| Exam Board | OCR |
|---|---|
| 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 C4 vectors question requiring finding the intersection of two lines (one in parametric form, one through two points) and then calculating a distance with unit conversion. The techniques are standard and the multi-step process is routine for this module, making it slightly easier than average. |
| Spec | 1.10c Magnitude and direction: of vectors4.04a Line equations: 2D and 3D, cartesian and vector forms4.04e Line intersections: parallel, skew, or intersecting |
| Answer | Marks |
|---|---|
| \(\overrightarrow{AB} = (4\mathbf{j} + \mathbf{k}) - (9\mathbf{i} - 8\mathbf{j} + 2\mathbf{k}) = (-9\mathbf{i} + 12\mathbf{j} - \mathbf{k})\) | M1 |
| \(\therefore \mathbf{r} = (9\mathbf{i} - 8\mathbf{j} + 2\mathbf{k}) + s(-9\mathbf{i} + 12\mathbf{j} - \mathbf{k})\) | A1 |
| at \(C\), \(2 - s = -1\), \(s = 3\) | 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 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\overrightarrow{AC} = 3(-9\mathbf{i} + 12\mathbf{j} - \mathbf{k})\), \(AC = 3\sqrt{81+144+1} = 45.10\) | M1 A1 | |
| \(\therefore \text{distance} = 200 \times 45.10 = 9020 \text{ m} = 9.02 \text{ km (3sf)}\) | M1 A1 | (9) |
## (i)
$\overrightarrow{AB} = (4\mathbf{j} + \mathbf{k}) - (9\mathbf{i} - 8\mathbf{j} + 2\mathbf{k}) = (-9\mathbf{i} + 12\mathbf{j} - \mathbf{k})$ | M1 |
$\therefore \mathbf{r} = (9\mathbf{i} - 8\mathbf{j} + 2\mathbf{k}) + s(-9\mathbf{i} + 12\mathbf{j} - \mathbf{k})$ | A1 |
at $C$, $2 - s = -1$, $s = 3$ | 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 |
## (ii)
$\overrightarrow{AC} = 3(-9\mathbf{i} + 12\mathbf{j} - \mathbf{k})$, $AC = 3\sqrt{81+144+1} = 45.10$ | M1 A1 |
$\therefore \text{distance} = 200 \times 45.10 = 9020 \text{ m} = 9.02 \text{ km (3sf)}$ | M1 A1 | **(9)** |
---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
$$\mathbf{r} = (2\mathbf{i} + 16\mathbf{j} - \mathbf{k}) + t(-5\mathbf{i} + 3\mathbf{j}),$$
where $t$ is a scalar parameter.
\begin{enumerate}[label=(\roman*)]
\item Find the position vector of $C$. [5]
\end{enumerate}
Given that 1 unit on each coordinate axis represents 200 metres,
\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{1}
\item find the distance, in kilometres, from the village at $A$ to the junction at $C$. [4]
\end{enumerate}
\hfill \mbox{\textit{OCR C4 Q7 [9]}}