| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2004 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors: Lines & Planes |
| Type | Angles between vectors |
| Difficulty | Moderate -0.3 This is a straightforward two-part vectors question requiring standard techniques: (i) applying the scalar product formula cos θ = (a·b)/(|a||b|) to find an angle, and (ii) using vector arithmetic to find point C then normalizing. Both parts are routine applications of basic vector methods with no conceptual challenges, making it slightly easier than average. |
| Spec | 1.10c Magnitude and direction: of vectors1.10g Problem solving with vectors: in geometry |
| Answer | Marks | Guidance |
|---|---|---|
| \((\mathbf{i}+7\mathbf{j}+2\mathbf{k}).(-5\mathbf{i}+5\mathbf{j}+6\mathbf{k})\) | M1 | Use of \(\rightarrow x_1x_2 + y_1y_2 + z_1z_2\) |
| \(\rightarrow -5+35+12 = 42\) | M1 | Modulus used in dot product |
| \(42 = \sqrt{54}\;\sqrt{86}\cos\theta\) | M1 | Everything linked correctly |
| \(\rightarrow\) angle \(AOB = 0.907\) | A1 [4] | Accept if more accuracy given; must be radians; any combination of OA/AO OB/BO ok for three M1 marks; if AB used with OA/OB max M1 M1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\mathbf{BC} = \frac{1}{2}(\mathbf{b}-\mathbf{a}) = -3\mathbf{i}-\mathbf{j}+2\mathbf{k}\) | ||
| \(\mathbf{OC} = \mathbf{OB}+\mathbf{BC} = -5\mathbf{i}+5\mathbf{j}+6\mathbf{k}-3\mathbf{i}-\mathbf{j}+2\mathbf{k} = -8\mathbf{i}+4\mathbf{j}+8\mathbf{k}\) | M1, A1 | Could be from \(\mathbf{OA}+\mathbf{AC}\); correct only |
| Unit Vector \(= (-8\mathbf{i}+4\mathbf{j}+8\mathbf{k}) \div 12\) | M1A1\(\sqrt{}\) [4] | Knowing to divide by length of vector; leaving as \(\sqrt{}\) is acceptable for both marks |
# Question 8:
## Part (i)
| $(\mathbf{i}+7\mathbf{j}+2\mathbf{k}).(-5\mathbf{i}+5\mathbf{j}+6\mathbf{k})$ | M1 | Use of $\rightarrow x_1x_2 + y_1y_2 + z_1z_2$ |
| $\rightarrow -5+35+12 = 42$ | M1 | Modulus used in dot product |
| $42 = \sqrt{54}\;\sqrt{86}\cos\theta$ | M1 | Everything linked correctly |
| $\rightarrow$ angle $AOB = 0.907$ | A1 [4] | Accept if more accuracy given; must be radians; any combination of **OA/AO OB/BO** ok for three M1 marks; if **AB** used with **OA/OB** max M1 M1 |
## Part (ii)
| $\mathbf{BC} = \frac{1}{2}(\mathbf{b}-\mathbf{a}) = -3\mathbf{i}-\mathbf{j}+2\mathbf{k}$ | | |
| $\mathbf{OC} = \mathbf{OB}+\mathbf{BC} = -5\mathbf{i}+5\mathbf{j}+6\mathbf{k}-3\mathbf{i}-\mathbf{j}+2\mathbf{k} = -8\mathbf{i}+4\mathbf{j}+8\mathbf{k}$ | M1, A1 | Could be from $\mathbf{OA}+\mathbf{AC}$; correct only |
| Unit Vector $= (-8\mathbf{i}+4\mathbf{j}+8\mathbf{k}) \div 12$ | M1A1$\sqrt{}$ [4] | Knowing to divide by length of vector; leaving as $\sqrt{}$ is acceptable for both marks |
---8 The points $A$ and $B$ have position vectors $\mathbf { i } + 7 \mathbf { j } + 2 \mathbf { k }$ and $- 5 \mathbf { i } + 5 \mathbf { j } + 6 \mathbf { k }$ respectively, relative to an origin $O$.\\
(i) Use a scalar product to calculate angle $A O B$, giving your answer in radians correct to 3 significant figures.\\
(ii) The point $C$ is such that $\overrightarrow { A B } = 2 \overrightarrow { B C }$. Find the unit vector in the direction of $\overrightarrow { O C }$.
\hfill \mbox{\textit{CAIE P1 2004 Q8 [8]}}