If \(|\overrightarrow{\mathbf{a}}|=2,|\overrightarrow{\mathbf{b}}|=3\) and \(\overrightarrow{\mathbf{a}}, \overrightarrow{\mathbf{b}}\) are mutually
perpendicular, then the area of the triangle whose vertices are \(\overrightarrow{\mathbf{0}}, \overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}}, \overrightarrow{\mathbf{a}}-\overrightarrow{\mathbf{b}}\) is

Let the position vectors of the points \(A, B, C\) are
\(\mathbf{0}, \mathbf{a}+\mathbf{b}, \mathbf{a}-\mathbf{b} \text { and } \theta=90^{\circ} . \)
\( \therefore \text {Area of triangle }=\frac{1}{2}|\overrightarrow{\mathbf{A B}} \times \overrightarrow{\mathbf{A C}}| \)
\( \left.=\frac{1}{2} \mid \overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}}\right) \times(\overrightarrow{\mathbf{a}}-\overrightarrow{\mathbf{b}}) \mid \)
\( =\frac{1}{2}|2 \overrightarrow{\mathbf{b}} \times \overrightarrow{\mathbf{a}}| \)
\( =|\overrightarrow{\mathbf{b}}||\overrightarrow{\mathbf{a}}| \sin \theta \)
\( =3 \times 2 \sin 90^{\circ} \)
\( =6\)