| Exam Board | Edexcel |
|---|---|
| Module | P4 (Pure Mathematics 4) |
| Year | 2024 |
| Session | January |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors: Lines & Planes |
| Type | Point on line satisfying condition |
| Difficulty | Standard +0.3 This is a standard multi-part vectors question covering routine techniques: finding intersection of lines (equating components, solving simultaneous equations), angle between lines (dot product formula), and perpendicular distance from point to line. All parts follow textbook methods with no novel insight required. Part (d) is slightly more involved but still algorithmic. Slightly easier than average due to straightforward application of standard procedures. |
| Spec | 1.10c Magnitude and direction: of vectors1.10h Vectors in kinematics: uniform acceleration in vector form |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Setting up equations from \(\mathbf{i}\), \(\mathbf{j}\), \(\mathbf{k}\) components: \(3+2\lambda=8+4\mu\), \(p-5\lambda=-2+\mu\), \(7+4\lambda=5+2\mu\) | M1 | |
| e.g. \((3)-2\times(1): 1=-11-6\mu \Rightarrow \mu=-2\) \(\left(\text{or } \lambda = -\frac{3}{2}\right)\) | M1 | |
| \(\mu=-2,\ \lambda=-\frac{3}{2} \Rightarrow p = -2-2-\frac{15}{2} = \ldots\) | M1 | |
| \(p = -\frac{23}{2}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Intersect at \((8\mathbf{i}-2\mathbf{j}+5\mathbf{k})+{-2}(4\mathbf{i}+\mathbf{j}+2\mathbf{k}) = \ldots\) | M1 | |
| \(= -4\mathbf{j}+\mathbf{k}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \((2\mathbf{i}-5\mathbf{j}+4\mathbf{k})\cdot(4\mathbf{i}+\mathbf{j}+2\mathbf{k}) = 2\times4 - 5\times1 + 4\times2 = \ldots\) | M1 | |
| \(\cos\theta = \frac{11}{\sqrt{2^2+5^2+4^2}\sqrt{4^2+1^2+2^2}} = \frac{11}{3\sqrt{105}} = 0.3578\ldots\) | M1 | |
| \(\theta = \cos^{-1}\frac{11}{3\sqrt{105}} = 69.0°\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\lambda=2 \Rightarrow \overrightarrow{OA} = \left(7\mathbf{i} - \frac{43}{2}\mathbf{j} + 15\mathbf{k}\right)\) | B1ft | Follow through |
| \(\overrightarrow{AB} = \pm\left((8\mathbf{i}-2\mathbf{j}+5\mathbf{k})+\mu(4\mathbf{i}+\mathbf{j}+2\mathbf{k}) - \left(7\mathbf{i}-\frac{43}{2}\mathbf{j}+15\mathbf{k}\right)\right)\) \(= \pm\left((1+4\mu)\mathbf{i}+\left(\frac{39}{2}+\mu\right)\mathbf{j}+(-10+2\mu)\mathbf{k}\right)\) | M1 | |
| \(\left((1+4\mu)\mathbf{i}+\left(\frac{39}{2}+\mu\right)\mathbf{j}+(-10+2\mu)\mathbf{k}\right)\cdot(4\mathbf{i}+\mathbf{j}+2\mathbf{k})=0 \Rightarrow \mu = -\frac{1}{6}\) | M1 | |
| \(\overrightarrow{OB} = (8\mathbf{i}-2\mathbf{j}+5\mathbf{k}) + \left(-\frac{1}{6}\right)(4\mathbf{i}+\mathbf{j}+2\mathbf{k}) = \ldots\) | dM1 | Dependent on previous M1 |
| \(= \frac{22}{3}\mathbf{i} - \frac{13}{6}\mathbf{j} + \frac{14}{3}\mathbf{k}\) or \(\left(\frac{22}{3}, -\frac{13}{6}, \frac{14}{3}\right)\) | A1 |
## Question 6:
### Part (a):
| Answer | Mark | Guidance |
|--------|------|----------|
| Setting up equations from $\mathbf{i}$, $\mathbf{j}$, $\mathbf{k}$ components: $3+2\lambda=8+4\mu$, $p-5\lambda=-2+\mu$, $7+4\lambda=5+2\mu$ | M1 | |
| e.g. $(3)-2\times(1): 1=-11-6\mu \Rightarrow \mu=-2$ $\left(\text{or } \lambda = -\frac{3}{2}\right)$ | M1 | |
| $\mu=-2,\ \lambda=-\frac{3}{2} \Rightarrow p = -2-2-\frac{15}{2} = \ldots$ | M1 | |
| $p = -\frac{23}{2}$ | A1 | |
### Part (b):
| Answer | Mark | Guidance |
|--------|------|----------|
| Intersect at $(8\mathbf{i}-2\mathbf{j}+5\mathbf{k})+{-2}(4\mathbf{i}+\mathbf{j}+2\mathbf{k}) = \ldots$ | M1 | |
| $= -4\mathbf{j}+\mathbf{k}$ | A1 | |
### Part (c):
| Answer | Mark | Guidance |
|--------|------|----------|
| $(2\mathbf{i}-5\mathbf{j}+4\mathbf{k})\cdot(4\mathbf{i}+\mathbf{j}+2\mathbf{k}) = 2\times4 - 5\times1 + 4\times2 = \ldots$ | M1 | |
| $\cos\theta = \frac{11}{\sqrt{2^2+5^2+4^2}\sqrt{4^2+1^2+2^2}} = \frac{11}{3\sqrt{105}} = 0.3578\ldots$ | M1 | |
| $\theta = \cos^{-1}\frac{11}{3\sqrt{105}} = 69.0°$ | A1 | |
### Part (d):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\lambda=2 \Rightarrow \overrightarrow{OA} = \left(7\mathbf{i} - \frac{43}{2}\mathbf{j} + 15\mathbf{k}\right)$ | B1ft | Follow through |
| $\overrightarrow{AB} = \pm\left((8\mathbf{i}-2\mathbf{j}+5\mathbf{k})+\mu(4\mathbf{i}+\mathbf{j}+2\mathbf{k}) - \left(7\mathbf{i}-\frac{43}{2}\mathbf{j}+15\mathbf{k}\right)\right)$ $= \pm\left((1+4\mu)\mathbf{i}+\left(\frac{39}{2}+\mu\right)\mathbf{j}+(-10+2\mu)\mathbf{k}\right)$ | M1 | |
| $\left((1+4\mu)\mathbf{i}+\left(\frac{39}{2}+\mu\right)\mathbf{j}+(-10+2\mu)\mathbf{k}\right)\cdot(4\mathbf{i}+\mathbf{j}+2\mathbf{k})=0 \Rightarrow \mu = -\frac{1}{6}$ | M1 | |
| $\overrightarrow{OB} = (8\mathbf{i}-2\mathbf{j}+5\mathbf{k}) + \left(-\frac{1}{6}\right)(4\mathbf{i}+\mathbf{j}+2\mathbf{k}) = \ldots$ | dM1 | Dependent on previous M1 |
| $= \frac{22}{3}\mathbf{i} - \frac{13}{6}\mathbf{j} + \frac{14}{3}\mathbf{k}$ or $\left(\frac{22}{3}, -\frac{13}{6}, \frac{14}{3}\right)$ | A1 | |\begin{enumerate}
\item Relative to a fixed origin $O$, the lines $l _ { 1 }$ and $l _ { 2 }$ are given by the equations
\end{enumerate}
$$\begin{aligned}
& l _ { 1 } : \mathbf { r } = ( 3 \mathbf { i } + p \mathbf { j } + 7 \mathbf { k } ) + \lambda ( 2 \mathbf { i } - 5 \mathbf { j } + 4 \mathbf { k } ) \\
& l _ { 2 } : \mathbf { r } = ( 8 \mathbf { i } - 2 \mathbf { j } + 5 \mathbf { k } ) + \mu ( 4 \mathbf { i } + \mathbf { j } + 2 \mathbf { k } )
\end{aligned}$$
where $\lambda$ and $\mu$ are scalar parameters and $p$ is a constant.\\
Given that $l _ { 1 }$ and $l _ { 2 }$ intersect,\\
(a) find the value of $p$,\\
(b) find the position vector of the point of intersection.\\
(c) Find the acute angle between $l _ { 1 }$ and $l _ { 2 }$
Give your answer in degrees to one decimal place.
The point $A$ lies on $l _ { 1 }$ with parameter $\lambda = 2$\\
The point $B$ lies on $l _ { 2 }$ with $\overrightarrow { A B }$ perpendicular to $l _ { 2 }$\\
(d) Find the coordinates of $B$
\hfill \mbox{\textit{Edexcel P4 2024 Q6 [14]}}