Question

Find the area of the parallelogram whose adjacent sides are $\overline{a}=2\hat{i}-2\hat{j}+\hat{k}$ and $\overline{b}=\hat{i}-3\hat{j}-3\hat{k}$

Answer 1

Given: $\overrightarrow{a}=2\hat{i}-2\hat{j}+\hat{k}$ and $\overrightarrow{b}=\hat{i}-3\hat{j}-3\hat{k}$
$\therefore \overrightarrow{a}\times \overrightarrow{b}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& -2& 1\\ 1& -3& -3\end{array}\right|$
$=(6+3)\hat{i}-(-6-1)\hat{j}+(-6+2)\hat{k}$
$=9\hat{i}+7\hat{j}-4\hat{k}$
$|\overrightarrow{a}\times \overrightarrow{b}|=\sqrt{9^2+7^2+(-4{)}^2}=\sqrt{81+49+16}=\sqrt{146}$
Area of the parallelogram whose adjacent sides are $\overrightarrow{a}$ and $\overrightarrow{b}=\sqrt{146}squnits$.