| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2017 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors 3D & Lines |
| Type | Parallel and perpendicular lines |
| Difficulty | Standard +0.3 This is a straightforward two-part vectors question requiring standard techniques: (i) computing a scalar product to find an angle using the cosine formula, and (ii) using the parallel vectors condition (one is a scalar multiple of the other). Both parts involve routine algebraic manipulation with no conceptual challenges beyond applying well-practiced formulas. Slightly easier than average due to the direct application of standard methods. |
| Spec | 1.10b Vectors in 3D: i,j,k notation4.04c Scalar product: calculate and use for angles |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Uses scalar product correctly: \(3\times6 + 2\times6 + (-4)\times3 = 18\) | M1 | Use of dot product with \(\overrightarrow{OA}\) or \(\overrightarrow{AO}\) & \(\overrightarrow{OB}\) or \(\overrightarrow{BO}\) only |
| \(\ | \overrightarrow{OA}\ | = \sqrt{29},\ \ |
| \(\sqrt{29}\times9\times\cos AOB = 18\) | M1 | All linked correctly |
| \(\rightarrow AOB = 68.2°\) or \(1.19^c\) | A1 | Multiples of \(\pi\) are acceptable (e.g. \(0.379\pi^c\)) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\overrightarrow{AB} = 3\mathbf{i} + 4\mathbf{j} + (3+2p)\mathbf{k}\) | *M1 | For use of \(\overrightarrow{OB}-\overrightarrow{OA}\), allow with \(p=2\) |
| Comparing "\(\mathbf{j}\)" | DM1 | For comparing, \(\overrightarrow{OC}\) must contain \(p\) & \(q\). Can be implied by \(\overrightarrow{AB}=2\overrightarrow{OC}\) |
| \(\rightarrow p=2\frac{1}{2}\) and \(q=4\) | A1 A1 | Accuracy marks only available if \(\overrightarrow{AB}\) is correct |
## Question 8(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Uses scalar product correctly: $3\times6 + 2\times6 + (-4)\times3 = 18$ | M1 | Use of dot product with $\overrightarrow{OA}$ or $\overrightarrow{AO}$ & $\overrightarrow{OB}$ or $\overrightarrow{BO}$ only |
| $\|\overrightarrow{OA}\| = \sqrt{29},\ \|\overrightarrow{OB}\| = 9$ | M1 | Correct method for any one of $\|\overrightarrow{OA}\|,\ \|\overrightarrow{AO}\|,\ \|\overrightarrow{OB}\|$ or $\|\overrightarrow{BO}\|$ |
| $\sqrt{29}\times9\times\cos AOB = 18$ | M1 | All linked correctly |
| $\rightarrow AOB = 68.2°$ or $1.19^c$ | A1 | Multiples of $\pi$ are acceptable (e.g. $0.379\pi^c$) |
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## Question 8(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\overrightarrow{AB} = 3\mathbf{i} + 4\mathbf{j} + (3+2p)\mathbf{k}$ | *M1 | For use of $\overrightarrow{OB}-\overrightarrow{OA}$, allow with $p=2$ |
| Comparing "$\mathbf{j}$" | DM1 | For comparing, $\overrightarrow{OC}$ must contain $p$ & $q$. Can be implied by $\overrightarrow{AB}=2\overrightarrow{OC}$ |
| $\rightarrow p=2\frac{1}{2}$ and $q=4$ | A1 A1 | Accuracy marks only available if $\overrightarrow{AB}$ is correct |
---8 Relative to an origin $O$, the position vectors of three points $A , B$ and $C$ are given by
$$\overrightarrow { O A } = 3 \mathbf { i } + p \mathbf { j } - 2 p \mathbf { k } , \quad \overrightarrow { O B } = 6 \mathbf { i } + ( p + 4 ) \mathbf { j } + 3 \mathbf { k } \quad \text { and } \quad \overrightarrow { O C } = ( p - 1 ) \mathbf { i } + 2 \mathbf { j } + q \mathbf { k }$$
where $p$ and $q$ are constants.\\
(i) In the case where $p = 2$, use a scalar product to find angle $A O B$.\\
(ii) In the case where $\overrightarrow { A B }$ is parallel to $\overrightarrow { O C }$, find the values of $p$ and $q$.\\
\hfill \mbox{\textit{CAIE P1 2017 Q8 [8]}}