| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2022 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors: Cross Product & Distances |
| Type | Angle between vectors using scalar product |
| Difficulty | Moderate -0.3 This is a straightforward two-part question requiring standard vector techniques: (a) finding angle using scalar product formula with direction vectors AB and AC, and (b) using the sine derived from part (a) in the triangle area formula. Both parts follow routine procedures taught in P3 with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.10c Magnitude and direction: of vectors1.10f Distance between points: using position vectors1.10g Problem solving with vectors: in geometry |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| State or imply \(\overrightarrow{AB}\) or \(\overrightarrow{AC}\) correctly in component form | B1 | \(\overrightarrow{AB} = 2\mathbf{i} - 2\mathbf{j} + \mathbf{k}\), \(\overrightarrow{AC} = 4\mathbf{i} - 3\mathbf{k}\) |
| Using the correct process with relevant vectors to evaluate the scalar product \(\overrightarrow{AB}.\overrightarrow{AC}\) | M1 | or \(\overrightarrow{BA}.\overrightarrow{CA}\) \((8-3=5)\). M0 for \(\overrightarrow{AB}.\overrightarrow{CA}\). |
| Using the correct process for the moduli, divide *their* scalar product by the product of *their* moduli to obtain \(\cos\theta\) or \(\theta\) | M1 | \(\left(\dfrac{5}{\sqrt{9}\sqrt{25}}\right)\) Independent of the first M1. |
| Obtain answer \(\dfrac{1}{3}\) | A1 | ISW. Need to see a value for \(\cos\theta\). Accept \(\frac{5}{15}\) or \(0.333\) (\(\cos^{-1}\frac{1}{3}\) alone is not sufficient) |
| 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Use correct method to find an exact value for the sine of angle \(BAC\) from *their* (a) | M1 | \(\left(\sqrt{1-\frac{1}{9}}\right)\) |
| Obtain answer \(\frac{2}{3}\sqrt{2}\), or equivalent | A1 | |
| Use correct area formula to find the area of triangle \(ABC\) with *their* versions of relevant vectors | M1 | \(\left(\frac{1}{2}\sqrt{9}\sqrt{25} \times \textit{their}\sin\theta\right)\) or \(\frac{1}{2}\sqrt{9}\sqrt{25}\times\sin\left(\cos^{-1}\frac{1}{3}\right)\) |
| Obtain answer \(5\sqrt{2}\) or \(\sqrt{50}\) | A1 | Only ISW |
| Alternative method 1: Use correct method to find perpendicular distance from \(A\) to \(BC\) (or \(B\) to \(AC\) or \(C\) to \(AB\)) | M1 | \(\begin{pmatrix}2+2\lambda\\-2+2\lambda\\1-4\lambda\end{pmatrix}.\begin{pmatrix}2\\2\\-4\end{pmatrix}=0 \Rightarrow \lambda=\frac{1}{6}\) |
| Obtain \(\frac{1}{3}\sqrt{75}\) | A1 | \(\left(\left |
| Use correct area formula to find the area of triangle \(ABC\) | M1 | \(\left(\frac{1}{2}\times \textit{their}\sqrt{24}\times \textit{their}\frac{1}{3}\sqrt{75}\right)\) The length they use for *their* base must be found correctly. |
| Obtain answer \(5\sqrt{2}\) or \(\sqrt{50}\) | A1 | |
| Alternative method 2: Correct method to find the semi-perimeter | M1 | |
| Obtain \(4+\sqrt{6}\) | A1 | |
| Correct application of Heron's formula | M1 | \(\sqrt{(4+\sqrt{6})(1+\sqrt{6})(-1+\sqrt{6})(4-\sqrt{6})}\) |
| Obtain answer \(5\sqrt{2}\) or \(\sqrt{50}\) | A1 | |
| 4 |
## Question 6(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| State or imply $\overrightarrow{AB}$ or $\overrightarrow{AC}$ correctly in component form | B1 | $\overrightarrow{AB} = 2\mathbf{i} - 2\mathbf{j} + \mathbf{k}$, $\overrightarrow{AC} = 4\mathbf{i} - 3\mathbf{k}$ |
| Using the correct process with relevant vectors to evaluate the scalar product $\overrightarrow{AB}.\overrightarrow{AC}$ | M1 | or $\overrightarrow{BA}.\overrightarrow{CA}$ $(8-3=5)$. M0 for $\overrightarrow{AB}.\overrightarrow{CA}$. |
| Using the correct process for the moduli, divide *their* scalar product by the product of *their* moduli to obtain $\cos\theta$ or $\theta$ | M1 | $\left(\dfrac{5}{\sqrt{9}\sqrt{25}}\right)$ Independent of the first M1. |
| Obtain answer $\dfrac{1}{3}$ | A1 | ISW. Need to see a value for $\cos\theta$. Accept $\frac{5}{15}$ or $0.333$ ($\cos^{-1}\frac{1}{3}$ alone is not sufficient) |
| | **4** | |
## Question 6(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| Use correct method to find an exact value for the sine of angle $BAC$ from *their* (a) | M1 | $\left(\sqrt{1-\frac{1}{9}}\right)$ |
| Obtain answer $\frac{2}{3}\sqrt{2}$, or equivalent | A1 | |
| Use correct area formula to find the area of triangle $ABC$ with *their* versions of relevant vectors | M1 | $\left(\frac{1}{2}\sqrt{9}\sqrt{25} \times \textit{their}\sin\theta\right)$ or $\frac{1}{2}\sqrt{9}\sqrt{25}\times\sin\left(\cos^{-1}\frac{1}{3}\right)$ |
| Obtain answer $5\sqrt{2}$ or $\sqrt{50}$ | A1 | Only ISW |
| **Alternative method 1:** Use correct method to find perpendicular distance from $A$ to $BC$ (or $B$ to $AC$ or $C$ to $AB$) | M1 | $\begin{pmatrix}2+2\lambda\\-2+2\lambda\\1-4\lambda\end{pmatrix}.\begin{pmatrix}2\\2\\-4\end{pmatrix}=0 \Rightarrow \lambda=\frac{1}{6}$ |
| Obtain $\frac{1}{3}\sqrt{75}$ | A1 | $\left(\left|\frac{7}{3}\mathbf{i}-\frac{5}{3}\mathbf{j}+\frac{1}{3}\mathbf{k}\right|\right)$ |
| Use correct area formula to find the area of triangle $ABC$ | M1 | $\left(\frac{1}{2}\times \textit{their}\sqrt{24}\times \textit{their}\frac{1}{3}\sqrt{75}\right)$ The length they use for *their* base must be found correctly. |
| Obtain answer $5\sqrt{2}$ or $\sqrt{50}$ | A1 | |
| **Alternative method 2:** Correct method to find the semi-perimeter | M1 | |
| Obtain $4+\sqrt{6}$ | A1 | |
| Correct application of Heron's formula | M1 | $\sqrt{(4+\sqrt{6})(1+\sqrt{6})(-1+\sqrt{6})(4-\sqrt{6})}$ |
| Obtain answer $5\sqrt{2}$ or $\sqrt{50}$ | A1 | |
| | **4** | |6 Relative to the origin $O$, the points $A , B$ and $C$ have position vectors given by
$$\overrightarrow { O A } = \left( \begin{array} { l }
1 \\
3 \\
1
\end{array} \right) , \quad \overrightarrow { O B } = \left( \begin{array} { l }
3 \\
1 \\
2
\end{array} \right) \quad \text { and } \quad \overrightarrow { O C } = \left( \begin{array} { r }
5 \\
3 \\
- 2
\end{array} \right)$$
\begin{enumerate}[label=(\alph*)]
\item Using a scalar product, find the cosine of angle $B A C$.
\item Hence find the area of triangle $A B C$. Give your answer in a simplified exact form.
\end{enumerate}
\hfill \mbox{\textit{CAIE P3 2022 Q6 [8]}}