31.

If $\left|\overrightarrow{\mathrm{a}}\right| = 2, \left|\overrightarrow{\mathrm{b}}\right| = 3$ and $\overrightarrow{\mathrm{a}}, \overrightarrow{\mathrm{b}}$ are mutually perpendicular, then the area of the triangle whose vertices are $\overrightarrow{0}, \overrightarrow{\mathrm{a}} + \overrightarrow{\mathrm{b}}, \overrightarrow{\mathrm{a}} - \overrightarrow{\mathrm{b}}$ is

• 5

• 1

• 6

• 8

$$\begin{gathered} \mathrm{Let} \mathrm{the} \mathrm{position} \mathrm{vectors} \mathrm{of} \mathrm{the} \mathrm{points} \mathrm{A}, \mathrm{B}, \mathrm{C} \mathrm{are} \overrightarrow{0}, \overrightarrow{\mathrm{a}} + \overrightarrow{\mathrm{b}}, \overrightarrow{\mathrm{a}} - \overrightarrow{\mathrm{b}} \mathrm{and} \mathrm{\theta } = 90° \\ \therefore \mathrm{Area} \mathrm{of} \mathrm{triangle} = \frac{1}{2}\left|\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}}\right| \\ = \frac{1}{2}\left|\left(\overrightarrow{\mathrm{a}} + \overrightarrow{\mathrm{b}}\right) \times \left(\overrightarrow{\mathrm{a}} - \overrightarrow{\mathrm{b}}\right)\right| \\ = \frac{1}{2}\left|2\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{a}}\right| \\ = \left|\overrightarrow{\mathrm{b}}\right|\left|\overrightarrow{\mathrm{a}}\right|\sin \left(\mathrm{\theta }\right) \\ = 3 \times 2 \sin \left(90°\right) \\ = 6 \end{gathered}$$