Q99 / 150 · MHT CET 2008 · Mathematics

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

  • A
    5
  • B
    1
  • Correct answer
    C
    6
  • D
    8

Solution

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

Hence, the correct answer is option: 'C'.

Relevant Questions