Question

The vectors from origin to the points $A$ and $B$ are $\overrightarrow{A}=3\hat{i}-6\hat{j}+2\hat{k}$ and $\overrightarrow{B}=2\hat{i}+\hat{j}-2\hat{k}$ respectively. The area of the triangle $OAB$ be

• A. $\frac{5}{2}\sqrt{17}sq.unit$
• B. $\frac{2}{5}\sqrt{17}sq.unit$
• C. $\frac{3}{5}\sqrt{17}sq.unit$
• D. $\frac{5}{3}\sqrt{17}sq.unit$

Answer 1

The correct option is A $\frac{5}{2}\sqrt{17}sq.unit$

Given $\overrightarrow{OA}=\overrightarrow{a}=3\hat{i}-6\hat{j}+2\hat{k}$ and $\overrightarrow{OB}=\overrightarrow{b}=2\hat{i}+\hat{j}-2\hat{k}$
$\therefore (\overrightarrow{a}\times \overrightarrow{b})=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 3& -6& 2\\ 2& 1& -2\end{array}\right|$
$=(12-2)\hat{i}+(4+6)\hat{j}+(3+12)\hat{k}$
$10\hat{i}+10\hat{j}+15\hat{k}\Rightarrow |\overrightarrow{a}\times \overrightarrow{b}|=\sqrt{{10}^2+{10}^2+{15}^2}$
$=\sqrt{425}=5\sqrt{17}$
Area of $\Delta OAB=\frac{1}{2}|\overrightarrow{a}\times \overrightarrow{b}|=\frac{5\sqrt{17}}{2}sq.unit.$