| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2014 |
| Session | November |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors 3D & Lines |
| Type | Area of triangle from given side vectors or coordinates |
| Difficulty | Moderate -0.3 This is a straightforward two-part question requiring standard vector techniques: (i) showing perpendicularity via dot product of vectors OA and AB equals zero, and (ii) using the formula ½|OA||AB| for a right-angled triangle. Both parts involve routine calculations with no conceptual challenges, making it slightly easier than average. |
| Spec | 1.10b Vectors in 3D: i,j,k notation4.04c Scalar product: calculate and use for angles4.04g Vector product: a x b perpendicular vector |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\vec{AB}\) or \(\vec{BA} = \pm[(7\mathbf{i} - 3\mathbf{j} + \mathbf{k}) - (3\mathbf{i} + 2\mathbf{j} - \mathbf{k})]\) \(= \pm(4\mathbf{i} - 5\mathbf{j} + 2\mathbf{k})\) | M1A1 | May be seen in part (ii) |
| \((\vec{AO}.\vec{AB}) = \pm(12 - 10 - 2)\) [allow as column if total given] | DM1 | OR \(AB^2 = 45,\ AO^2 = 14,\ OB^2 = 59\) |
| \(= 0\) hence \(OAB = 90°\) | A1 | Hence \(AB^2 + AO^2 = OB^2\); Hence \(OAB = 90°\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\ | OA\ | = \sqrt{9+4+1} = \sqrt{14}\), \(\ |
| Area \(\Delta = \frac{1}{2}\sqrt{14}\left(\sqrt{45}\right) = 12.5\) | M1A1 |
## Question 6:
### Part (i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\vec{AB}$ or $\vec{BA} = \pm[(7\mathbf{i} - 3\mathbf{j} + \mathbf{k}) - (3\mathbf{i} + 2\mathbf{j} - \mathbf{k})]$ $= \pm(4\mathbf{i} - 5\mathbf{j} + 2\mathbf{k})$ | **M1A1** | May be seen in part **(ii)** |
| $(\vec{AO}.\vec{AB}) = \pm(12 - 10 - 2)$ [allow as column if total given] | **DM1** | **OR** $AB^2 = 45,\ AO^2 = 14,\ OB^2 = 59$ |
| $= 0$ hence $OAB = 90°$ | **A1** | Hence $AB^2 + AO^2 = OB^2$; Hence $OAB = 90°$ |
### Part (ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\|OA\| = \sqrt{9+4+1} = \sqrt{14}$, $\|AB\| = \sqrt{16+25+4} = \sqrt{45}$ | **B1** | At least one magnitude correct in **(i)** or **(ii)**; Accept $12.6,\ \frac{3\sqrt{70}}{2}$ oe |
| Area $\Delta = \frac{1}{2}\sqrt{14}\left(\sqrt{45}\right) = 12.5$ | **M1A1** | |
---6 Relative to an origin $O$, the position vector of $A$ is $3 \mathbf { i } + 2 \mathbf { j } - \mathbf { k }$ and the position vector of $B$ is $7 \mathbf { i } - 3 \mathbf { j } + \mathbf { k }$.\\
(i) Show that angle $O A B$ is a right angle.\\
(ii) Find the area of triangle $O A B$.
\hfill \mbox{\textit{CAIE P1 2014 Q6 [7]}}