Question

If $\left|a\right|=2,\left|b\right|=3$ and $a,b$ are mutually perpendicular, then the area of the triangle whose vertices are$0,a+b,a–b$is?

• A. $5$
• B. $1$
• C. $6$
• D. $8$

Answer 1

The correct option is C

$6$

Find the area of the triangle whose vertices are$\mathbf{}\mathbf{0}\mathbf{,}\mathbf{}\mathit{a}\mathbf{}\mathbf{+}\mathbf{}\mathit{b}\mathbf{,}\mathbf{}\mathit{a}\mathbf{}\mathbf{–}\mathbf{}\mathit{b}$:

Suppose,$A,B$ and $C$are the position vectors of vertices $0,a+b$and $a-b$.

Area of triangle ,

$$\begin{gathered} =\frac{1}{2}\left|\right(a+b)x(a-b\left)\right| \\ =\frac{1}{2}|2bxa| \\ =\left|b\right|\left|a\right|\sin \theta \\ =3\times 2\sin {90}^0 \\ =6\mathbf{}\mathbf{(}\mathbf{\because }\mathit{s}\mathit{i}\mathit{n}{\mathbf{90}}^{\mathbf{0}}\mathbf{=}\mathbf{1}\mathbf{)} \end{gathered}$$$\mathbf{}\mathit{f}\mathit{o}\mathit{r}\mathit{m}\mathit{u}\mathit{l}\mathit{a}=\mathbf{}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{|}\mathit{A}\mathit{B}\mathbf{}\mathit{x}\mathbf{}\mathit{A}\mathit{C}\mathbf{|}$

Hence, the correct option is (C).