Question

Volume of the parallelepiped whose coterminous edges are \[2\widehat{i}-3\widehat{j}+4\widehat{k}\], \[\widehat{i}+2\widehat{j}-2\widehat{k}\] and \[3\widehat{i}-\widehat{j}+\widehat{k}\] is
A. 5 cubic units
B. 6 cubic units
C. 7 cubic units
D. 8 cubic units

Answer 1

Hint: In the given question, we are to find the volume of the parallelepiped whose edges from the same corner are given in the form of vector equations. We can calculate the volume of the parallelepiped using the concept of determinant calculation using the matrix method.

Formula used: Scalar triple product of three vectors:
We have the vectors \[\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}\] as
\[\begin{align}
  & \overrightarrow{a}={{a}_{1}}\overrightarrow{i}+{{a}_{2}}\overrightarrow{j}+{{a}_{3}}\overrightarrow{k} \\
 & \overrightarrow{b}={{b}_{1}}\overrightarrow{i}+{{b}_{2}}\overrightarrow{j}+{{b}_{3}}\overrightarrow{k} \\
 & \overrightarrow{c}={{c}_{1}}\overrightarrow{i}+{{c}_{2}}\overrightarrow{j}+{{c}_{3}}\overrightarrow{k} \\
\end{align}\]
Then, the triple product is calculated by,
\[[\overrightarrow{a}\text{ }\overrightarrow{b}\text{ }\overrightarrow{c}]=\left| \begin{matrix}
   {{a}_{1}} & {{a}_{2}} & {{a}_{3}} \\
   {{b}_{1}} & {{b}_{2}} & {{b}_{3}} \\
   {{c}_{1}} & {{c}_{2}} & {{c}_{3}} \\
\end{matrix} \right|\]
Thus, the volume of a 3D structure is calculated by $V=[\overrightarrow{a}\text{ }\overrightarrow{b}\text{ }\overrightarrow{c}]$

Complete step by step solution: Here in the above question, we are given three vectors that represent the coterminous edges of the parallelepiped form. The three edges are coterminous which literally means they are passing through the same point on the plane. They are:
\[\begin{align}
  & \overrightarrow{a}=2\widehat{i}-3\widehat{j}+4\widehat{k} \\
 & \overrightarrow{b}=\widehat{i}+2\widehat{j}-2\widehat{k} \\
 & \overrightarrow{c}=3\widehat{i}-\widehat{j}+\widehat{k} \\
\end{align}\]
Then, the required volume of the parallelepiped is
$\begin{align}
  & V=[\overrightarrow{a}\text{ }\overrightarrow{b}\text{ }\overrightarrow{c}] \\
 & \Rightarrow V=[\overrightarrow{a}\text{ }\overrightarrow{b}\text{ }\overrightarrow{c}]=\left| \begin{matrix}
   {{a}_{1}} & {{a}_{2}} & {{a}_{3}} \\
   {{b}_{1}} & {{b}_{2}} & {{b}_{3}} \\
   {{c}_{1}} & {{c}_{2}} & {{c}_{3}} \\
\end{matrix} \right| \\
\end{align}$
On substituting,
\[\begin{align}
  & V=\left| \begin{matrix}
   2 & -3 & 4 \\
   1 & 2 & -2 \\
   3 & -1 & 1 \\
\end{matrix} \right| \\
 & \text{ }=2(2-2)+3(1+6)+4(-1-6) \\
 & \text{ }=0+21-28 \\
 & \text{ }=\left| -7 \right| \\
 & \text{ }=7 \\
\end{align}\]

Thus, Option (C) is correct.

Note: Here we may go wrong with the vector identities and scalar triple product. Here are the simple formulas used for solving the given vector. By applying appropriate vector products, the given vector equation is evaluated.