Question

Find the area of the parallelogram determined by the vectors $\hat{i}+2\hat{j}+3\hat{k}$ and $3\hat{i}-2\hat{j}+\hat{k}$.

Answer 1

Let $\overrightarrow{a}=\hat{i}+2\hat{j}+3\hat{k}$ and $\overrightarrow{b}=3\hat{i}-2\hat{j}+\hat{k}$

Therefore,

$\overrightarrow{a}\times \overrightarrow{b}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& 2& 3\\ 3& -2& 1\end{array}\right|$

$\overrightarrow{a}\times \overrightarrow{b}=8\hat{i}+8\hat{j}-8\hat{k}$

$Areaofparallelogram=|\overrightarrow{a}\times \overrightarrow{b}|=\sqrt{8^2+8^2+(-8{)}^2}$

$Areaofparallelogram=8\sqrt{3}sq.units$