Question

Find the volume of a parallelopiped whose sides are given by $-3\hat{i}+7\hat{j}+5\hat{k},-5\hat{i}+7\hat{j}-3\hat{k}$ and $7\hat{i}-5\hat{j}-3\hat{k}$.

Answer 1

Let $\overrightarrow{a}=-3\hat{i}+7\hat{j}+5\hat{k},\overrightarrow{b}=-5\hat{i}+7\hat{j}-3\hat{k}$ and $\overrightarrow{c}=7\hat{i}-5\hat{j}-3\hat{k}$.

We know, the volume of a parallelopiped whose three adjacent edges are $\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}$ is equal to $\left[\begin{array}{ccc}\overrightarrow{a}& \overrightarrow{b}& \overrightarrow{c}\end{array}\right]$.

Here, $\left[\begin{array}{ccc}\overrightarrow{a}& \overrightarrow{b}& \overrightarrow{c}\end{array}\right]=\left|\begin{array}{ccc}-3& 7& 5\\ -5& 7& -3\\ 7& -5& -3\end{array}\right|$

$=|-3(-21-15)-7(15+21)+5(25-49)|$

$=|108-252-120|=|-264|=264$

Hence, volume of the parallelopiped = $264cu.units$