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If $|\mathbf{a}|=2,|\mathbf{b}|=3$ and $\mathbf{a}, \mathbf{b}$ are mutually perpendicular, then the area of the triangle whose vertices are $\mathbf{0}, \mathbf{a}+\mathbf{b}, \mathbf{a}-\mathbf{b}$ is
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The correct answer is:
6
Let the position vectors of the points $A, B, C$ are $\mathbf{0}, \mathbf{a + b}, \mathbf{a}-\mathbf{b}$ and $\theta=90^{\circ}$.
$\begin{aligned}
\text { Area of triangle } & =\frac{1}{2}|\overrightarrow{A B} \times \overrightarrow{A C}| \\
& =\frac{1}{2}|(\mathbf{a}+\mathbf{b}) \times(\mathbf{a}-\mathbf{b})|=\frac{1}{2}|2 \mathbf{b} \times \mathbf{a}| \\
& =b a \sin \theta=3 \times 2 \sin 90^{\circ}=6 .
\end{aligned}$
$\begin{aligned}
\text { Area of triangle } & =\frac{1}{2}|\overrightarrow{A B} \times \overrightarrow{A C}| \\
& =\frac{1}{2}|(\mathbf{a}+\mathbf{b}) \times(\mathbf{a}-\mathbf{b})|=\frac{1}{2}|2 \mathbf{b} \times \mathbf{a}| \\
& =b a \sin \theta=3 \times 2 \sin 90^{\circ}=6 .
\end{aligned}$
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