| Exam Board | Edexcel |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Marks | 9 |
| 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 straightforward line intersection problem requiring students to find the vector equation of the first line through A and B, equate it to the second line, solve for parameters, and then calculate distance. While it involves multiple steps and 3D vectors, the techniques are standard C4 material with no novel insight required—slightly easier than average due to the routine nature of the procedure. |
| Spec | 1.10a Vectors in 2D: i,j notation and column vectors1.10f Distance between points: using position vectors4.04a Line equations: 2D and 3D, cartesian and vector forms4.04e Line intersections: parallel, skew, or intersecting |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(\overrightarrow{AB} = (4\mathbf{i} + \mathbf{k}) - (9\mathbf{i} - 8\mathbf{j} + 2\mathbf{k}) = (-9\mathbf{i} + 12\mathbf{j} - \mathbf{k})\) | M1 | |
| \(\therefore \mathbf{r} = (9\mathbf{i} - 8\mathbf{j} + 2\mathbf{k}) + \lambda(-9\mathbf{i} + 12\mathbf{j} - \mathbf{k})\) | A1 | |
| At \(C\): \(2 - \lambda = -1, \lambda = 3\) | M1 A1 | |
| \(\therefore \overrightarrow{OC} = (9\mathbf{i} - 8\mathbf{j} + 2\mathbf{k}) + 3(-9\mathbf{i} + 12\mathbf{j} - \mathbf{k}) = (-18\mathbf{i} + 28\mathbf{j} - \mathbf{k})\) | A1 | |
| (b) \(\overrightarrow{AC} = 3(-9\mathbf{i} + 12\mathbf{j} - \mathbf{k}), \quad AC = 3\sqrt{81 + 144 + 1} = 45.10\) | M1 A1 | |
| \(\therefore \text{distance} = 200 \times 45.10 = 9020 \text{ m} = 9.02 \text{ km}\) (3sf) | M1 A1 | (9 marks) |
**(a)** $\overrightarrow{AB} = (4\mathbf{i} + \mathbf{k}) - (9\mathbf{i} - 8\mathbf{j} + 2\mathbf{k}) = (-9\mathbf{i} + 12\mathbf{j} - \mathbf{k})$ | M1 |
$\therefore \mathbf{r} = (9\mathbf{i} - 8\mathbf{j} + 2\mathbf{k}) + \lambda(-9\mathbf{i} + 12\mathbf{j} - \mathbf{k})$ | A1 |
At $C$: $2 - \lambda = -1, \lambda = 3$ | M1 A1 |
$\therefore \overrightarrow{OC} = (9\mathbf{i} - 8\mathbf{j} + 2\mathbf{k}) + 3(-9\mathbf{i} + 12\mathbf{j} - \mathbf{k}) = (-18\mathbf{i} + 28\mathbf{j} - \mathbf{k})$ | A1 |
**(b)** $\overrightarrow{AC} = 3(-9\mathbf{i} + 12\mathbf{j} - \mathbf{k}), \quad AC = 3\sqrt{81 + 144 + 1} = 45.10$ | M1 A1 |
$\therefore \text{distance} = 200 \times 45.10 = 9020 \text{ m} = 9.02 \text{ km}$ (3sf) | M1 A1 | (9 marks)5. A straight road passes through villages at the points $A$ and $B$ with position vectors ( $9 \mathbf { i } - 8 \mathbf { j } + 2 \mathbf { k }$ ) and ( $4 \mathbf { j } + \mathbf { k }$ ) respectively, relative to a fixed origin.
The road ends at a junction at the point $C$ with another straight road which lies along the line with equation
$$\mathbf { r } = ( 2 \mathbf { i } + 16 \mathbf { j } - \mathbf { k } ) + \mu ( - 5 \mathbf { i } + 3 \mathbf { j } ) ,$$
where $\mu$ is a scalar parameter.
\begin{enumerate}[label=(\alph*)]
\item Find the position vector of $C$.
Given that 1 unit on each coordinate axis represents 200 metres,
\item find the distance, in kilometres, from the village at $A$ to the junction at $C$.\\
5. continued
\end{enumerate}
\hfill \mbox{\textit{Edexcel C4 Q5 [9]}}