| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2017 |
| Session | March |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors: Cross Product & Distances |
| Type | Angle between vectors using scalar product |
| Difficulty | Moderate -0.3 This is a straightforward two-part vector question requiring standard techniques: (i) finding an angle using the scalar product formula (compute vectors, dot product, magnitudes, then arccos), and (ii) finding triangle area using ½|a||b|sin(θ) or ½|a×b|. Both parts are direct applications of formulas with routine calculation, making it slightly easier than average but not trivial due to the computational steps involved. |
| Spec | 1.10c Magnitude and direction: of vectors1.10g Problem solving with vectors: in geometry |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\mathbf{BA} = \mathbf{OA} - \mathbf{OB} = -5\mathbf{i} - \mathbf{j} + 2\mathbf{k}\) | B1 | Allow vector reversed. Ignore label BA or AB |
| \(\mathbf{OA}.\mathbf{BA} = -10 - 3 + 10 = -3\) | M1 | soi by \(\pm 3\) |
| \( | \mathbf{OA} | \times |
| \(\cos OAB = \dfrac{+/-3}{\sqrt{38} \times \sqrt{30}}\) | M1 | |
| \(OAB = 95.1°\) (or \(1.66^c\)) | A1 | |
| Total: | 5 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\Delta OAB = \frac{1}{2}\sqrt{38} \times \sqrt{30}\sin 95.1\). Allow \(\frac{1}{2}\sqrt{38} \times \sqrt{74}\sin 39.4\) | M1 | Allow their moduli product from (i) |
| \(= 16.8\) | A1 | cao but NOT from \(\sin 84.9\ (1.482^c)\) |
| Total: | 2 |
## Question 6(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\mathbf{BA} = \mathbf{OA} - \mathbf{OB} = -5\mathbf{i} - \mathbf{j} + 2\mathbf{k}$ | B1 | Allow vector reversed. Ignore label **BA** or **AB** |
| $\mathbf{OA}.\mathbf{BA} = -10 - 3 + 10 = -3$ | M1 | soi by $\pm 3$ |
| $|\mathbf{OA}| \times |\mathbf{BA}| = \sqrt{2^2+3^2+5^2} \times \sqrt{5^2+1^2+2^2}$ | M1 | Prod. of mods for at least 1 correct vector or reverse |
| $\cos OAB = \dfrac{+/-3}{\sqrt{38} \times \sqrt{30}}$ | M1 | |
| $OAB = 95.1°$ (or $1.66^c$) | A1 | |
| **Total:** | **5** | |
## Question 6(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\Delta OAB = \frac{1}{2}\sqrt{38} \times \sqrt{30}\sin 95.1$. Allow $\frac{1}{2}\sqrt{38} \times \sqrt{74}\sin 39.4$ | M1 | Allow their moduli product from **(i)** |
| $= 16.8$ | A1 | cao but NOT from $\sin 84.9\ (1.482^c)$ |
| **Total:** | **2** | |6 Relative to an origin $O$, the position vectors of the points $A$ and $B$ are given by
$$\overrightarrow { O A } = 2 \mathbf { i } + 3 \mathbf { j } + 5 \mathbf { k } \quad \text { and } \quad \overrightarrow { O B } = 7 \mathbf { i } + 4 \mathbf { j } + 3 \mathbf { k }$$
(i) Use a scalar product to find angle $O A B$.\\
(ii) Find the area of triangle $O A B$.\\
\hfill \mbox{\textit{CAIE P1 2017 Q6 [7]}}