda$ into $l_2$; dependent on previous M mark. Can be implied by at least 2 (out of 6) correct $x,y,z$ ordinates |
| $\{\overrightarrow{OE}\} = \begin{pmatrix}3\\\frac{17}{5}\\\frac{4}{5}\end{pmatrix}$ or $\begin{pmatrix}3\\3.4\\0.8\end{pmatrix}$, $\{\overrightarrow{OE}\} = \begin{pmatrix}-1\\\frac{33}{5}\\\frac{16}{5}\end{pmatrix}$ or $\begin{pmatrix}-1\\6.6\\3.2\end{pmatrix}$ | A1 | At least one set of coordinates correct |
| Both sets of coordinates correct | A1 | |
| **Total: (5)** | | |
| **Question Total: (15 marks)** | | |\begin{enumerate}
\item With respect to a fixed origin $O$, the line $l _ { 1 }$ is given by the equation
\end{enumerate}
$$\mathbf { r } = \left( \begin{array} { r }
8 \\
1 \\
- 3
\end{array} \right) + \mu \left( \begin{array} { r }
- 5 \\
4 \\
3
\end{array} \right)$$
where $\mu$ is a scalar parameter.\\
The point $A$ lies on $l _ { 1 }$ where $\mu = 1$\\
(a) Find the coordinates of $A$.
The point $P$ has position vector $\left( \begin{array} { l } 1 \\ 5 \\ 2 \end{array} \right)$\\
The line $l _ { 2 }$ passes through the point $P$ and is parallel to the line $l _ { 1 }$\\
(b) Write down a vector equation for the line $l _ { 2 }$\\
(c) Find the exact value of the distance $A P$.
Give your answer in the form $k \sqrt { 2 }$, where $k$ is a constant to be found.
The acute angle between $A P$ and $l _ { 2 }$ is $\theta$\\
(d) Find the value of $\cos \theta$
A point $E$ lies on the line $l _ { 2 }$\\
Given that $A P = P E$,\\
(e) find the area of triangle $A P E$,\\
(f) find the coordinates of the two possible positions of $E$.\\
\hfill \mbox{\textit{Edexcel P4 2018 Q9 [15]}}