| Exam Board | Edexcel |
|---|---|
| Module | P4 (Pure Mathematics 4) |
| Year | 2020 |
| Session | October |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors: Lines & Planes |
| Type | Point on line satisfying condition |
| Difficulty | Standard +0.3 This is a straightforward vectors question requiring standard techniques: (a) equating components to find intersection point (routine 3-equation system), and (b) using perpendicularity condition (dot product = 0) to find a point on a line. Both parts are direct applications of A-level methods with no novel insight required, making it slightly easier than average. |
| Spec | 1.10e Position vectors: and displacement4.04c Scalar product: calculate and use for angles4.04e Line intersections: parallel, skew, or intersecting |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Write equations from equal coordinates: \(4+3\lambda=2+2\mu\), \(-3-2\lambda=0-1\mu\), \(2-1\lambda=-9-3\mu\) — any two of these | M1 | Must attempt to set coordinates equal; condone one slip |
| Full method to find either \(\lambda\) or \(\mu\), e.g. \((1)+3(3) \Rightarrow \mu = \ldots\) | M1 | Can be implied if either value correct for their equations |
| Either \(\lambda=-4\) or \(\mu=-5\) | A1 | Correct value(s) following correct equations implies M1 A1 |
| Substitutes \(\lambda\) into \(l_1\): \(\begin{pmatrix}4\\-3\\2\end{pmatrix}-4\begin{pmatrix}3\\-2\\-1\end{pmatrix}\) OR substitutes \(\mu\) into \(l_2\): \(\begin{pmatrix}2\\0\\-9\end{pmatrix}-5\begin{pmatrix}2\\-1\\-3\end{pmatrix}\) | dM1 | Dependent on second M; can be implied by two correct coordinates |
| \(=\begin{pmatrix}-8\\5\\6\end{pmatrix}\) | A1 | Accept \(-8\mathbf{i}+5\mathbf{j}+6\mathbf{k}\) but not coordinate form \((-8,5,6)\); ISW after correct vector |
| (5) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| States general point on \(l_2=\begin{pmatrix}2+2\mu\\0-1\mu\\-9-3\mu\end{pmatrix}\) and attempts \(\overrightarrow{PQ}=\begin{pmatrix}2+2\mu\\0-1\mu\\-9-3\mu\end{pmatrix}-\begin{pmatrix}10\\-7\\0\end{pmatrix}=\begin{pmatrix}2\mu-8\\7-1\mu\\-9-3\mu\end{pmatrix}\) | M1 | Condone slips |
| Uses \(\overrightarrow{PQ}\cdot\begin{pmatrix}2\\-1\\-3\end{pmatrix}=0 \Rightarrow 4\mu-16-7+\mu+27+9\mu=0 \Rightarrow \mu=-\dfrac{2}{7}\) | dM1 A1 | Scalar product of \(\overrightarrow{PQ}\) with direction vector \(=0\); condone slips |
| Substitutes \(\mu=-\dfrac{2}{7}\) into \(\begin{pmatrix}2+2\mu\\0-1\mu\\-9-3\mu\end{pmatrix} \Rightarrow Q=\left(\dfrac{10}{7},\dfrac{2}{7},-\dfrac{57}{7}\right)\) | ddM1 A1 | Dependent on both previous M marks; accept coordinate form here |
| (5) | ||
| (10 marks total) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\overrightarrow{PQ}\) as above | M1 | |
| \(PQ^2+QX^2=PX^2 \Rightarrow 28\mu^2+148\mu+544=504 \Rightarrow \mu=-\dfrac{2}{7}\) | dM1 A1 | |
| Substitutes to find \(Q=\left(\dfrac{10}{7},\dfrac{2}{7},-\dfrac{57}{7}\right)\) | ddM1 A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\overrightarrow{PQ}\) as above | M1 | |
| \(PQ^2=(2\mu-8)^2+(7-\mu)^2+(-9-3\mu)^2\); \(\dfrac{d}{d\mu}PQ^2=0 \Rightarrow 4(2\mu-8)-2(7-\mu)-6(-9-3\mu)=0 \Rightarrow \mu=-\dfrac{2}{7}\) | dM1 A1 | |
| Substitutes to find \(Q=\left(\dfrac{10}{7},\dfrac{2}{7},-\dfrac{57}{7}\right)\) | ddM1 A1 |
## Question 8:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Write equations from equal coordinates: $4+3\lambda=2+2\mu$, $-3-2\lambda=0-1\mu$, $2-1\lambda=-9-3\mu$ — any two of these | M1 | Must attempt to set coordinates equal; condone one slip |
| Full method to find either $\lambda$ **or** $\mu$, e.g. $(1)+3(3) \Rightarrow \mu = \ldots$ | M1 | Can be implied if either value correct for their equations |
| Either $\lambda=-4$ or $\mu=-5$ | A1 | Correct value(s) following correct equations implies M1 A1 |
| Substitutes $\lambda$ into $l_1$: $\begin{pmatrix}4\\-3\\2\end{pmatrix}-4\begin{pmatrix}3\\-2\\-1\end{pmatrix}$ OR substitutes $\mu$ into $l_2$: $\begin{pmatrix}2\\0\\-9\end{pmatrix}-5\begin{pmatrix}2\\-1\\-3\end{pmatrix}$ | dM1 | Dependent on second M; can be implied by two correct coordinates |
| $=\begin{pmatrix}-8\\5\\6\end{pmatrix}$ | A1 | Accept $-8\mathbf{i}+5\mathbf{j}+6\mathbf{k}$ but **not** coordinate form $(-8,5,6)$; ISW after correct vector |
| | **(5)** | |
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| States general point on $l_2=\begin{pmatrix}2+2\mu\\0-1\mu\\-9-3\mu\end{pmatrix}$ and attempts $\overrightarrow{PQ}=\begin{pmatrix}2+2\mu\\0-1\mu\\-9-3\mu\end{pmatrix}-\begin{pmatrix}10\\-7\\0\end{pmatrix}=\begin{pmatrix}2\mu-8\\7-1\mu\\-9-3\mu\end{pmatrix}$ | M1 | Condone slips |
| Uses $\overrightarrow{PQ}\cdot\begin{pmatrix}2\\-1\\-3\end{pmatrix}=0 \Rightarrow 4\mu-16-7+\mu+27+9\mu=0 \Rightarrow \mu=-\dfrac{2}{7}$ | dM1 A1 | Scalar product of $\overrightarrow{PQ}$ with direction vector $=0$; condone slips |
| Substitutes $\mu=-\dfrac{2}{7}$ into $\begin{pmatrix}2+2\mu\\0-1\mu\\-9-3\mu\end{pmatrix} \Rightarrow Q=\left(\dfrac{10}{7},\dfrac{2}{7},-\dfrac{57}{7}\right)$ | ddM1 A1 | Dependent on both previous M marks; accept coordinate form here |
| | **(5)** | |
| | **(10 marks total)** | |
**Alt (b) using Pythagoras:**
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\overrightarrow{PQ}$ as above | M1 | |
| $PQ^2+QX^2=PX^2 \Rightarrow 28\mu^2+148\mu+544=504 \Rightarrow \mu=-\dfrac{2}{7}$ | dM1 A1 | |
| Substitutes to find $Q=\left(\dfrac{10}{7},\dfrac{2}{7},-\dfrac{57}{7}\right)$ | ddM1 A1 | |
**Alt (b) using minimum distance/differentiation:**
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\overrightarrow{PQ}$ as above | M1 | |
| $PQ^2=(2\mu-8)^2+(7-\mu)^2+(-9-3\mu)^2$; $\dfrac{d}{d\mu}PQ^2=0 \Rightarrow 4(2\mu-8)-2(7-\mu)-6(-9-3\mu)=0 \Rightarrow \mu=-\dfrac{2}{7}$ | dM1 A1 | |
| Substitutes to find $Q=\left(\dfrac{10}{7},\dfrac{2}{7},-\dfrac{57}{7}\right)$ | ddM1 A1 | |
---8. Relative to a fixed origin $O$, the lines $l _ { 1 }$ and $l _ { 2 }$ are given by the equations
$$\begin{aligned}
& l _ { 1 } : \quad \mathbf { r } = \left( \begin{array} { r }
4 \\
- 3 \\
2
\end{array} \right) + \lambda \left( \begin{array} { r }
3 \\
- 2 \\
- 1
\end{array} \right) \quad \text { where } \lambda \text { is a scalar parameter } \\
& l _ { 2 } : \quad \mathbf { r } = \left( \begin{array} { r }
2 \\
0 \\
- 9
\end{array} \right) + \mu \left( \begin{array} { r }
2 \\
- 1 \\
- 3
\end{array} \right) \quad \text { where } \mu \text { is a scalar parameter }
\end{aligned}$$
Given that $l _ { 1 }$ and $l _ { 2 }$ meet at the point $X$,
\begin{enumerate}[label=(\alph*)]
\item find the position vector of $X$.
The point $P ( 10 , - 7,0 )$ lies on $l _ { 1 }$\\
The point $Q$ lies on $l _ { 2 }$\\
Given that $\overrightarrow { P Q }$ is perpendicular to $l _ { 2 }$
\item calculate the coordinates of $Q$.\\
\begin{center}
\end{center}
\begin{center}
\end{center}
\end{enumerate}
\hfill \mbox{\textit{Edexcel P4 2020 Q8 [10]}}