| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2019 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors 3D & Lines |
| Type | Area of parallelogram or trapezium using vectors |
| Difficulty | Standard +0.3 This is a straightforward multi-part vector question requiring standard techniques: dot product for perpendicularity, parallel vector comparison for trapezium, and area calculation using cross product magnitude. All steps are routine applications of formulas with no novel insight required, making it slightly easier than average. |
| Spec | 1.10a Vectors in 2D: i,j notation and column vectors1.10b Vectors in 3D: i,j,k notation1.10c Magnitude and direction: of vectors1.10d Vector operations: addition and scalar multiplication1.10g Problem solving with vectors: in geometry |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\mathbf{AB} = \begin{pmatrix} 2 \\ -3 \\ 5 \end{pmatrix} - \begin{pmatrix} -1 \\ 3 \\ -4 \end{pmatrix} = \begin{pmatrix} 3 \\ -6 \\ 9 \end{pmatrix}\), \(\mathbf{BC} = \begin{pmatrix} 4 \\ -2 \\ 5 \end{pmatrix} - \begin{pmatrix} 2 \\ -3 \\ 5 \end{pmatrix} = \begin{pmatrix} 2 \\ 1 \\ 0 \end{pmatrix}\) | B1B1 | Condone reversal of labels |
| \(\mathbf{AB} \cdot \mathbf{BC} = 6 - 6 \to 0\) (hence perpendicular) | B1 | AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\mathbf{DC} = \begin{pmatrix} 4 \\ -2 \\ 5 \end{pmatrix} - \begin{pmatrix} 2 \\ 2 \\ -1 \end{pmatrix} = \begin{pmatrix} 2 \\ -4 \\ 6 \end{pmatrix}\) | B1 | Or: \(\mathbf{CD} = \begin{pmatrix} -2 \\ 4 \\ -6 \end{pmatrix}\) |
| \(\mathbf{AB} = k\mathbf{DC}\) | M1 | OE; Expect \(k = \frac{3}{2}\); Or: \(\mathbf{DC} \cdot \mathbf{BC} = 4 - 4 = 0\) hence \(BC\) is also perpendicular to \(DC\); Or: \(\mathbf{AB} \cdot \mathbf{DC} = 1\) or \(\mathbf{AB} \cdot \mathbf{CD} = -1\), angle between lines is 0 or 180 |
| \(AB\) is parallel to \(DC\), hence \(ABCD\) is a trapezium | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \( | \mathbf{AB} | = \sqrt{9+36+81} = \sqrt{126} = 11.22\) \( |
| Area \(= \frac{1}{2}(their AB + their DC) \times their BC = 20.92\) | M1A1 | OE |
## Question 10:
### Part 10(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\mathbf{AB} = \begin{pmatrix} 2 \\ -3 \\ 5 \end{pmatrix} - \begin{pmatrix} -1 \\ 3 \\ -4 \end{pmatrix} = \begin{pmatrix} 3 \\ -6 \\ 9 \end{pmatrix}$, $\mathbf{BC} = \begin{pmatrix} 4 \\ -2 \\ 5 \end{pmatrix} - \begin{pmatrix} 2 \\ -3 \\ 5 \end{pmatrix} = \begin{pmatrix} 2 \\ 1 \\ 0 \end{pmatrix}$ | B1B1 | Condone reversal of labels |
| $\mathbf{AB} \cdot \mathbf{BC} = 6 - 6 \to 0$ (hence perpendicular) | B1 | AG |
### Part 10(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\mathbf{DC} = \begin{pmatrix} 4 \\ -2 \\ 5 \end{pmatrix} - \begin{pmatrix} 2 \\ 2 \\ -1 \end{pmatrix} = \begin{pmatrix} 2 \\ -4 \\ 6 \end{pmatrix}$ | B1 | Or: $\mathbf{CD} = \begin{pmatrix} -2 \\ 4 \\ -6 \end{pmatrix}$ |
| $\mathbf{AB} = k\mathbf{DC}$ | M1 | OE; Expect $k = \frac{3}{2}$; Or: $\mathbf{DC} \cdot \mathbf{BC} = 4 - 4 = 0$ hence $BC$ is also perpendicular to $DC$; Or: $\mathbf{AB} \cdot \mathbf{DC} = 1$ or $\mathbf{AB} \cdot \mathbf{CD} = -1$, angle between lines is 0 or 180 |
| $AB$ is parallel to $DC$, hence $ABCD$ is a trapezium | A1 | |
### Part 10(iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $|\mathbf{AB}| = \sqrt{9+36+81} = \sqrt{126} = 11.22$ $|\mathbf{DC}| = \sqrt{4+16+36} = \sqrt{56} = 7.483$ $|\mathbf{BC}| = \sqrt{4+1+0} = \sqrt{5} = 2.236$ | M1 | Method for finding at least 2 magnitudes |
| Area $= \frac{1}{2}(their AB + their DC) \times their BC = 20.92$ | M1A1 | OE |
---10\\
\includegraphics[max width=\textwidth, alt={}, center]{0e4a249a-9e6a-49d4-996c-fe07b7730f59-16_318_1006_260_568}
Relative to an origin $O$, the position vectors of the points $A , B , C$ and $D$, shown in the diagram, are given by
$$\overrightarrow { O A } = \left( \begin{array} { r }
- 1 \\
3 \\
- 4
\end{array} \right) , \quad \overrightarrow { O B } = \left( \begin{array} { r }
2 \\
- 3 \\
5
\end{array} \right) , \quad \overrightarrow { O C } = \left( \begin{array} { r }
4 \\
- 2 \\
5
\end{array} \right) \quad \text { and } \quad \overrightarrow { O D } = \left( \begin{array} { r }
2 \\
2 \\
- 1
\end{array} \right) .$$
(i) Show that $A B$ is perpendicular to $B C$.\\
(ii) Show that $A B C D$ is a trapezium.\\
(iii) Find the area of $A B C D$, giving your answer correct to 2 decimal places.\\
\hfill \mbox{\textit{CAIE P1 2019 Q10 [9]}}