| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2015 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors: Lines & Planes |
| Type | Vector operations and magnitudes |
| Difficulty | Standard +0.3 This is a straightforward vectors question from CAIE P1 involving basic position vector operations. The question appears incomplete but typically involves finding vectors between points (e.g., AB, BC) and possibly magnitudes or angles. These are standard textbook exercises requiring only direct application of vector arithmetic formulas with no problem-solving insight needed, making it slightly easier than average. |
| Spec | 1.10c Magnitude and direction: of vectors1.10g Problem solving with vectors: in geometry |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\overrightarrow{AB} = \begin{pmatrix}5\\-1\\-2\end{pmatrix} - \begin{pmatrix}3\\2\\-3\end{pmatrix} = \begin{pmatrix}2\\-3\\1\end{pmatrix}\) | B1 | Or \(\overrightarrow{BA}\), \(\overrightarrow{CB}\). Allow any combination. Ignore labels. |
| \(\overrightarrow{BC} = \begin{pmatrix}6\\1\\2\end{pmatrix} - \begin{pmatrix}5\\-1\\-2\end{pmatrix} = \begin{pmatrix}1\\2\\4\end{pmatrix}\) | B1 | |
| \(\overrightarrow{AB}\cdot\overrightarrow{BC} = 2 - 6 + 4\) oe — must be seen \(= 0\), hence \(ABC = 90°\) | M1 A1 [4] | Could be part of calculation for angle \(ABC\). AG Alt methods Pythag, Cosine Rule |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \( | \overrightarrow{AB} | = \sqrt{14}\), \( |
| Area \(= \frac{1}{2}\sqrt{14}\sqrt{21}\) | M1 | Reasonable attempt at vectors and their magnitudes |
| \(8.6\) oe | A1 [3] | Allow \(\frac{7\sqrt{6}}{2}\) |
## Question 5(i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\overrightarrow{AB} = \begin{pmatrix}5\\-1\\-2\end{pmatrix} - \begin{pmatrix}3\\2\\-3\end{pmatrix} = \begin{pmatrix}2\\-3\\1\end{pmatrix}$ | B1 | Or $\overrightarrow{BA}$, $\overrightarrow{CB}$. Allow any combination. Ignore labels. |
| $\overrightarrow{BC} = \begin{pmatrix}6\\1\\2\end{pmatrix} - \begin{pmatrix}5\\-1\\-2\end{pmatrix} = \begin{pmatrix}1\\2\\4\end{pmatrix}$ | B1 | |
| $\overrightarrow{AB}\cdot\overrightarrow{BC} = 2 - 6 + 4$ oe — must be seen $= 0$, hence $ABC = 90°$ | M1 A1 [4] | Could be part of calculation for angle $ABC$. AG Alt methods Pythag, Cosine Rule |
## Question 5(ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $|\overrightarrow{AB}| = \sqrt{14}$, $|\overrightarrow{BC}| = \sqrt{21}$ oe | B1 | At least one correct |
| Area $= \frac{1}{2}\sqrt{14}\sqrt{21}$ | M1 | Reasonable attempt at vectors and their magnitudes |
| $8.6$ oe | A1 [3] | Allow $\frac{7\sqrt{6}}{2}$ |
---5 Relative to an origin $O$, the position vectors of the points $A , B$ and $C$ are given by
$$\overrightarrow { O A } = \left( \begin{array} { r }
3 \\
2 \\
- 3
\end{array} \right) , \quad \overrightarrow { O B } = \left( \begin{array} { r }
5 \\
- 1 \\
- 2
\end{array} \right) \quad \text { and } \quad \overrightarrow { O C } = \left( \begin{array} { l }
6 \\
1 \\
2
\end{array} \right)$$
(i) Show that angle $A B C$ is $90 ^ { \circ }$.\\
(ii) Find the area of triangle $A B C$, giving your answer correct to 1 decimal place.
\hfill \mbox{\textit{CAIE P1 2015 Q5 [7]}}