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Question:
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Show that the area of the triangle contained between the vectors $\vec{a}$ and $\vec{b}$ is one half of the magnitude of $\vec{a} \times \vec{b}$.
Solution:
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Verified Answer
Let $\overrightarrow{O A}=\vec{a}$ and $\overrightarrow{O B}=\vec{b}$ and $\angle A O B=\theta$. Complete the parallelogram $O A C B$. Join $B A$. Draw $B N \perp O A$.
In $\triangle O B N$,
$$
\begin{aligned}
&\sin \theta=\frac{B N}{O B}=\frac{B N}{b} \quad \therefore B N=b \sin \theta \\
&|\vec{a} \times \vec{b}|=a b \sin \theta=O A \times B N=\frac{2(O A \times B N)}{2} \\
&=2(\text { area of } \triangle O A B)
\end{aligned}
$$
$\therefore$ Area of $\triangle O A B=\frac{1}{2}|\vec{a} \times \vec{b}|$
In $\triangle O B N$,
$$
\begin{aligned}
&\sin \theta=\frac{B N}{O B}=\frac{B N}{b} \quad \therefore B N=b \sin \theta \\
&|\vec{a} \times \vec{b}|=a b \sin \theta=O A \times B N=\frac{2(O A \times B N)}{2} \\
&=2(\text { area of } \triangle O A B)
\end{aligned}
$$
$\therefore$ Area of $\triangle O A B=\frac{1}{2}|\vec{a} \times \vec{b}|$
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