| Exam Board | Edexcel |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2005 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors 3D & Lines |
| Type | Show lines intersect and find intersection point |
| Difficulty | Standard +0.3 This is a standard multi-part vectors question covering routine techniques: finding intersection points by equating parameters, calculating angles between lines using dot product, and parallelogram properties. All parts follow textbook methods with no novel insight required. The calculations are straightforward, making it slightly easier than average for C4. |
| Spec | 1.10d Vector operations: addition and scalar multiplication1.10f Distance between points: using position 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 |
|---|---|
| (a) k component: \(2 + 4\lambda = -2 \Rightarrow \lambda = -1\) | M1 A1 |
| Note: \(\mu = 2\) | |
| Substituting their \(\lambda\) (or \(\mu\)) into equation of line and obtaining \(B\) | M1 |
| \(B: (2, 2, -2)\) | Accept vector forms A1 |
| Answer | Marks |
|---|---|
| (b) \(\begin{vmatrix} 1 \\ -1 \\ 4 \end{vmatrix} = \sqrt{18}; \quad \begin{vmatrix} 1 \\ -1 \\ 0 \end{vmatrix} = \sqrt{2}\) | both B1 |
| \(\begin{vmatrix} 1 \\ -1 \\ 4 \end{vmatrix} \cdot \begin{vmatrix} 1 \\ -1 \\ 0 \end{vmatrix} = 1 + 1 + 0 (= 2)\) | B1 |
| \(\cos\theta = \frac{2}{\sqrt{18}\sqrt{2}} = \frac{1}{3}\) | cao M1 A1 |
| Answer | Marks | Guidance |
|---|---|---|
| (c) \(\overline{AB} = -\mathbf{i} + \mathbf{j} - 4\mathbf{k} \Rightarrow | \overline{AB} | ^2 = 18\) or \( |
| \(\overline{BC} = 3\mathbf{i} - 3\mathbf{j} \Rightarrow | \overline{BC} | ^2 = 18\) or \( |
| Hence \( | \overline{AB} | = |
| Answer | Marks |
|---|---|
| (d) \(\overline{OD} = 6\mathbf{i} - 2\mathbf{j} + 2\mathbf{k}\) | Allow first B1 for any two correct; Accept column form or coordinates B1 B1 |
**(a)** **k component:** $2 + 4\lambda = -2 \Rightarrow \lambda = -1$ | M1 A1 |
Note: $\mu = 2$ | |
Substituting their $\lambda$ (or $\mu$) into equation of line and obtaining $B$ | M1 |
$B: (2, 2, -2)$ | Accept vector forms A1 |
**Total for (a): [4]**
**(b)** $\begin{vmatrix} 1 \\ -1 \\ 4 \end{vmatrix} = \sqrt{18}; \quad \begin{vmatrix} 1 \\ -1 \\ 0 \end{vmatrix} = \sqrt{2}$ | both B1 |
$\begin{vmatrix} 1 \\ -1 \\ 4 \end{vmatrix} \cdot \begin{vmatrix} 1 \\ -1 \\ 0 \end{vmatrix} = 1 + 1 + 0 (= 2)$ | B1 |
$\cos\theta = \frac{2}{\sqrt{18}\sqrt{2}} = \frac{1}{3}$ | cao M1 A1 |
**Total for (b): [4]**
**(c)** $\overline{AB} = -\mathbf{i} + \mathbf{j} - 4\mathbf{k} \Rightarrow |\overline{AB}|^2 = 18$ or $|\overline{AB}| = \sqrt{18}$ | ignore direction of vector M1 |
$\overline{BC} = 3\mathbf{i} - 3\mathbf{j} \Rightarrow |\overline{BC}|^2 = 18$ or $|\overline{BC}| = \sqrt{18}$ | ignore direction of vector M1 |
Hence $|\overline{AB}| = |\overline{BC}|$ ★ | A1 |
**Total for (c): [3]**
**(d)** $\overline{OD} = 6\mathbf{i} - 2\mathbf{j} + 2\mathbf{k}$ | Allow first B1 for any two correct; Accept column form or coordinates B1 B1 |
**Total for (d): [2]**
**Total: [13]**
---\begin{enumerate}
\item The line $l _ { 1 }$ has vector equation
\end{enumerate}
$$\mathbf { r } = \left( \begin{array} { l }
3 \\
1 \\
2
\end{array} \right) + \lambda \left( \begin{array} { r }
1 \\
- 1 \\
4
\end{array} \right)$$
and the line $l _ { 2 }$ has vector equation
$$\mathbf { r } = \left( \begin{array} { r }
0 \\
4 \\
- 2
\end{array} \right) + \mu \left( \begin{array} { r }
1 \\
- 1 \\
0
\end{array} \right) ,$$
where $\lambda$ and $\mu$ are parameters.\\
The lines $l _ { 1 }$ and $l _ { 2 }$ intersect at the point $B$ and the acute angle between $l _ { 1 }$ and $l _ { 2 }$ is $\theta$.\\
(a) Find the coordinates of $B$.\\
(b) Find the value of $\cos \theta$, giving your answer as a simplified fraction.
The point $A$, which lies on $l _ { 1 }$, has position vector $\mathbf { a } = 3 \mathbf { i } + \mathbf { j } + 2 \mathbf { k }$.\\
The point $C$, which lies on $l _ { 2 }$, has position vector $\mathbf { c } = 5 \mathbf { i } - \mathbf { j } - 2 \mathbf { k }$.\\
The point $D$ is such that $A B C D$ is a parallelogram.\\
(c) Show that $| \overrightarrow { A B } | = | \overrightarrow { B C } |$.\\
(d) Find the position vector of the point $D$.
\hfill \mbox{\textit{Edexcel C4 2005 Q7 [13]}}