Question

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

• A. $15\sqrt{3}$
• B. $15\sqrt{2}$
• C. $31\sqrt{2}$
• D. $34\sqrt{3}$

Answer 1

Answer: (B)

The area of the parallelogram whose adjacent sides are a and $b$, is $|a\times b|$.
Adjacent sides are given as $a=\hat{i}-\hat{j}+3\hat{k}$ and
$b=-2\hat{i}-7\hat{j}+\hat{k}$
$\therefore a\times b=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& -1& 3\\ 2& -7& 1\end{array}\right|$
$=\hat{i}(-1+21)-\hat{j}(1-6)+\hat{k}(-7+2)$
$=20\hat{i}+5\hat{j}-5\hat{k}$
On comparing with $X=x\hat{i}+y\hat{j}+z\hat{k}$, 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 units
Hence, the area of the given parallelogram is $15\sqrt{2}$ sq units.