Question

If the volume of a Tetrahedron formed by coterminous edges $\vec{a},\vec{b},\vec{c}$ is 2,then the volume of parallelepiped formed by coterminous edges $\vec{a}\times \vec{b},\vec{b}\times \vec{c},\vec{c}\times \vec{a}$ is
(a) 72
(b) 144
(c) 36
(d) 108

Answer 1

Hint: First, we will draw the figure of a tetrahedron. Then we should know the formula of volume of tetrahedron formed by coterminous edges is given as $\dfrac{1}{6}\left[ \begin{matrix}
   {\vec{a}} & {\vec{b}} & {\vec{c}} \\
\end{matrix} \right]$ . Using this, we will get value of $\left[ \begin{matrix}
   {\vec{a}} & {\vec{b}} & {\vec{c}} \\
\end{matrix} \right]$ . Then for finding value of $\vec{a}\times \vec{b},\vec{b}\times \vec{c},\vec{c}\times \vec{a}$ , formula to be used is given as ${{\left[ \begin{matrix}
   {\vec{a}} & {\vec{b}} & {\vec{c}} \\
\end{matrix} \right]}^{2}}$ . Thus, on substituting the values and solving we will get the answer. The general formula of volume of parallelepiped having coterminous edges $\vec{a},\vec{b},\vec{c}$ is given as $\left| \left( \vec{a}\times \vec{b} \right)\cdot \vec{c} \right|$ .

Complete step-by-step answer:
Here, we will draw a tetrahedron figure.

We should know that volume of Tetrahedron formed by coterminous edges $\vec{a},\vec{b},\vec{c}$ is given as $\dfrac{1}{6}\left[ \begin{matrix}
   {\vec{a}} & {\vec{b}} & {\vec{c}} \\
\end{matrix} \right]$ .
We are also given that this volume is equal to 2. So, we can write it as
$\dfrac{1}{6}\left[ \begin{matrix}
   {\vec{a}} & {\vec{b}} & {\vec{c}} \\
\end{matrix} \right]=2$
On solving this, we get as
$\left[ \begin{matrix}
   {\vec{a}} & {\vec{b}} & {\vec{c}} \\
\end{matrix} \right]=2\times 6=12$ ………………….(1)
Now, we have to find value of parallelepiped formed by coterminous edges $\vec{a}\times \vec{b},\vec{b}\times \vec{c},\vec{c}\times \vec{a}$ . So, we can write this as by replacing $\vec{a}$ as $\vec{a}\times \vec{b}$ , $\vec{b}$ as $\vec{b}\times \vec{c}$ , $\vec{c}$ as $\vec{c}\times \vec{a}$ in the general formula $\left| \left( \vec{a}\times \vec{b} \right)\cdot \vec{c} \right|$ which is basically written as $\left[ \begin{matrix}
   {\vec{a}} & {\vec{b}} & {\vec{c}} \\
\end{matrix} \right]$ . So, after replacing the values we will get as
$\left[ \begin{matrix}
   \vec{a}\times \vec{b} & \vec{b}\times \vec{c} & \vec{c}\times \vec{a} \\
\end{matrix} \right]$
So, we can write it as $\left( \vec{a}\times \vec{b} \right).\left( \left( \vec{b}\times \vec{c} \right)\times \left( \vec{c}\times \vec{a} \right) \right)$ which is known as quadrupole product of four product.
We have formula i.e. $\left( \vec{a}\times \vec{b} \right)\times \left( \vec{c}\times \vec{d} \right)=\left[ \begin{matrix}
   a & b & d \\
\end{matrix} \right]c-\left[ \begin{matrix}
   a & b & c \\
\end{matrix} \right]d$ . So, on applying this we get as
$\left( \vec{a}\times \vec{b} \right).\left( \left( \vec{b}\times \vec{c} \right)\times \left( \vec{c}\times \vec{a} \right) \right)=\left( \vec{a}\times \vec{b} \right)\cdot \left( \left[ \begin{matrix}
   b & c & a \\
\end{matrix} \right]c-\left[ \begin{matrix}
   b & c & c \\
\end{matrix} \right]a \right)$
We also know that if two out of three vectors are the same then that is equal to zero. So, we can write it as
$\left( \vec{a}\times \vec{b} \right).\left( \left( \vec{b}\times \vec{c} \right)\times \left( \vec{c}\times \vec{a} \right) \right)=\left( \vec{a}\times \vec{b} \right)\cdot \left( \left[ \begin{matrix}
   b & c & a \\
\end{matrix} \right]c \right)=\left( \left( \vec{a}\times \vec{b} \right)\cdot c \right)\left[ \begin{matrix}
   b & c & a \\
\end{matrix} \right]$
So, we get as $=\left[ \begin{matrix}
   a & b & c \\
\end{matrix} \right]\left[ \begin{matrix}
   b & c & a \\
\end{matrix} \right]$ . We know the cumulative rule of box i.e. given as $\left[ \begin{matrix}
   a & b & c \\
\end{matrix} \right]=\left[ \begin{matrix}
   b & c & a \\
\end{matrix} \right]=\left[ \begin{matrix}
   c & a & b \\
\end{matrix} \right]$
So, we will get as
${{\left[ \begin{matrix}
   {\vec{a}} & {\vec{b}} & {\vec{c}} \\
\end{matrix} \right]}^{2}}$ ………………..(2)
So, we will directly substitute the value in above equation, and we get as
$\left[ \begin{matrix}
   \vec{a}\times \vec{b} & \vec{b}\times \vec{c} & {\vec{c}} \\
\end{matrix}\times \vec{a} \right]={{\left[ \begin{matrix}
   {\vec{a}} & {\vec{b}} & {\vec{c}} \\
\end{matrix} \right]}^{2}}={{\left[ 12 \right]}^{2}}$
Thus, on solving we get as
$\left[ \begin{matrix}
   \vec{a}\times \vec{b} & \vec{b}\times \vec{c} & {\vec{c}} \\
\end{matrix}\times \vec{a} \right]=144$
Hence, option (b) is the correct answer.

Note: Students sometimes do not read question carefully and end up finding the value of $\left[ \begin{matrix}
   \vec{a}+\vec{b} & \vec{b}+\vec{c} & {\vec{c}} \\
\end{matrix}+\vec{a} \right]$ . For finding this formula is almost similar as compare to finding value of $\left[ \begin{matrix}
   \vec{a}\times \vec{b} & \vec{b}\times \vec{c} & \vec{c}\times \vec{a} \\
\end{matrix} \right]$ i.e. $2\left[ \begin{matrix}
   {\vec{a}} & {\vec{b}} & {\vec{c}} \\
\end{matrix} \right]$ . By placing value in the formula answer will be 24 which is wrong. So, do not solve it in a hurry and please read the question carefully and then attempt it.