| Exam Board | OCR |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2006 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors 3D & Lines |
| Type | Area of triangle from given side vectors or coordinates |
| Difficulty | Standard +0.3 This is a straightforward C4 vectors question requiring standard techniques: finding direction vectors AB and AC, computing their dot product and magnitudes for the angle, then using the cross product (or ½|AB||AC|sin θ) for area. All steps are routine applications of formulas with no conceptual challenges beyond careful arithmetic. |
| Spec | 1.10c Magnitude and direction: of vectors1.10d Vector operations: addition and scalar multiplication4.04c Scalar product: calculate and use for angles4.04g Vector product: a x b perpendicular vector |
| Answer | Marks | Guidance |
|---|---|---|
| Working out \(b - a\) or \(a - b\) or \(c - a\) or \(a - c\) | M1 | Irrespective of label |
| Method for finding magnitude of any vector | A1 | If not scored, these 1st 3 marks can be |
| M1 | awarded in part (ii) | |
| M1 | ||
| Method for finding scalar product of any 2 vectors | ||
| Using \(\cos\theta = \frac{a \cdot b}{ | a | |
| [Alternative cosine rule method] \(\overline{BC} = \sqrt{6}\) | B1 | |
| Cosine rule used | M1 | 'Recognisable' form |
| \(45.3°, 0.79(0), \frac{\pi}{3.97}\) | A1 | 6 Do not accept supplement (134.7 etc) |
| Answer | Marks | Guidance |
|---|---|---|
| Use of \(\frac{1}{2} | \overrightarrow{AB} | |
| \(3.54 (3.5355)\) or \(\frac{5\sqrt{2}}{2}\) | A1 | 2 Accept from correct supp (134.7 etc) |
### (i)
Working out $b - a$ or $a - b$ or $c - a$ or $a - c$ | M1 | Irrespective of label
Method for finding magnitude of any vector | A1 | If not scored, these 1st 3 marks can be
| M1 | awarded in part (ii)
| M1 |
Method for finding scalar product of any 2 vectors | |
Using $\cos\theta = \frac{a \cdot b}{|a||b|}$ AEF for any 2 vectors | M1 |
[Alternative cosine rule method] $\overline{BC} = \sqrt{6}$ | B1 |
Cosine rule used | M1 | 'Recognisable' form
$45.3°, 0.79(0), \frac{\pi}{3.97}$ | A1 | 6 Do not accept supplement (134.7 etc)
---
### (ii)
Use of $\frac{1}{2}|\overrightarrow{AB}||\overrightarrow{AC}|\sin\theta$ | M1 | Accept $\frac{1}{2}\overrightarrow{AB} \times \overrightarrow{AC}$
$3.54 (3.5355)$ or $\frac{5\sqrt{2}}{2}$ | A1 | 2 Accept from correct supp (134.7 etc)
---The position vectors of three points $A$, $B$ and $C$ relative to an origin $O$ are given respectively by
$$\overrightarrow{OA} = 7\mathbf{i} + 3\mathbf{j} - 3\mathbf{k},$$
$$\overrightarrow{OB} = 4\mathbf{i} + 2\mathbf{j} - 4\mathbf{k}$$
and
$$\overrightarrow{OC} = 5\mathbf{i} + 4\mathbf{j} - 5\mathbf{k}.$$
\begin{enumerate}[label=(\roman*)]
\item Find the angle between $AB$ and $AC$. [6]
\item Find the area of triangle $ABC$. [2]
\end{enumerate}
\hfill \mbox{\textit{OCR C4 2006 Q4 [8]}}