| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2010 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors 3D & Lines |
| Type | Perpendicularity conditions |
| Difficulty | Standard +0.3 This is a straightforward 3D vectors question requiring standard techniques: (i) angle between vectors using dot product formula, (ii) vector subtraction with scalar multiplication, (iii) perpendicularity condition using dot product equals zero. All steps are routine applications of basic vector operations with no novel insight required, making it slightly easier than average. |
| Spec | 1.10a Vectors in 2D: i,j notation and column vectors1.10b 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 |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\overrightarrow{OA}\cdot\overrightarrow{OB} = -6 + 2 + 12 = 8\) | M1 | Use of \(x_1x_2 + y_1y_2 + z_1z_2\) |
| \(\cos AOB = \frac{8}{\sqrt{14}\sqrt{29}}\) | M1, M1 | Mod worked correctly for either one; Division of "8" by product of mods |
| \(AOB = 66.6°\) | A1 | |
| [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(3\mathbf{i} - 2\mathbf{j} + 4\mathbf{k} + p(2\mathbf{i} + \mathbf{j} - 3\mathbf{k})\) | B1 | In any unsimplified form |
| [1] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\overrightarrow{BC} = \mathbf{i}(3+2p) + \mathbf{j}(-2+p) + \mathbf{k}(4-3p)\) | M1 | |
| Their \(\overrightarrow{BC}\cdot[2\mathbf{i} + \mathbf{j} - 3\mathbf{k}] = 0\) | M1 | Scalar product \(= 0\) used |
| \(2(3+2p) + (p-2) - 3(4-3p) = 0\) | A1\(\sqrt{}\) | ft from their \(BC\) |
| \(p = 4/7 \approx 0.571\) | A1 | cao |
| [4] |
## Question 10:
### Part (i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\overrightarrow{OA}\cdot\overrightarrow{OB} = -6 + 2 + 12 = 8$ | M1 | Use of $x_1x_2 + y_1y_2 + z_1z_2$ |
| $\cos AOB = \frac{8}{\sqrt{14}\sqrt{29}}$ | M1, M1 | Mod worked correctly for either one; Division of "8" by product of mods |
| $AOB = 66.6°$ | A1 | |
| **[4]** | | |
### Part (ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $3\mathbf{i} - 2\mathbf{j} + 4\mathbf{k} + p(2\mathbf{i} + \mathbf{j} - 3\mathbf{k})$ | B1 | In any unsimplified form |
| **[1]** | | |
### Part (iii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\overrightarrow{BC} = \mathbf{i}(3+2p) + \mathbf{j}(-2+p) + \mathbf{k}(4-3p)$ | M1 | |
| Their $\overrightarrow{BC}\cdot[2\mathbf{i} + \mathbf{j} - 3\mathbf{k}] = 0$ | M1 | Scalar product $= 0$ used |
| $2(3+2p) + (p-2) - 3(4-3p) = 0$ | A1$\sqrt{}$ | ft from their $BC$ |
| $p = 4/7 \approx 0.571$ | A1 | cao |
| **[4]** | | |
---10\\
\includegraphics[max width=\textwidth, alt={}, center]{32a57386-2696-4fda-a3cb-ca0c5c3be432-4_561_599_744_774}
The diagram shows triangle $O A B$, in which the position vectors of $A$ and $B$ with respect to $O$ are given by
$$\overrightarrow { O A } = 2 \mathbf { i } + \mathbf { j } - 3 \mathbf { k } \quad \text { and } \quad \overrightarrow { O B } = - 3 \mathbf { i } + 2 \mathbf { j } - 4 \mathbf { k } .$$
$C$ is a point on $O A$ such that $\overrightarrow { O C } = p \overrightarrow { O A }$, where $p$ is a constant.\\
(i) Find angle $A O B$.\\
(ii) Find $\overrightarrow { B C }$ in terms of $p$ and vectors $\mathbf { i } , \mathbf { j }$ and $\mathbf { k }$.\\
(iii) Find the value of $p$ given that $B C$ is perpendicular to $O A$.
\hfill \mbox{\textit{CAIE P1 2010 Q10 [9]}}