Question

If three coterminous edges of a parallelopiped are represented by\[\overrightarrow{a}-\overrightarrow{b},\overrightarrow{b}-\overrightarrow{c},\overrightarrow{c}-\overrightarrow{a}\]. Then its volume is
A. \[[\overrightarrow{a}\text{ }\overrightarrow{b}\text{ }\overrightarrow{c}]\]
B. \[2[\overrightarrow{a}\text{ }\overrightarrow{b}\text{ }\overrightarrow{c}]\]
C. \[{{[\overrightarrow{a}\text{ }\overrightarrow{b}\text{ }\overrightarrow{c}]}^{2}}\]
D. \[0\]

Answer 1

Hint: In this question, we are to find the volume. Here, the dot and cross products of vectors are applied to find the required volume. The dot product is said to be a scalar product and the cross product is said to be a skew product or vector product. By using appropriate formulae, the required vector product is calculated.



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, we can use this triple product for finding the volume of the parallelopiped with the coterminous edges represented by \[\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}\].
Thus,
Volume \[V=[\overrightarrow{a}\text{ }\overrightarrow{b}\text{ }\overrightarrow{c}]\]
In vector triple product is 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:
\[\overrightarrow{a}\times \overrightarrow{a}=0\]
\[[\overrightarrow{a}\text{ }\overrightarrow{a}\text{ }\overrightarrow{b}]=[\overrightarrow{a}\text{ }\overrightarrow{b}\text{ }\overrightarrow{a}]=[\overrightarrow{b}\text{ }\overrightarrow{a}\text{ }\overrightarrow{a}]=0\]
\[\begin{align}
  & \overrightarrow{i}\cdot \overrightarrow{i}=\overrightarrow{j}\cdot \overrightarrow{j}=\overrightarrow{k}\cdot \overrightarrow{k}=1 \\
 & \overrightarrow{i}\times \overrightarrow{j}=\overrightarrow{k} \\
 & \overrightarrow{j}\times \overrightarrow{k}=\overrightarrow{i} \\
 & \overrightarrow{k}\times \overrightarrow{i}=\overrightarrow{j} \\
\end{align}\]



Complete step by step solution:It is given that,
The coterminous edges of a parallelopiped are \[\overrightarrow{a}-\overrightarrow{b},\overrightarrow{b}-\overrightarrow{c},\overrightarrow{c}-\overrightarrow{a}\]
Then, its volume is calculated by,
\[V=[\overrightarrow{a}\text{ }\overrightarrow{b}\text{ }\overrightarrow{c}]\]
On substituting the given edges, we get
\[V=[\overrightarrow{a}-\overrightarrow{b}\text{ }\overrightarrow{b}-\overrightarrow{c}\text{ }\overrightarrow{c}-\overrightarrow{a}]\]
\[\begin{align}
  & \Rightarrow V=(\overrightarrow{a}\times \overrightarrow{b})\cdot \overrightarrow{c} \\
 & \text{ }=(\overrightarrow{a}-\overrightarrow{b})\times (\overrightarrow{b}-\overrightarrow{c})\cdot (\overrightarrow{c}-\overrightarrow{a}) \\
 & \text{ }=(\overrightarrow{a}\times \overrightarrow{b}-\overrightarrow{a}\times \overrightarrow{c}-\overrightarrow{b}\times \overrightarrow{b}+\overrightarrow{b}\times \overrightarrow{c})\cdot (\overrightarrow{c}-\overrightarrow{a}) \\
 & \text{ }=(\overrightarrow{a}\times \overrightarrow{b}-\overrightarrow{a}\times \overrightarrow{c}-0+\overrightarrow{b}\times \overrightarrow{c})\cdot (\overrightarrow{c}-\overrightarrow{a}) \\
\end{align}\]
\[\begin{align}
  & =(\overrightarrow{a}\times \overrightarrow{b}-\overrightarrow{a}\times \overrightarrow{c}+\overrightarrow{b}\times \overrightarrow{c})\cdot (\overrightarrow{c}-\overrightarrow{a}) \\
 & =(\overrightarrow{a}\times \overrightarrow{b})\cdot \overrightarrow{c}-(\overrightarrow{a}\times \overrightarrow{c})\cdot \overrightarrow{c}+(\overrightarrow{b}\times \overrightarrow{c})\cdot \overrightarrow{c}-(\overrightarrow{a}\times \overrightarrow{b})\cdot \overrightarrow{a}+(\overrightarrow{a}\times \overrightarrow{c})\cdot \overrightarrow{a}-(\overrightarrow{b}\times \overrightarrow{c})\cdot \overrightarrow{a} \\
 & =[\overrightarrow{a}\text{ }\overrightarrow{b}\text{ }\overrightarrow{c}]-[\overrightarrow{a}\text{ }\overrightarrow{c}\text{ }\overrightarrow{c}]+[\overrightarrow{b}\text{ }\overrightarrow{c}\text{ }\overrightarrow{c}]-[\overrightarrow{a}\text{ }\overrightarrow{b}\text{ }\overrightarrow{a}]+[\overrightarrow{a}\text{ }\overrightarrow{c}\text{ }\overrightarrow{a}]-[\overrightarrow{b}\text{ }\overrightarrow{c}\text{ }\overrightarrow{a}] \\
 & =[\overrightarrow{a}\text{ }\overrightarrow{b}\text{ }\overrightarrow{c}]-[\overrightarrow{b}\text{ }\overrightarrow{c}\text{ }\overrightarrow{a}] \\
\end{align}\]
\[\begin{align}
  & =[\overrightarrow{a}\text{ }\overrightarrow{b}\text{ }\overrightarrow{c}]-[\overrightarrow{a}\text{ }\overrightarrow{b}\text{ }\overrightarrow{c}] \\
 & =0 \\
\end{align}\]



Option ‘D’ is correct

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