question_answer

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:

Let \[\vec{a}=-\,3\hat{i}+7\hat{j}+5k,\] \[\vec{b}=-\,5\hat{i}+7\hat{j}-3k\] and \[\vec{c}=7\hat{i}-5\hat{j}-3k.\] We know that, the volume of a parallelepiped whose three adjacent edges are \[\vec{a},\,\,\vec{b},\,\,\vec{c}\] is equal to \[\left| [\vec{a},\,\,\vec{b},\,\,\vec{c}] \right|.\] Here, \[\Rightarrow \]   \[[\begin{matrix}    {\vec{a}} & {\vec{b}} & {\vec{c}}  \\ \end{matrix}]=-\,3(-2\,1-15)-7(15+21)\] \[+5(25-49)\] \[\Rightarrow \]   \[[\begin{matrix}    {\vec{a}} & {\vec{b}} & {\vec{c}}  \\ \end{matrix}]=108-252-120=-\,264\] Hence, volume of the parallelepiped \[|[\begin{matrix}    {\vec{a}} & {\vec{b}} & {\vec{c}}  \\ \end{matrix}]|\,\,=\,\,|-\,264|\,\,=264\] cu units.