Question

Find the area of the parallelogram whose adjacent sides are determined by the vectors $\overrightarrow{a}=\hat{i}-\hat{j}+3\hat{k}\text{and}\overrightarrow{b}=2\hat{i}-7\hat{j}+\hat{k}$.

Answer 1

Given $\overrightarrow{a}=\hat{i}-\hat{j}+3\hat{k}\text{and}\overrightarrow{b}=2\hat{i}-7\hat{j}+\hat{k}$
$\text{Area}=\frac{1}{2}|\overline{a}\times \overline{b}|$
$\overline{a}\times \overline{b}=$$\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& -1& 3\\ 2& -7& 1\end{array}\right|$
$=20\overline{i}+5\overline{j}-5\overline{k}$
$\Rightarrow |\overline{a}\times \overline{b}|=\sqrt{(20{)}^2+(5{)}^2+(-5{)}^2}$
$=\sqrt{400+25+25}$
$=\sqrt{450}$
$\therefore |\overline{a}\times \overline{b}|=15\sqrt{2}$
$\text{Area}=\frac{1}{2}\left(15\sqrt{2}\right)=\frac{15\sqrt{2}}{2}$square units,