| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2016 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors: Lines & Planes |
| Type | Perpendicularity conditions |
| Difficulty | Moderate -0.3 This is a straightforward multi-part vectors question testing standard techniques: scalar product for angles, unit vector scaling, and perpendicularity condition. All parts are routine applications of formulas with no conceptual challenges or novel problem-solving 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 |
|---|---|---|
| \(-4 - 6 - 6 = -16\) | M1 | Use of \(x_1x_2 + y_1y_2 + z_1z_2\) on their \(\overrightarrow{OA}\) & \(\overrightarrow{OB}\) |
| \(\sqrt{x_1^2 + y_1^2 + z_1^2}\) or \(\sqrt{x_2^2 + y_2^2 + z_2^2}\) | M1 | Modulus once on either their \(\overrightarrow{OA}\) or \(\overrightarrow{OB}\) |
| \(3 \times 7 \times \cos\theta = -16 \rightarrow \theta = 139.6°\) or \(2.44^c\) or \(0.776\pi\) | M1 A1 | All linked using their \(\overrightarrow{OA}\) & \(\overrightarrow{OB}\) [4] |
| Answer | Marks | Guidance |
|---|---|---|
| \(\overrightarrow{AC} = c - a = \begin{pmatrix} 0 \\ 8 \\ 6 \end{pmatrix}\) | B1 | |
| Magnitude \(= 10\) | ||
| Scaling \(\rightarrow \frac{15}{their\ 10} \times \begin{pmatrix} 0 \\ 8 \\ 6 \end{pmatrix} = \begin{pmatrix} 0 \\ 12 \\ 9 \end{pmatrix}\) | M1 A1 | For \(15 \times\) *their* unit vector [3] |
| Answer | Marks | Guidance |
|---|---|---|
| \(\begin{pmatrix} 2+2p \\ 6-2p \\ 5-p \end{pmatrix}\) | B1 | Single vector soi by scalar product |
| \(\rightarrow -2(2+2p) + 3(6-2p) + 6(5-p) = 0 \rightarrow p = 2\frac{3}{4}\) | M1 A1 | Dot product of \((p\overrightarrow{OA} + \overrightarrow{OC})\) and \(\overrightarrow{OB} = 0\) [3] |
# Question 9:
## Part (i):
$-4 - 6 - 6 = -16$ | M1 | Use of $x_1x_2 + y_1y_2 + z_1z_2$ on their $\overrightarrow{OA}$ & $\overrightarrow{OB}$
$\sqrt{x_1^2 + y_1^2 + z_1^2}$ or $\sqrt{x_2^2 + y_2^2 + z_2^2}$ | M1 | Modulus once on either their $\overrightarrow{OA}$ or $\overrightarrow{OB}$
$3 \times 7 \times \cos\theta = -16 \rightarrow \theta = 139.6°$ or $2.44^c$ or $0.776\pi$ | M1 A1 | All linked using their $\overrightarrow{OA}$ & $\overrightarrow{OB}$ [4]
## Part (ii):
$\overrightarrow{AC} = c - a = \begin{pmatrix} 0 \\ 8 \\ 6 \end{pmatrix}$ | B1 |
Magnitude $= 10$ | |
Scaling $\rightarrow \frac{15}{their\ 10} \times \begin{pmatrix} 0 \\ 8 \\ 6 \end{pmatrix} = \begin{pmatrix} 0 \\ 12 \\ 9 \end{pmatrix}$ | M1 A1 | For $15 \times$ *their* unit vector [3]
## Part (iii):
$\begin{pmatrix} 2+2p \\ 6-2p \\ 5-p \end{pmatrix}$ | B1 | Single vector soi by scalar product
$\rightarrow -2(2+2p) + 3(6-2p) + 6(5-p) = 0 \rightarrow p = 2\frac{3}{4}$ | M1 A1 | Dot product of $(p\overrightarrow{OA} + \overrightarrow{OC})$ and $\overrightarrow{OB} = 0$ [3]
---9 Relative to an origin $O$, the position vectors of the points $A , B$ and $C$ are given by
$$\overrightarrow { O A } = \left( \begin{array} { r }
2 \\
- 2 \\
- 1
\end{array} \right) , \quad \overrightarrow { O B } = \left( \begin{array} { r }
- 2 \\
3 \\
6
\end{array} \right) \quad \text { and } \quad \overrightarrow { O C } = \left( \begin{array} { l }
2 \\
6 \\
5
\end{array} \right)$$
(i) Use a scalar product to find angle $A O B$.\\
(ii) Find the vector which is in the same direction as $\overrightarrow { A C }$ and of magnitude 15 units.\\
(iii) Find the value of the constant $p$ for which $p \overrightarrow { O A } + \overrightarrow { O C }$ is perpendicular to $\overrightarrow { O B }$.
\hfill \mbox{\textit{CAIE P1 2016 Q9 [10]}}