| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2010 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors: Lines & Planes |
| Type | Vector operations and magnitudes |
| Difficulty | Standard +0.3 This is a straightforward multi-part vectors question requiring standard techniques: vector addition for the diagonal, finding a unit vector (magnitude then division), angle between vectors using the scalar product formula, and perimeter calculation using magnitudes. All steps are routine applications of basic vector operations with no novel insight required, making it slightly easier than average. |
| Spec | 1.10c Magnitude and direction: of vectors1.10d Vector operations: addition and scalar multiplication1.10f Distance between points: using position vectors |
| Answer | Marks | Guidance |
|---|---|---|
| \(\overrightarrow{OB} = \overrightarrow{OA} + \overrightarrow{OC} = 4\mathbf{i} + 2\mathbf{j} + 4\mathbf{k}\) | B1 | co |
| Unit vector \(= \frac{1}{6}(4\mathbf{i} + 2\mathbf{j} + 4\mathbf{k})\) | M1 A1\(\sqrt{}\) [3] | Divides by the modulus. \(\sqrt{}\) on \(\overrightarrow{OB}\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\overrightarrow{AC} = \overrightarrow{OC} - \overrightarrow{OA} = 2\mathbf{i} - 4\mathbf{j} - 2\mathbf{k}\) | B1 | co |
| \(\overrightarrow{AC} \cdot \overrightarrow{OB} = 8 - 8 - 8 = -8\) | M1 | Use of \(x_1x_2 + y_1y_2 + z_1z_2\) |
| \( | \overrightarrow{OB} | = 6\); \( |
| \(-8 = 6 \times \sqrt{24} \times \cos\theta\); \(\theta = 105.8° \rightarrow 74.2°\) | M1 A1 [5] | Connected correctly provided \(\overrightarrow{OB}\), \(\overrightarrow{AC}\) used; co (accept acute or obtuse) |
| Answer | Marks | Guidance |
|---|---|---|
| \(OA = \sqrt{19}\) or \(OC = \sqrt{11}\); Perimeter \(= 2(\sqrt{} + \sqrt{}) \rightarrow 15.4\) | B1, M1, A1 [3] | Used as a length; co (accept 15.3) |
## Question 10:
$\overrightarrow{OA} = \mathbf{i} + 3\mathbf{j} + 3\mathbf{k}$, $\overrightarrow{OC} = 3\mathbf{i} - \mathbf{j} + \mathbf{k}$
**Part (i):**
$\overrightarrow{OB} = \overrightarrow{OA} + \overrightarrow{OC} = 4\mathbf{i} + 2\mathbf{j} + 4\mathbf{k}$ | B1 | co
Unit vector $= \frac{1}{6}(4\mathbf{i} + 2\mathbf{j} + 4\mathbf{k})$ | M1 A1$\sqrt{}$ [3] | Divides by the modulus. $\sqrt{}$ on $\overrightarrow{OB}$
**Part (ii):**
$\overrightarrow{AC} = \overrightarrow{OC} - \overrightarrow{OA} = 2\mathbf{i} - 4\mathbf{j} - 2\mathbf{k}$ | B1 | co
$\overrightarrow{AC} \cdot \overrightarrow{OB} = 8 - 8 - 8 = -8$ | M1 | Use of $x_1x_2 + y_1y_2 + z_1z_2$
$|\overrightarrow{OB}| = 6$; $|\overrightarrow{AC}| = \sqrt{24}$ | M1 | Correct method for a modulus
$-8 = 6 \times \sqrt{24} \times \cos\theta$; $\theta = 105.8° \rightarrow 74.2°$ | M1 A1 [5] | Connected correctly provided $\overrightarrow{OB}$, $\overrightarrow{AC}$ used; co (accept acute or obtuse)
**Part (iii):**
$OA = \sqrt{19}$ or $OC = \sqrt{11}$; Perimeter $= 2(\sqrt{} + \sqrt{}) \rightarrow 15.4$ | B1, M1, A1 [3] | Used as a length; co (accept 15.3)10\\
\includegraphics[max width=\textwidth, alt={}, center]{56d4d40a-32f5-4f2d-938e-a24312cd42e7-4_552_629_842_758}
The diagram shows the parallelogram $O A B C$. Given that $\overrightarrow { O A } = \mathbf { i } + 3 \mathbf { j } + 3 \mathbf { k }$ and $\overrightarrow { O C } = 3 \mathbf { i } - \mathbf { j } + \mathbf { k }$, find\\
(i) the unit vector in the direction of $\overrightarrow { O B }$,\\
(ii) the acute angle between the diagonals of the parallelogram,\\
(iii) the perimeter of the parallelogram, correct to 1 decimal place.
\hfill \mbox{\textit{CAIE P1 2010 Q10 [11]}}