Question

Let $\overrightarrow{a}=\hat{i}+\hat{j}-\hat{k}$ and $\overrightarrow{b}=\hat{i}-\hat{j}+\hat{k}.$
Then area of the triangle whose adjacent sides are $\overrightarrow{a}$ and $\overrightarrow{b}$ is

[2 marks]

• A. $2\sqrt{2}$ sq. units
• B. $\sqrt{2}$ sq. units
• C. $2$ sq. units
• D. $\frac{1}{\sqrt{2}}$ sq. units

Answer 1

The correct option is B $\sqrt{2}$ sq. units

Given : $\overrightarrow{a}=\hat{i}+\hat{j}-\hat{k}$
and $\overrightarrow{b}=\hat{i}-\hat{j}+\hat{k}$
$\overrightarrow{a}\times \overrightarrow{b}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& 1& -1\\ 1& -1& 1\end{array}\right|$
$\Rightarrow \overrightarrow{a}\times \overrightarrow{b}=\hat{i}(1-1)-\hat{j}(1+1)+\hat{k}(-1-1)$
$\Rightarrow \overrightarrow{a}\times \overrightarrow{b}=-2\hat{j}-2\hat{k}$
$\Rightarrow |\overrightarrow{a}\times \overrightarrow{b}|=\sqrt{(-2{)}^2+(-2{)}^2}=2\sqrt{2}$
Area of the triangle whose adjacent sides are $\overrightarrow{a}$ and $\overrightarrow{b}$ is
$\frac{1}{2}|\overrightarrow{a}\times \overrightarrow{b}|=\frac{1}{2}\times 2\sqrt{2}$
$\therefore$ Area of triangle $=\sqrt{2}$ sq. units