Question

if $\overrightarrow{a}=2\hat{i}-3\hat{j}+\hat{k}$, $\overrightarrow{b}=\hat{i}+\hat{k}$, $\overrightarrow{c}=2\hat{j}-\hat{k}$ are three vectors, the area $A$ of the parallelogram having diagonals $(\overrightarrow{a}+\overrightarrow{b})$ and $(\overrightarrow{b}+\overrightarrow{c})$
find $4A^2$

Answer 1

$\overrightarrow{a}=2\hat{i}-3\hat{j}+\hat{k}$, $\overrightarrow{b}=\hat{i}+\hat{k}$, $\overrightarrow{c}=2\hat{j}-\hat{k}$ are three vectors
$\overrightarrow{a}+\overrightarrow{b}=3\hat{i}-3\hat{j}+2\hat{k}$
$\overrightarrow{b}+\overrightarrow{c}=\hat{i}+2\hat{j}$
If $\overrightarrow{d_1},\overrightarrow{d_2}$ are two diagonals of a parallelogram
then area is $\frac{1}{2}|\overrightarrow{d_1}\times \overrightarrow{d_2}|$
since $\overrightarrow{a}+\overrightarrow{b},\overrightarrow{b}+\overrightarrow{c}$ are diagonals
$\frac{1}{2}\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 3& -3& 2\\ 1& 2& 0\end{array}\right|=\frac{1}{2}\left|(-4\hat{i}+2\hat{j}+9\hat{k})\right|$
$\therefore$ Area = $\frac{1}{2}\sqrt{101}$