| Exam Board | Edexcel |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2007 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors 3D & Lines |
| Type | Line intersection verification |
| Difficulty | Standard +0.3 This is a standard C4 vectors question requiring routine techniques: equating parametric equations to show lines don't meet (part a), then finding a vector and using the scalar product formula for angles (part b). While it involves multiple steps, each step follows textbook procedures with no novel insight required, making it slightly easier than average. |
| Spec | 1.10c Magnitude and direction: of vectors4.04a Line equations: 2D and 3D, cartesian and vector forms4.04c Scalar product: calculate and use for angles4.04e Line intersections: parallel, skew, or intersecting |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\begin{pmatrix}1\\0\\-1\end{pmatrix}+\lambda\begin{pmatrix}1\\1\\0\end{pmatrix} = \begin{pmatrix}1\\3\\6\end{pmatrix}+\mu\begin{pmatrix}2\\1\\-1\end{pmatrix}\) | ||
| i: \(1+\lambda = 1+2\mu\) (1); j: \(\lambda = 3+\mu\) (2); k: \(-1 = 6-\mu\) (3) | M1 | Writes down any two of these equations correctly |
| (1)&(2): \(\lambda=6, \mu=3\); (1)&(3): \(\lambda=14, \mu=7\); (2)&(3): \(\lambda=10, \mu=7\) | A1 | Either one of \(\lambda\) or \(\mu\) correct |
| Both \(\lambda\) and \(\mu\) correct | A1 | |
| Checking eqn (3): \(-1 \neq 3\), or checking eqn (2): \(14\neq 10\) etc. — contradiction shown | B1\(\checkmark\) | Complete method of putting values of \(\lambda\) and \(\mu\) into a third equation to show contradiction |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\lambda=1 \Rightarrow \overrightarrow{OA}=\begin{pmatrix}2\\1\\-1\end{pmatrix}\), \(\mu=2 \Rightarrow \overrightarrow{OB}=\begin{pmatrix}5\\5\\4\end{pmatrix}\) | B1 | Only one of either \(\overrightarrow{OA}\) or \(\overrightarrow{OB}\) correct (can be implied) |
| \(\overrightarrow{AB} = \overrightarrow{OB}-\overrightarrow{OA} = \begin{pmatrix}3\\4\\5\end{pmatrix}\) | M1\(\checkmark\) | Finding the difference between \(\overrightarrow{OB}\) and \(\overrightarrow{OA}\) |
| \(\overrightarrow{AB} = 3\mathbf{i}+4\mathbf{j}+5\mathbf{k}\), \(\mathbf{d}_1 = \mathbf{i}+\mathbf{j}+0\mathbf{k}\), \(\theta\) is angle | M1 | Applying dot product formula between "allowable" vectors |
| \(\cos\theta = \frac{\overrightarrow{AB}\cdot\mathbf{d}_1}{ | \overrightarrow{AB} | |
| A1 | Correct expression | |
| \(\cos\theta = \frac{7}{10}\) | A1 cao | \(\frac{7}{10}\) or \(0.7\) or \(\frac{7}{\sqrt{100}}\), but not \(\frac{7}{\sqrt{50}\cdot\sqrt{2}}\) |
## Question 5(a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\begin{pmatrix}1\\0\\-1\end{pmatrix}+\lambda\begin{pmatrix}1\\1\\0\end{pmatrix} = \begin{pmatrix}1\\3\\6\end{pmatrix}+\mu\begin{pmatrix}2\\1\\-1\end{pmatrix}$ | | |
| **i**: $1+\lambda = 1+2\mu$ (1); **j**: $\lambda = 3+\mu$ (2); **k**: $-1 = 6-\mu$ (3) | M1 | Writes down any two of these equations correctly |
| (1)&(2): $\lambda=6, \mu=3$; (1)&(3): $\lambda=14, \mu=7$; (2)&(3): $\lambda=10, \mu=7$ | A1 | Either one of $\lambda$ or $\mu$ correct |
| Both $\lambda$ and $\mu$ correct | A1 | |
| Checking eqn (3): $-1 \neq 3$, or checking eqn (2): $14\neq 10$ etc. — contradiction shown | B1$\checkmark$ | Complete method of putting values of $\lambda$ and $\mu$ into a third equation to show contradiction |
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## Question 5(b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\lambda=1 \Rightarrow \overrightarrow{OA}=\begin{pmatrix}2\\1\\-1\end{pmatrix}$, $\mu=2 \Rightarrow \overrightarrow{OB}=\begin{pmatrix}5\\5\\4\end{pmatrix}$ | B1 | Only one of either $\overrightarrow{OA}$ or $\overrightarrow{OB}$ correct (can be implied) |
| $\overrightarrow{AB} = \overrightarrow{OB}-\overrightarrow{OA} = \begin{pmatrix}3\\4\\5\end{pmatrix}$ | M1$\checkmark$ | Finding the difference between $\overrightarrow{OB}$ and $\overrightarrow{OA}$ |
| $\overrightarrow{AB} = 3\mathbf{i}+4\mathbf{j}+5\mathbf{k}$, $\mathbf{d}_1 = \mathbf{i}+\mathbf{j}+0\mathbf{k}$, $\theta$ is angle | M1 | Applying dot product formula between "allowable" vectors |
| $\cos\theta = \frac{\overrightarrow{AB}\cdot\mathbf{d}_1}{|\overrightarrow{AB}||\mathbf{d}_1|} = \pm\left(\frac{3+4+0}{\sqrt{50}\cdot\sqrt{2}}\right)$ | M1$\checkmark$ | Applies dot product formula between $\mathbf{d}_1$ and $\pm\overrightarrow{AB}$; correct expression |
| | A1 | Correct expression |
| $\cos\theta = \frac{7}{10}$ | A1 cao | $\frac{7}{10}$ or $0.7$ or $\frac{7}{\sqrt{100}}$, but not $\frac{7}{\sqrt{50}\cdot\sqrt{2}}$ |5.
The line $l _ { 1 }$ has equation $\mathbf { r } = \left( \begin{array} { r } 1 \\ 0 \\ - 1 \end{array} \right) + \lambda \left( \begin{array} { l } 1 \\ 1 \\ 0 \end{array} \right)$.
The line $l _ { 2 }$ has equation $\mathbf { r } = \left( \begin{array} { l } 1 \\ 3 \\ 6 \end{array} \right) + \mu \left( \begin{array} { r } 2 \\ 1 \\ - 1 \end{array} \right)$.
\begin{enumerate}[label=(\alph*)]
\item Show that $l _ { 1 }$ and $l _ { 2 }$ do not meet.
The point $A$ is on $l _ { 1 }$ where $\lambda = 1$, and the point $B$ is on $l _ { 2 }$ where $\mu = 2$.
\item Find the cosine of the acute angle between $A B$ and $l _ { 1 }$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C4 2007 Q5 [10]}}