Question

Find the area of parallelogram whose diagonals are represented by the vectors $(2\hat{i}-3\hat{j}+4\hat{k})$ and $(2\hat{i}-\hat{j}+2\hat{k})$.

Answer 1

Area of parallelogram = $\frac{1}{2}|({\overrightarrow{d}}_1\times {\overrightarrow{d}}_2)|$

Here,

${\overrightarrow{d}}_1=2\hat{i}-3\hat{j}+4\hat{k}$

${\overrightarrow{d}}_2=2\hat{i}-\hat{j}+2\hat{k}$

${\overrightarrow{d}}_1\times {\overrightarrow{d}}_2=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& -3& 4\\ 2& -1& 2\end{array}\right|$

$=\hat{i}(-6+4)+\hat{j}(4-8)+\hat{k}(-2+6)$

$=-2\hat{i}-4\hat{j}+4\hat{k}$

Now,

$|{\overrightarrow{d}}_1\times {\overrightarrow{d}}_2|$ = $\sqrt{(-2{)}^2+(-4{)}^2+(4{)}^2}$ = $\sqrt{4+16+16}$ = $\sqrt{36}=6$

Therefore, area of parallelogram = $\frac{1}{2}\times 6=3$ sq. units.