Question

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

Answer 1

The area of the parallelogram whose adjacent sides are a and b is $|a\times b|$.
Adjacent sides are given as $a=\hat{i}-\hat{j}+3\hat{k}andb=2\hat{i}-7\hat{j}+\hat{k}\therefore a\times b=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& -1& 3\\ 2& -7& 1\end{array}\right|=\hat{i}(-1+21)-\hat{j}(1-6)+\hat{k}(-7+2)=20\hat{i}+5\hat{j}-5\hat{k}$
Comparing with $X=x\hat{i}+y\hat{j}+z\hat{k},\text{we get}x=20,y=5,z=-5$
$\therefore$ Area of the parallelogram $=|A\times B|$
$\Rightarrow |a\times b|=\sqrt{x^2+y^2+z^2}=\sqrt{(20{)}^2+5^2+(-5{)}^2}=\sqrt{450}=\sqrt{225\times 2}=15\sqrt{2}sq.unit$
Hence, the area of the given parallelogram is $15\sqrt{2}sq.unit$.