| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2013 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors 3D & Lines |
| Type | Triangle and parallelogram problems |
| Difficulty | Moderate -0.3 This is a straightforward application of standard vector techniques: (i) uses the scalar product formula for angles with simple 3D vectors requiring basic arithmetic, and (ii) involves finding a unit vector and scaling it. Both parts are routine textbook exercises with no problem-solving insight required, making it slightly easier than average. |
| Spec | 1.10c Magnitude and direction: of vectors1.10d Vector operations: addition and scalar multiplication1.10g Problem solving with vectors: in geometry |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\mathbf{OC} = \mathbf{AB} = \mathbf{b} - \mathbf{a} = \begin{pmatrix}2\\-3\\6\end{pmatrix}\) | M1 | Knowing how to find OC |
| Uses OC and OB | B1 | Using OC.OB or CO.BO |
| \(\mathbf{OC.OB} = 22 = 7 \times \sqrt{29} \cos BOC\) | M1 M1 | M1 use of \(x_1x_2 + \ldots\) M1 for modulus |
| \(\rightarrow\) Angle \(BOC = 54.3°\) (or \(0.948\) rad) | M1 A1 [6] | M1 everything linked. (nb uses BO.OC loses B1 A1) (nb uses other vectors – max M1M1) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Modulus of \(\mathbf{OC} = 7\); Vector \(= 35 \div 7 \times \mathbf{OC}\) | M1 | Knows to scale by factor of \(35 \div\) Mod |
| \(\rightarrow \pm 5\begin{pmatrix}2\\-3\\6\end{pmatrix}\) | A1\(\sqrt{}\) [2] | For their OC |
## Question 8:
$\overrightarrow{OA} = \begin{pmatrix}3\\3\\-4\end{pmatrix}$ and $\overrightarrow{OB} = \begin{pmatrix}5\\0\\2\end{pmatrix}$
### Part (i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\mathbf{OC} = \mathbf{AB} = \mathbf{b} - \mathbf{a} = \begin{pmatrix}2\\-3\\6\end{pmatrix}$ | M1 | Knowing how to find **OC** |
| Uses **OC** and **OB** | B1 | Using **OC.OB** or **CO.BO** |
| $\mathbf{OC.OB} = 22 = 7 \times \sqrt{29} \cos BOC$ | M1 M1 | M1 use of $x_1x_2 + \ldots$ M1 for modulus |
| $\rightarrow$ Angle $BOC = 54.3°$ (or $0.948$ rad) | M1 A1 [6] | M1 everything linked. (nb uses **BO.OC** loses B1 A1) (nb uses other vectors – max M1M1) |
### Part (ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Modulus of $\mathbf{OC} = 7$; Vector $= 35 \div 7 \times \mathbf{OC}$ | M1 | Knows to scale by factor of $35 \div$ Mod |
| $\rightarrow \pm 5\begin{pmatrix}2\\-3\\6\end{pmatrix}$ | A1$\sqrt{}$ [2] | For their **OC** |
---8\\
\includegraphics[max width=\textwidth, alt={}, center]{13cfb59a-7781-4786-a625-919b01a2a4f0-3_716_437_1137_854}
The diagram shows a parallelogram $O A B C$ in which
$$\overrightarrow { O A } = \left( \begin{array} { r }
3 \\
3 \\
- 4
\end{array} \right) \quad \text { and } \quad \overrightarrow { O B } = \left( \begin{array} { l }
5 \\
0 \\
2
\end{array} \right)$$
(i) Use a scalar product to find angle $B O C$.\\
(ii) Find a vector which has magnitude 35 and is parallel to the vector $\overrightarrow { O C }$.
\hfill \mbox{\textit{CAIE P1 2013 Q8 [8]}}