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Q.
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
Vector Algebra
Solution:
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 =\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)$
$=20 \hat{i}+5 \hat{j}-5 \hat{k}$
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.