| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2005 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors 3D & Lines |
| Type | Equal length conditions |
| Difficulty | Standard +0.3 This is a straightforward multi-part vectors question requiring standard techniques: scalar product for angle (routine formula application), unit vector calculation (normalize a vector), and equal length condition leading to a quadratic equation. All parts are textbook exercises with no novel insight required, making it slightly easier than average. |
| Spec | 1.10b Vectors in 3D: i,j,k notation1.10c Magnitude and direction: of vectors1.10d Vector operations: addition and scalar multiplication1.10e Position vectors: and displacement1.10f Distance between points: using position vectors |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(\overrightarrow{OA} \cdot \overrightarrow{OB} = 8 - 9 - 2 = -3\) | M1 | Correct use of \(a_1a_2 + b_1b_2 + c_1c_2\) |
| \(\overrightarrow{OA} \cdot \overrightarrow{OB} = \sqrt{14} \times \sqrt{29}\cos AOB \Rightarrow AOB = 99°\) | M1 M1 A1 [4] | Modulus. Correct use of abcos \(\theta\). CAO |
| (ii) \(\overrightarrow{AB} = \mathbf{b} - \mathbf{a} = 2\mathbf{i} - 6\mathbf{j} + 3\mathbf{k}\) | B1 | CAO |
| Magnitude of \(\overrightarrow{AB} = \sqrt{49} = 7\) | M1 | Use of Pythagoras + division. |
| \(\Rightarrow\) Unit vector \(= \frac{1}{7}(2\mathbf{i} - 6\mathbf{j} + 3\mathbf{k})\) | A1 \(\checkmark\) [3] | CAO (use of \(\overrightarrow{BA}\) for \(\overrightarrow{AB}\) has max 2/3). |
| (iii) \(\overrightarrow{AC} = -2\mathbf{i} + 3\mathbf{j} + (p+1)\mathbf{k}\) | B1 | CAO - condone \(a - c\) here. |
| \(4 + 9 + (p+1)^2 = 49\) | M1 A1 \(\checkmark\) | Correct method for forming an equation |
| \(\Rightarrow p = 5\) or \(-7\) | A1 [4] | CAO |
$\overrightarrow{OA} = 2\mathbf{i} + 3\mathbf{j} - \mathbf{k}$, $\overrightarrow{OB} = 4\mathbf{i} - 3\mathbf{j} + 2\mathbf{k}$
(i) $\overrightarrow{OA} \cdot \overrightarrow{OB} = 8 - 9 - 2 = -3$ | M1 | Correct use of $a_1a_2 + b_1b_2 + c_1c_2$
$\overrightarrow{OA} \cdot \overrightarrow{OB} = \sqrt{14} \times \sqrt{29}\cos AOB \Rightarrow AOB = 99°$ | M1 M1 A1 [4] | Modulus. Correct use of abcos $\theta$. CAO
(ii) $\overrightarrow{AB} = \mathbf{b} - \mathbf{a} = 2\mathbf{i} - 6\mathbf{j} + 3\mathbf{k}$ | B1 | CAO
Magnitude of $\overrightarrow{AB} = \sqrt{49} = 7$ | M1 | Use of Pythagoras + division.
$\Rightarrow$ Unit vector $= \frac{1}{7}(2\mathbf{i} - 6\mathbf{j} + 3\mathbf{k})$ | A1 $\checkmark$ [3] | CAO (use of $\overrightarrow{BA}$ for $\overrightarrow{AB}$ has max 2/3).
(iii) $\overrightarrow{AC} = -2\mathbf{i} + 3\mathbf{j} + (p+1)\mathbf{k}$ | B1 | CAO - condone $a - c$ here.
$4 + 9 + (p+1)^2 = 49$ | M1 A1 $\checkmark$ | Correct method for forming an equation
$\Rightarrow p = 5$ or $-7$ | A1 [4] | CAO11 Relative to an origin $O$, the position vectors of the points $A$ and $B$ are given by
$$\overrightarrow { O A } = 2 \mathbf { i } + 3 \mathbf { j } - \mathbf { k } \quad \text { and } \quad \overrightarrow { O B } = 4 \mathbf { i } - 3 \mathbf { j } + 2 \mathbf { k }$$
(i) Use a scalar product to find angle $A O B$, correct to the nearest degree.\\
(ii) Find the unit vector in the direction of $\overrightarrow { A B }$.\\
(iii) The point $C$ is such that $\overrightarrow { O C } = 6 \mathbf { j } + p \mathbf { k }$, where $p$ is a constant. Given that the lengths of $\overrightarrow { A B }$ and $\overrightarrow { A C }$ are equal, find the possible values of $p$.
\hfill \mbox{\textit{CAIE P1 2005 Q11 [11]}}