| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2017 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors 3D & Lines |
| Type | Perpendicularity conditions |
| Difficulty | Standard +0.3 This is a straightforward perpendicularity question requiring students to apply the dot product condition (vectors perpendicular when dot product equals zero) and solve a resulting linear equation. It's slightly easier than average as it involves standard vector operations with no geometric insight or multi-step problem-solving required. |
| Spec | 1.10b Vectors in 3D: i,j,k notation1.10c Magnitude and direction: of vectors1.10e Position vectors: and displacement1.10f Distance between points: using position vectors |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Angle \(AOB = 90° \rightarrow 6 + 36 - 7p = 0\) | M1 | Use of \(x_1x_2 + y_1y_2 + z_1z_2 = 0\) or Pythagoras |
| \(\rightarrow p = 6\) | A1 | |
| Total: 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\overrightarrow{OC} = \frac{2}{3}\begin{pmatrix}3\\-6\\p\end{pmatrix} = \begin{pmatrix}2\\-4\\4\end{pmatrix}\) | B1 FT | CAO FT on their value of \(p\) |
| \(\overrightarrow{BC} = \mathbf{c} - \mathbf{b} = \begin{pmatrix}0\\2\\11\end{pmatrix}\); magnitude \(= \sqrt{125}\) | M1 M1 | Use of \(\mathbf{c} - \mathbf{b}\). Allow magnitude of \(\mathbf{b} + \mathbf{c}\) or \(\mathbf{b} - \mathbf{c}\). Allow first M1 in terms of \(p\) |
| Unit vector \(= \frac{1}{\sqrt{125}}\begin{pmatrix}0\\2\\11\end{pmatrix}\) | A1 | OE Allow \(\pm\) and decimal equivalent |
## Question 2(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Angle $AOB = 90° \rightarrow 6 + 36 - 7p = 0$ | **M1** | Use of $x_1x_2 + y_1y_2 + z_1z_2 = 0$ or Pythagoras |
| $\rightarrow p = 6$ | **A1** | |
| **Total: 2** | | |
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## Question 2(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\overrightarrow{OC} = \frac{2}{3}\begin{pmatrix}3\\-6\\p\end{pmatrix} = \begin{pmatrix}2\\-4\\4\end{pmatrix}$ | **B1 FT** | CAO FT on their value of $p$ |
| $\overrightarrow{BC} = \mathbf{c} - \mathbf{b} = \begin{pmatrix}0\\2\\11\end{pmatrix}$; magnitude $= \sqrt{125}$ | **M1 M1** | Use of $\mathbf{c} - \mathbf{b}$. Allow magnitude of $\mathbf{b} + \mathbf{c}$ or $\mathbf{b} - \mathbf{c}$. Allow first **M1** in terms of $p$ |
| Unit vector $= \frac{1}{\sqrt{125}}\begin{pmatrix}0\\2\\11\end{pmatrix}$ | **A1** | OE Allow $\pm$ and decimal equivalent |
---2 Relative to an origin $O$, the position vectors of points $A$ and $B$ are given by
$$\overrightarrow { O A } = \left( \begin{array} { r }
3 \\
- 6 \\
p
\end{array} \right) \quad \text { and } \quad \overrightarrow { O B } = \left( \begin{array} { r }
2 \\
- 6 \\
- 7
\end{array} \right)$$
and angle $A O B = 90 ^ { \circ }$.\\
(i) Find the value of $p$.\\
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The point $C$ is such that $\overrightarrow { O C } = \frac { 2 } { 3 } \overrightarrow { O A }$.\\
(ii) Find the unit vector in the direction of $\overrightarrow { B C }$.\\
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\hfill \mbox{\textit{CAIE P1 2017 Q2 [6]}}