| Exam Board | Edexcel |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2008 |
| Session | January |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors 3D & Lines |
| Type | Angle between two lines |
| Difficulty | Standard +0.3 This is a standard C4 vectors question requiring routine techniques: finding a direction vector by subtraction, writing a vector equation, calculating angle between lines using the scalar product formula, and finding an intersection point by equating parametric forms. All steps are textbook exercises with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.10b Vectors in 3D: i,j,k notation1.10d Vector operations: addition and scalar multiplication1.10f Distance between points: using position vectors4.04c Scalar product: calculate and use for angles |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = \begin{pmatrix}3\\4\\1\end{pmatrix} - \begin{pmatrix}2\\6\\-1\end{pmatrix} = \begin{pmatrix}1\\-2\\2\end{pmatrix}\) | M1± | Finding the difference between \(\overrightarrow{OB}\) and \(\overrightarrow{OA}\) |
| Correct answer | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(l_1: \mathbf{r} = \begin{pmatrix}2\\6\\-1\end{pmatrix} + \lambda\begin{pmatrix}1\\-2\\2\end{pmatrix}\) or \(\mathbf{r} = \begin{pmatrix}3\\4\\1\end{pmatrix} + \lambda\begin{pmatrix}1\\-2\\2\end{pmatrix}\) | M1 | An expression of the form (vector) \(\pm \lambda\)(vector); \(\mathbf{r} = \overrightarrow{OA} \pm \lambda(\text{their } \overrightarrow{AB})\) or \(\mathbf{r} = \overrightarrow{OB} \pm \lambda(\text{their } \overrightarrow{AB})\) or using \(\overrightarrow{BA}\) (r is needed) |
| Correct equation | A1\(\sqrt{}\) aef |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\cos\theta = \frac{\overrightarrow{AB} \cdot \mathbf{d}_2}{\left(\vert\overrightarrow{AB}\vert \cdot \vert\mathbf{d}_2\vert\right)} = \frac{\begin{pmatrix}1\\-2\\2\end{pmatrix}\cdot\begin{pmatrix}1\\0\\1\end{pmatrix}}{\sqrt{1^2+(-2)^2+2^2}\cdot\sqrt{1^2+0^2+1^2}}\) | M1\(\sqrt{}\) | Considers dot product between \(\mathbf{d}_2\) and their \(\overrightarrow{AB}\) |
| \(\cos\theta = \frac{1+0+2}{\sqrt{9}\cdot\sqrt{2}}\) | A1\(\sqrt{}\) | Correct followed through expression or equation |
| \(\cos\theta = \frac{3}{3\sqrt{2}} \Rightarrow \theta = 45°\) or \(\frac{\pi}{4}\) or awrt \(0.79\) | A1 cao | \(\theta = 45°\) or \(\frac{\pi}{4}\) or awrt \(0.79\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Setting \(l_1\) and \(l_2\) equal: \(\begin{pmatrix}2\\6\\-1\end{pmatrix}+\lambda\begin{pmatrix}1\\-2\\2\end{pmatrix} = \mu\begin{pmatrix}1\\0\\1\end{pmatrix}\) | — | |
| i: \(2+\lambda = \mu\) (1); j: \(6-2\lambda=0\) (2); k: \(-1+2\lambda=\mu\) (3) | M1\(\sqrt{}\) | Either seeing equation (2) written down correctly with or without any other equation, or seeing equations (1) and (3) written down correctly |
| (2) yields \(\lambda=3\); any two yield \(\lambda=3\), \(\mu=5\) | dM1 | Attempt to solve either equation (2) or simultaneously solve any two of the three equations |
| Either one of \(\lambda\) or \(\mu\) correct | A1 | |
| \(l_1: \mathbf{r} = \begin{pmatrix}2\\6\\-1\end{pmatrix}+3\begin{pmatrix}1\\-2\\2\end{pmatrix} = \begin{pmatrix}5\\0\\5\end{pmatrix}\) or \(\mathbf{r}=5\begin{pmatrix}1\\0\\1\end{pmatrix}=\begin{pmatrix}5\\0\\5\end{pmatrix}\) | A1 cso | \(\begin{pmatrix}5\\0\\5\end{pmatrix}\) or \(5\mathbf{i}+5\mathbf{k}\); fully correct solution & no incorrect values of \(\lambda\) or \(\mu\) seen earlier |
## Question 6(a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = \begin{pmatrix}3\\4\\1\end{pmatrix} - \begin{pmatrix}2\\6\\-1\end{pmatrix} = \begin{pmatrix}1\\-2\\2\end{pmatrix}$ | M1± | Finding the difference between $\overrightarrow{OB}$ and $\overrightarrow{OA}$ |
| Correct answer | A1 | |
---
## Question 6(b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $l_1: \mathbf{r} = \begin{pmatrix}2\\6\\-1\end{pmatrix} + \lambda\begin{pmatrix}1\\-2\\2\end{pmatrix}$ or $\mathbf{r} = \begin{pmatrix}3\\4\\1\end{pmatrix} + \lambda\begin{pmatrix}1\\-2\\2\end{pmatrix}$ | M1 | An expression of the form (vector) $\pm \lambda$(vector); $\mathbf{r} = \overrightarrow{OA} \pm \lambda(\text{their } \overrightarrow{AB})$ or $\mathbf{r} = \overrightarrow{OB} \pm \lambda(\text{their } \overrightarrow{AB})$ or using $\overrightarrow{BA}$ (**r** is needed) |
| Correct equation | A1$\sqrt{}$ aef | |
---
## Question 6(c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\cos\theta = \frac{\overrightarrow{AB} \cdot \mathbf{d}_2}{\left(\vert\overrightarrow{AB}\vert \cdot \vert\mathbf{d}_2\vert\right)} = \frac{\begin{pmatrix}1\\-2\\2\end{pmatrix}\cdot\begin{pmatrix}1\\0\\1\end{pmatrix}}{\sqrt{1^2+(-2)^2+2^2}\cdot\sqrt{1^2+0^2+1^2}}$ | M1$\sqrt{}$ | Considers dot product between $\mathbf{d}_2$ and their $\overrightarrow{AB}$ |
| $\cos\theta = \frac{1+0+2}{\sqrt{9}\cdot\sqrt{2}}$ | A1$\sqrt{}$ | Correct followed through expression or equation |
| $\cos\theta = \frac{3}{3\sqrt{2}} \Rightarrow \theta = 45°$ or $\frac{\pi}{4}$ or awrt $0.79$ | A1 cao | $\theta = 45°$ or $\frac{\pi}{4}$ or awrt $0.79$ |
---
## Question 6(d):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Setting $l_1$ and $l_2$ equal: $\begin{pmatrix}2\\6\\-1\end{pmatrix}+\lambda\begin{pmatrix}1\\-2\\2\end{pmatrix} = \mu\begin{pmatrix}1\\0\\1\end{pmatrix}$ | — | |
| **i**: $2+\lambda = \mu$ (1); **j**: $6-2\lambda=0$ (2); **k**: $-1+2\lambda=\mu$ (3) | M1$\sqrt{}$ | Either seeing equation (2) written down correctly with or without any other equation, **or** seeing equations (1) and (3) written down correctly |
| (2) yields $\lambda=3$; any two yield $\lambda=3$, $\mu=5$ | dM1 | Attempt to solve either equation (2) or simultaneously solve any two of the three equations |
| Either one of $\lambda$ or $\mu$ correct | A1 | |
| $l_1: \mathbf{r} = \begin{pmatrix}2\\6\\-1\end{pmatrix}+3\begin{pmatrix}1\\-2\\2\end{pmatrix} = \begin{pmatrix}5\\0\\5\end{pmatrix}$ or $\mathbf{r}=5\begin{pmatrix}1\\0\\1\end{pmatrix}=\begin{pmatrix}5\\0\\5\end{pmatrix}$ | A1 cso | $\begin{pmatrix}5\\0\\5\end{pmatrix}$ or $5\mathbf{i}+5\mathbf{k}$; fully correct solution & no incorrect values of $\lambda$ or $\mu$ seen earlier |6. The points $A$ and $B$ have position vectors $2 \mathbf { i } + 6 \mathbf { j } - \mathbf { k }$ and $3 \mathbf { i } + 4 \mathbf { j } + \mathbf { k }$ respectively.
The line $l _ { 1 }$ passes through the points $A$ and $B$.
\begin{enumerate}[label=(\alph*)]
\item Find the vector $\overrightarrow { A B }$.
\item Find a vector equation for the line $l _ { 1 }$.
A second line $l _ { 2 }$ passes through the origin and is parallel to the vector $\mathbf { i } + \mathbf { k }$. The line $l _ { 1 }$ meets the line $l _ { 2 }$ at the point $C$.
\item Find the acute angle between $l _ { 1 }$ and $l _ { 2 }$.
\item Find the position vector of the point $C$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C4 2008 Q6 [11]}}