Q14 (c) (i)

Criteria	Marks
Provides correct solution	3
Recognise that the dot product of $\overrightarrow{OB}$ with the vector perpendicular to $\overrightarrow{OA}$, using the given information, is related to the area, or vice versa, or equivalent merit	2
Attempts to use a vector perpendicular to $\overrightarrow{OA}$ or $\overrightarrow{OB}$, or equivalent merit	1

Sample answer:

Let $\theta$ be the angle between $\overrightarrow{OA}$ and $\overrightarrow{OB}$.

Area of triangle $=\frac12ab\sin\theta$, where $a=|\overrightarrow{OA}|$ and $b=|\overrightarrow{OB}|$.

$$\overrightarrow{OC}=\begin{pmatrix}a_2\\-a_1\end{pmatrix}$$

We are told that $\overrightarrow{OC}$ is perpendicular to $\overrightarrow{OA}$ and with the same magnitude.

$$\overrightarrow{OB}\cdot\overrightarrow{OC}=\begin{pmatrix}b_1\\b_2\end{pmatrix}\cdot\begin{pmatrix}a_2\\-a_1\end{pmatrix}=a_2b_1-a_1b_2\tag{1}$$

We can also express $\overrightarrow{OB}\cdot\overrightarrow{OC}$ in terms of $\theta$:

$$\overrightarrow{OB}\cdot\overrightarrow{OC}=|\overrightarrow{OB}|\,|\overrightarrow{OC}|\cos(\angle BOC)=b\times a\times\cos(\angle BOC)$$

Case 1: $\angle BOC=\frac\pi2-\theta$, so $\overrightarrow{OB}\cdot\overrightarrow{OC}=ab\sin\theta$.

Case 2: $\angle BOC=\frac\pi2+\theta$, so $\overrightarrow{OB}\cdot\overrightarrow{OC}=-ab\sin\theta$.

In both cases,

$$\left|\overrightarrow{OB}\cdot\overrightarrow{OC}\right|=|ab\sin\theta|=2\times(\text{Area of triangle})$$

Therefore,

$$\text{area of triangle}=\frac12\left|\overrightarrow{OB}\cdot\overrightarrow{OC}\right|=\frac12|a_2b_1-a_1b_2|=\frac12|a_1b_2-a_2b_1|$$