| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2023 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors: Lines & Planes |
| Type | Triangle and parallelogram areas |
| Difficulty | Moderate -0.3 This is a straightforward multi-part vectors question requiring standard techniques: finding the fourth vertex of a parallelogram using vector addition, calculating an angle using the dot product formula, and finding area using |a||b|sin(θ). All steps are routine applications of A-level formulas with no novel insight required, making it slightly easier than average. |
| Spec | 1.10b Vectors in 3D: i,j,k notation1.10d Vector operations: addition and scalar multiplication4.04c Scalar product: calculate and use for angles |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Obtain a vector for one side of the parallelogram | B1 | e.g. \(\overrightarrow{AB} = \begin{pmatrix}2\\2\\-1\end{pmatrix}\) or \(\overrightarrow{BC} = \begin{pmatrix}-1\\-5\\-6\end{pmatrix}\) |
| Correct method to obtain \(\pm\overrightarrow{OD}\) | M1 | e.g. \(\overrightarrow{OD} = \overrightarrow{OA} + \overrightarrow{BC}\); MO if use \(\overrightarrow{AB} = \overrightarrow{CD}\) or \(\overrightarrow{BC} = \overrightarrow{DA}\) |
| Obtain \(\overrightarrow{OD} = \mathbf{i} - 4\mathbf{j} - 3\mathbf{k}\) | A1 | Any equivalent form. Accept coordinates |
| 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Using the correct process, evaluate the scalar product \(\overrightarrow{BA}\cdot\overrightarrow{BC}\) | M1 | \((2+10-6)\) Scalar product of two relevant vectors. OE |
| Using the correct process for the moduli, divide the scalar product by the product of the moduli | M1 | \(\frac{2+10-6}{\sqrt{9}\times\sqrt{62}}\) |
| Obtain answer \(\frac{-2}{\sqrt{62}}\) | A1 | ISW. Or simplified equivalent i.e. \(\frac{\sqrt{62}}{31}\) |
| 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| State or imply \(\sin\theta = \sqrt{\frac{58}{62}}\) | B1 FT | Follow *their* \(\cos\theta\) |
| Use correct method to find the area of \(ABCD\) | M1 | e.g. \(2 \times \frac{1}{2} BA \times BC \sin\theta\). Condone decimals. |
| Correct unsimplified expression for the area | A1 FT | e.g. \(2 \times \frac{1}{2} \times 3 \times \sqrt{62} \times \sin\theta\). Condone decimals. Follow *their* sides and angle. |
| Obtain answer \(3\sqrt{58}\) | A1 | Correct only. |
| Total | 4 |
## Question 6(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Obtain a vector for one side of the parallelogram | B1 | e.g. $\overrightarrow{AB} = \begin{pmatrix}2\\2\\-1\end{pmatrix}$ or $\overrightarrow{BC} = \begin{pmatrix}-1\\-5\\-6\end{pmatrix}$ |
| Correct method to obtain $\pm\overrightarrow{OD}$ | M1 | e.g. $\overrightarrow{OD} = \overrightarrow{OA} + \overrightarrow{BC}$; MO if use $\overrightarrow{AB} = \overrightarrow{CD}$ or $\overrightarrow{BC} = \overrightarrow{DA}$ |
| Obtain $\overrightarrow{OD} = \mathbf{i} - 4\mathbf{j} - 3\mathbf{k}$ | A1 | Any equivalent form. Accept coordinates |
| | **3** | |
---
## Question 6(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Using the correct process, evaluate the scalar product $\overrightarrow{BA}\cdot\overrightarrow{BC}$ | M1 | $(2+10-6)$ Scalar product of two relevant vectors. OE |
| Using the correct process for the moduli, divide the scalar product by the product of the moduli | M1 | $\frac{2+10-6}{\sqrt{9}\times\sqrt{62}}$ |
| Obtain answer $\frac{-2}{\sqrt{62}}$ | A1 | ISW. Or simplified equivalent i.e. $\frac{\sqrt{62}}{31}$ |
| | **3** | |
## Question 6(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| State or imply $\sin\theta = \sqrt{\frac{58}{62}}$ | B1 FT | Follow *their* $\cos\theta$ |
| Use correct method to find the area of $ABCD$ | M1 | e.g. $2 \times \frac{1}{2} BA \times BC \sin\theta$. Condone decimals. |
| Correct unsimplified expression for the area | A1 FT | e.g. $2 \times \frac{1}{2} \times 3 \times \sqrt{62} \times \sin\theta$. Condone decimals. Follow *their* sides and angle. |
| Obtain answer $3\sqrt{58}$ | A1 | Correct only. |
| **Total** | **4** | |
---6 Relative to the origin $O$, the points $A , B$ and $C$ have position vectors given by
$$\overrightarrow { O A } = \left( \begin{array} { l }
2 \\
1 \\
3
\end{array} \right) , \quad \overrightarrow { O B } = \left( \begin{array} { l }
4 \\
3 \\
2
\end{array} \right) \quad \text { and } \quad \overrightarrow { O C } = \left( \begin{array} { r }
3 \\
- 2 \\
- 4
\end{array} \right) .$$
The quadrilateral $A B C D$ is a parallelogram.
\begin{enumerate}[label=(\alph*)]
\item Find the position vector of $D$.
\item The angle between $B A$ and $B C$ is $\theta$.
Find the exact value of $\cos \theta$.
\item Hence find the area of $A B C D$, giving your answer in the form $p \sqrt { q }$, where $p$ and $q$ are integers.
\end{enumerate}
\hfill \mbox{\textit{CAIE P3 2023 Q6 [10]}}