| Exam Board | Edexcel |
|---|---|
| Module | CP AS (Core Pure AS) |
| Year | 2024 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors 3D & Lines |
| Type | Line-plane intersection and related angle/perpendicularity |
| Difficulty | Moderate -0.3 This is a standard multi-part vectors question covering routine techniques: finding vector equation from two points, intersection with a plane, angle between lines using scalar product, and shortest distance between skew lines. All parts follow textbook methods with no novel insight required, making it slightly easier than average for A-level Further Maths. |
| Spec | 4.04a Line equations: 2D and 3D, cartesian and vector forms4.04c Scalar product: calculate and use for angles4.04f Line-plane intersection: find point4.04h Shortest distances: between parallel lines and between skew lines |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Direction: \(\pm(3\mathbf{i}+4\mathbf{j}-2\mathbf{k}-(-2\mathbf{i}-8\mathbf{j}-3\mathbf{k}))\) | M1 | Subtracts the given coordinates either way round; 2 correct values implies method |
| e.g. \(\mathbf{r} = 3\mathbf{i}+4\mathbf{j}-2\mathbf{k}+\lambda(5\mathbf{i}+12\mathbf{j}+\mathbf{k})\) or \(\mathbf{r} = -2\mathbf{i}-8\mathbf{j}-3\mathbf{k}+\lambda(5\mathbf{i}+12\mathbf{j}+\mathbf{k})\) | A1 | Correct equation; must use \(\mathbf{r}=\ldots\); allow column vectors |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(z=0 \Rightarrow -2+\lambda=0 \Rightarrow \lambda=2 \Rightarrow C = \ldots\) | M1 | Uses \(z=0\) in \(P_1\) equation to find parameter value |
| \(\lambda=2 \Rightarrow C\) is \((13, 28, 0)\) | A1 | Correct coordinates; condone written as a vector |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \((5\mathbf{i}+12\mathbf{j}+\mathbf{k})\cdot(2\mathbf{i}+4\mathbf{j}-2\mathbf{k}) = 10+48-2\) | M1 | Scalar product between direction from (a) and direction vector of \(P_2\); calculates value |
| \(56 = \sqrt{5^2+12^2+1^2}\sqrt{2^2+4^2+2^2}\cos\theta \Rightarrow \cos\theta = \dfrac{56}{\sqrt{170}\sqrt{24}}\) | M1 | Completes method; reaches value for cosine |
| \(\theta = \text{awrt } 28.8°\) | A1 | Correct acute angle |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\mathbf{P}_1 - \mathbf{P}_2 = 3\mathbf{i}+4\mathbf{j}-2\mathbf{k}+\lambda(5\mathbf{i}+12\mathbf{j}+\mathbf{k})-(\mathbf{i}+3\mathbf{j}-\mathbf{k}+\mu(2\mathbf{i}+4\mathbf{j}-2\mathbf{k}))\) | M1 | Forms general vector connecting two lines; condone sign slips if intention clear; condone same parameter for this mark only |
| \(\begin{pmatrix}2+5\lambda-2\mu\\1+12\lambda-4\mu\\-1+\lambda+2\mu\end{pmatrix}\cdot\begin{pmatrix}5\\12\\1\end{pmatrix}=0 \Rightarrow 170\lambda-56\mu=-21\) AND \(\begin{pmatrix}2+5\lambda-2\mu\\1+12\lambda-4\mu\\-1+\lambda+2\mu\end{pmatrix}\cdot\begin{pmatrix}2\\4\\-2\end{pmatrix}=0 \Rightarrow 56\lambda-24\mu=-10\) | M1 | Scalar product of general vector with both direction vectors \(= 0\); forms 2 simultaneous equations |
| \(\lambda = \ldots\left(\dfrac{7}{118}\right),\ \mu = \ldots\left(\dfrac{131}{236}\right)\) | M1 | Attempts to solve simultaneously; value for each parameter |
| \(\mathbf{P}_1-\mathbf{P}_2 = \ldots\left(\dfrac{70}{59}\mathbf{i}-\dfrac{30}{59}\mathbf{j}+\dfrac{10}{59}\mathbf{k}\right)\) | — | Substitutes parameter values |
| \( | \mathbf{P}_1-\mathbf{P}_2 | = \sqrt{\left(\dfrac{70}{59}\right)^2+\left(\dfrac{30}{59}\right)^2+\left(\dfrac{10}{59}\right)^2}\) |
| \(\text{awrt } 1.3\{0\}\) m or \(\dfrac{10\sqrt{59}}{59}\) m | A1 | Units required |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Find \((\mathbf{a}-\mathbf{c})\) between position vectors | M1 | |
| Find cross product \((\mathbf{b}\times\mathbf{d})\) and dot product \((\mathbf{a}-\mathbf{c})\cdot(\mathbf{b}\times\mathbf{d})\) | M1 | |
| Find \( | \mathbf{b}\times\mathbf{d} | \) |
| Use \(\left\ | \dfrac{(\mathbf{a}-\mathbf{c})\cdot(\mathbf{b}\times\mathbf{d})}{ | \mathbf{b}\times\mathbf{d} |
| \(\text{awrt } 1.3\{0\}\) m | A1 |
# Question 6:
## Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Direction: $\pm(3\mathbf{i}+4\mathbf{j}-2\mathbf{k}-(-2\mathbf{i}-8\mathbf{j}-3\mathbf{k}))$ | M1 | Subtracts the given coordinates either way round; 2 correct values implies method |
| e.g. $\mathbf{r} = 3\mathbf{i}+4\mathbf{j}-2\mathbf{k}+\lambda(5\mathbf{i}+12\mathbf{j}+\mathbf{k})$ or $\mathbf{r} = -2\mathbf{i}-8\mathbf{j}-3\mathbf{k}+\lambda(5\mathbf{i}+12\mathbf{j}+\mathbf{k})$ | A1 | Correct equation; must use $\mathbf{r}=\ldots$; allow column vectors |
## Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $z=0 \Rightarrow -2+\lambda=0 \Rightarrow \lambda=2 \Rightarrow C = \ldots$ | M1 | Uses $z=0$ in $P_1$ equation to find parameter value |
| $\lambda=2 \Rightarrow C$ is $(13, 28, 0)$ | A1 | Correct coordinates; condone written as a vector |
## Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $(5\mathbf{i}+12\mathbf{j}+\mathbf{k})\cdot(2\mathbf{i}+4\mathbf{j}-2\mathbf{k}) = 10+48-2$ | M1 | Scalar product between direction from (a) and direction vector of $P_2$; calculates value |
| $56 = \sqrt{5^2+12^2+1^2}\sqrt{2^2+4^2+2^2}\cos\theta \Rightarrow \cos\theta = \dfrac{56}{\sqrt{170}\sqrt{24}}$ | M1 | Completes method; reaches value for cosine |
| $\theta = \text{awrt } 28.8°$ | A1 | Correct acute angle |
## Part (d):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\mathbf{P}_1 - \mathbf{P}_2 = 3\mathbf{i}+4\mathbf{j}-2\mathbf{k}+\lambda(5\mathbf{i}+12\mathbf{j}+\mathbf{k})-(\mathbf{i}+3\mathbf{j}-\mathbf{k}+\mu(2\mathbf{i}+4\mathbf{j}-2\mathbf{k}))$ | M1 | Forms general vector connecting two lines; condone sign slips if intention clear; condone same parameter for this mark only |
| $\begin{pmatrix}2+5\lambda-2\mu\\1+12\lambda-4\mu\\-1+\lambda+2\mu\end{pmatrix}\cdot\begin{pmatrix}5\\12\\1\end{pmatrix}=0 \Rightarrow 170\lambda-56\mu=-21$ AND $\begin{pmatrix}2+5\lambda-2\mu\\1+12\lambda-4\mu\\-1+\lambda+2\mu\end{pmatrix}\cdot\begin{pmatrix}2\\4\\-2\end{pmatrix}=0 \Rightarrow 56\lambda-24\mu=-10$ | M1 | Scalar product of general vector with both direction vectors $= 0$; forms 2 simultaneous equations |
| $\lambda = \ldots\left(\dfrac{7}{118}\right),\ \mu = \ldots\left(\dfrac{131}{236}\right)$ | M1 | Attempts to solve simultaneously; value for each parameter |
| $\mathbf{P}_1-\mathbf{P}_2 = \ldots\left(\dfrac{70}{59}\mathbf{i}-\dfrac{30}{59}\mathbf{j}+\dfrac{10}{59}\mathbf{k}\right)$ | — | Substitutes parameter values |
| $|\mathbf{P}_1-\mathbf{P}_2| = \sqrt{\left(\dfrac{70}{59}\right)^2+\left(\dfrac{30}{59}\right)^2+\left(\dfrac{10}{59}\right)^2}$ | dM1 | Finds modulus of shortest vector; dependent on previous M mark |
| $\text{awrt } 1.3\{0\}$ m or $\dfrac{10\sqrt{59}}{59}$ m | A1 | Units required |
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**Alternative 2 (outside spec):**
| Answer/Working | Marks | Guidance |
|---|---|---|
| Find $(\mathbf{a}-\mathbf{c})$ between position vectors | M1 | |
| Find cross product $(\mathbf{b}\times\mathbf{d})$ and dot product $(\mathbf{a}-\mathbf{c})\cdot(\mathbf{b}\times\mathbf{d})$ | M1 | |
| Find $|\mathbf{b}\times\mathbf{d}|$ | M1 | |
| Use $\left\|\dfrac{(\mathbf{a}-\mathbf{c})\cdot(\mathbf{b}\times\mathbf{d})}{|\mathbf{b}\times\mathbf{d}|}\right\|$ | M1 | |
| $\text{awrt } 1.3\{0\}$ **m** | A1 | |\begin{enumerate}
\item The drainage system for a sports field consists of underground pipes.
\end{enumerate}
This situation is modelled with respect to a fixed origin $O$.\\
According to the model,
\begin{itemize}
\item the surface of the sports field is a plane with equation $z = 0$
\item the pipes are straight lines
\item one of the pipes, $P _ { 1 }$, passes through the points $A ( 3,4 , - 2 )$ and $B ( - 2 , - 8 , - 3 )$
\item a different pipe, $P _ { 2 }$, has equation $\frac { x - 1 } { 2 } = \frac { y - 3 } { 4 } = \frac { z + 1 } { - 2 }$
\item the units are metres\\
(a) Determine a vector equation of the line representing the pipe $P _ { 1 }$\\
(b) Determine the coordinates of the point at which the pipe $P _ { 1 }$ meets the surface of the playing field, according to the model.
\end{itemize}
Determine, according to the model,\\
(c) the acute angle between pipes $P _ { 1 }$ and $P _ { 2 }$, giving your answer in degrees to 3 significant figures,\\
(d) the shortest distance between pipes $P _ { 1 }$ and $P _ { 2 }$
\hfill \mbox{\textit{Edexcel CP AS 2024 Q6 [12]}}