Question

Three concurrent edges \[OA\], \[OB\], and \[OC\] of a parallelepiped are represented by three vectors \[2\widehat{i}+\widehat{j}-\widehat{k}\], \[\widehat{i}+2\widehat{j}+3\widehat{k}\], and \[-3\widehat{i}-\widehat{j}+\widehat{k}\], the volume of the solid so formed in the cubic unit is
A. $5$
B. $6$
C. $7$
D. $8$

Answer 1

Hint: In the above question, we are to find the volume of the solid formed by the concurrent edges of a parallelepiped which are represented by three vectors. In order to get the required volume of the given solid, we have to calculate the determinant formed by the coefficients of the direction cosines of the three vectors.

Formula used: The dot product of two vectors is
$\overrightarrow{a}\cdot \overrightarrow{b}=\left| \overrightarrow{a} \right|\left| \overrightarrow{b} \right|\cos (\overrightarrow{a},\overrightarrow{b})$
The cross product of two vectors is
$\overrightarrow{a}\times \overrightarrow{b}=\left| \overrightarrow{a} \right|\left| \overrightarrow{b} \right|\sin (\overrightarrow{a},\overrightarrow{b})\overrightarrow{n}$
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: We are given a parallelepiped that has three concurrent edges named \[OA\], \[OB\], and \[OC\]. They are represented by three vectors which can be given as follows:
\[\overrightarrow{a}=2\widehat{i}+\widehat{j}-\widehat{k}\]
\[\overrightarrow{b}=\widehat{i}+2\widehat{j}+3\widehat{k}\]
\[\overrightarrow{c}=-3\widehat{i}-\widehat{j}+\widehat{k}\]
Then, the volume of the given parallelepiped is $V=[\overrightarrow{a}\text{ }\overrightarrow{b}\text{ }\overrightarrow{c}]$
$\begin{align}
  & V=\left| \begin{matrix}
   {{a}_{1}} & {{a}_{2}} & {{a}_{3}} \\
   {{b}_{1}} & {{b}_{2}} & {{b}_{3}} \\
   {{c}_{1}} & {{c}_{2}} & {{c}_{3}} \\
\end{matrix} \right| \\
 & \text{ }=\left| \begin{matrix}
   2 & 1 & -1 \\
   1 & 2 & 3 \\
   -3 & -1 & 1 \\
\end{matrix} \right| \\
 & \text{ }=\left| 2(2+3)-1(1+9)-1(-1+6) \right| \\
 & \text{ }=\left| 10-10-5 \right| \\
 & \text{ }=5 \\
\end{align}$

Thus, Option (A) is the correct value.

Additional Information: In vector triple products are cross and dot products are interchangeable. i.e.,
\[\begin{align}
  & [\overrightarrow{a}\text{ }\overrightarrow{b}\text{ }\overrightarrow{c}]=\overrightarrow{a}\cdot \overrightarrow{b}\times \overrightarrow{c}=\overrightarrow{a}\times \overrightarrow{b}\cdot \overrightarrow{c}=\overrightarrow{b}\times \overrightarrow{c}\cdot \overrightarrow{a}=\overrightarrow{c}\times \overrightarrow{a}\cdot \overrightarrow{b} \\
 & [\overrightarrow{a}\text{ }\overrightarrow{b}\text{ }\overrightarrow{c}]=[\overrightarrow{b}\text{ }\overrightarrow{c}\text{ }\overrightarrow{a}]=[\overrightarrow{c}\text{ }\overrightarrow{a}\text{ }\overrightarrow{b}] \\
 & [\overrightarrow{a}\text{ }\overrightarrow{b}\text{ }\overrightarrow{c}]=-[\overrightarrow{b}\text{ }\overrightarrow{a}\text{ }\overrightarrow{c}]=-[\overrightarrow{c}\text{ }\overrightarrow{b}\text{ }\overrightarrow{a}]=-[\overrightarrow{a}\text{ }\overrightarrow{c}\text{ }\overrightarrow{b}] \\
\end{align}\]
Important vector identities for solving vector equations are:
\[\overrightarrow{a}\times \overrightarrow{a}=0\]
\[[\overrightarrow{a}\text{ }\overrightarrow{a}\text{ }\overrightarrow{b}]=[\overrightarrow{a}\text{ }\overrightarrow{b}\text{ }\overrightarrow{a}]=[\overrightarrow{b}\text{ }\overrightarrow{a}\text{ }\overrightarrow{a}]=0\]
\[\begin{align}
  & \overrightarrow{i}\cdot \overrightarrow{i}=\overrightarrow{j}\cdot \overrightarrow{j}=\overrightarrow{k}\cdot \overrightarrow{k}=1 \\
 & \overrightarrow{i}\times \overrightarrow{j}=\overrightarrow{k} \\
 & \overrightarrow{j}\times \overrightarrow{k}=\overrightarrow{i} \\
 & \overrightarrow{k}\times \overrightarrow{i}=\overrightarrow{j} \\
\end{align}\]

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.