Question

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

Answer 1

The area of the parallelogram whose adjacent sides are $\vec{a}$ and $\vec{b}$ is |$\vec{a}\times\vec{b}$|
Adjacent sides are given as:
$\vec{a}=\hat{i}-\hat{j}+3\hat{k}$ and $\vec{b}=2\hat{i}-7\hat{j}+\hat{k}.$
∴$\vec{a}\times\vec{b}$ =\begin{vmatrix} \hat{ i}& \hat{j} & \hat{k}\\\ 1 & -1 & 3\\\2&-7&1 \end{vmatrix}$$=\hat{i}(-1+21)-\hat{j}(1-6)+\hat{k}(-7+2)=20\hat{i}+5\hat{j}-5\hat{k}
|$\vec{a}\times\vec{b}$|$=\sqrt{20^{2}+5^{2}+5^{2}}=\sqrt{400+25+25}=\sqrt{450}=15\sqrt{2}$
Hence,the area of the given parallelogram is 15√2square units.