If $|\bar{a}|=2,|\bar{b}|=3$ and $\bar{a}, \bar{b}$ are mutually perpendicular vectors, then the area of the…

If $|\bar{a}|=2,|\bar{b}|=3$ and $\bar{a}, \bar{b}$ are mutually perpendicular vectors, then the area of the triangle whose vertices are $0, a+2 b, a-2 b$ is
  1. 6 sq.units
  2. 12 sq.units
  3. $24$ sq.units
  4. 8 sq.units

Solution

Let position vectors of $\mathrm{A}, \mathrm{B}, \mathrm{C}$ be $0, a+2 b, a-2 b$ $\begin{aligned} & \text {Area of } \triangle \mathrm{ABC}=\frac{1}{2}|\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}}| \\ & =\frac{1}{2}|(\overline{\mathrm{a}}+2 \overline{\mathrm{~b}})(\overline{\mathrm{a}}-2 \overline{\mathrm{~b}})| \\ & =\frac{1}{2}|\overline{\mathrm{a}} \times \overline{\mathrm{a}}-\overline{\mathrm{a}} \times 2 \overline{\mathrm{~b}}+2 \overline{\mathrm{~b}} \times \overline{\mathrm{a}}=2 \overline{\mathrm{~b}} \times \overline{\mathrm{b}}| \\ & =\frac{1}{2}|2 \overline{\mathrm{~b}} \times \overline{\mathrm{a}}+2 \overline{\mathrm{~b}} \times \overline{\mathrm{a}}| \\ & =\frac{1}{2} \times 4|\overline{\mathrm{~b}} \times \overline{\mathrm{a}}| \\ & =2 \times 2 \times 3 \\ & =12 \text { sq. units. } \end{aligned}$

Asked in: MHT CET 2024 (10 May Shift 1)

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