| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2018 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors: Lines & Planes |
| Type | Angles between vectors |
| Difficulty | Moderate -0.8 This is a straightforward multi-part vectors question testing basic operations: vector subtraction, midpoint formula, unit vectors, and dot product for angles. All parts are routine calculations with no problem-solving insight required, making it easier than average but not trivial due to the computational steps involved. |
| Spec | 1.10b Vectors in 3D: i,j,k notation1.10e Position vectors: and displacement4.04c Scalar product: calculate and use for angles |
| Answer | Marks | Guidance |
|---|---|---|
| \(\overrightarrow{AC} = \begin{pmatrix} 2 \\ 4 \\ -4 \end{pmatrix}\) | B1 | B1 for \(\overrightarrow{AC}\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\overrightarrow{OM} = \overrightarrow{OA} + \overrightarrow{AM} = \begin{pmatrix} 2 \\ -1 \\ 0 \end{pmatrix}\) or \(\frac{1}{2}\left[\begin{pmatrix} 1 \\ -3 \\ 2 \end{pmatrix} + \begin{pmatrix} 3 \\ 1 \\ -2 \end{pmatrix}\right]\) | M1 | M1 for their \(\overrightarrow{OM} = \overrightarrow{OA} + \overrightarrow{AM}\) oe |
| Unit vector in direction of \(\overrightarrow{OM} = \frac{1}{\sqrt{5}}(\overrightarrow{OM})\) | M1 A1 | M1 for dividing their \(\overrightarrow{OM}\) by their modulus |
| Answer | Marks | Guidance |
|---|---|---|
| \(\overrightarrow{AB} = \begin{pmatrix} -2 \\ 6 \\ 3 \end{pmatrix}\), Allow \(\pm\) | B1 | |
| \( | \overrightarrow{AB} | = 7,\ |
| \(7 \times 6\cos\theta = 8 \rightarrow \theta = 79.(0)°\) | A1 | 1.38 radians ok |
## Question 7(i):
| $\overrightarrow{AC} = \begin{pmatrix} 2 \\ 4 \\ -4 \end{pmatrix}$ | B1 | B1 for $\overrightarrow{AC}$ |
|---|---|---|
---
## Question 7(ii):
| $\overrightarrow{OM} = \overrightarrow{OA} + \overrightarrow{AM} = \begin{pmatrix} 2 \\ -1 \\ 0 \end{pmatrix}$ or $\frac{1}{2}\left[\begin{pmatrix} 1 \\ -3 \\ 2 \end{pmatrix} + \begin{pmatrix} 3 \\ 1 \\ -2 \end{pmatrix}\right]$ | M1 | M1 for their $\overrightarrow{OM} = \overrightarrow{OA} + \overrightarrow{AM}$ oe |
|---|---|---|
| Unit vector in direction of $\overrightarrow{OM} = \frac{1}{\sqrt{5}}(\overrightarrow{OM})$ | M1 A1 | M1 for dividing their $\overrightarrow{OM}$ by their modulus |
---
## Question 7(iii):
| $\overrightarrow{AB} = \begin{pmatrix} -2 \\ 6 \\ 3 \end{pmatrix}$, Allow $\pm$ | B1 | |
|---|---|---|
| $|\overrightarrow{AB}| = 7,\ |\overrightarrow{AC}| = 6$; $\begin{pmatrix} -2 \\ 6 \\ 3 \end{pmatrix} \cdot \begin{pmatrix} 2 \\ 4 \\ -4 \end{pmatrix} = -4 + 24 - 12 = 8$ | M1 M1 | Product of both moduli, scalar product of $\pm$ their $AB$ and $AC$ |
| $7 \times 6\cos\theta = 8 \rightarrow \theta = 79.(0)°$ | A1 | 1.38 radians ok |7 Relative to an origin $O$, the position vectors of the points $A , B$ and $C$ are given by
$$\overrightarrow { O A } = \left( \begin{array} { r }
1 \\
- 3 \\
2
\end{array} \right) , \quad \overrightarrow { O B } = \left( \begin{array} { r }
- 1 \\
3 \\
5
\end{array} \right) \quad \text { and } \quad \overrightarrow { O C } = \left( \begin{array} { r }
3 \\
1 \\
- 2
\end{array} \right)$$
(i) Find $\overrightarrow { A C }$.\\
(ii) The point $M$ is the mid-point of $A C$. Find the unit vector in the direction of $\overrightarrow { O M }$.\\
(iii) Evaluate $\overrightarrow { A B } \cdot \overrightarrow { A C }$ and hence find angle $B A C$.\\
\hfill \mbox{\textit{CAIE P1 2018 Q7 [8]}}