If the volume of the tetrahedron formed by the coterminous edges $\mathbf{a}, \mathbf{b}$ and $\mathbf{c}$…
If the volume of the tetrahedron formed by the coterminous edges $\mathbf{a}, \mathbf{b}$ and $\mathbf{c}$ is 4 , then the volume of the parallelopiped formed by the coterminous edges $\mathbf{a} \times \mathbf{b}, \mathbf{b} \times \mathbf{c}$ and $\mathbf{c} \times \mathbf{a}$ is
576
48
16
144
Solution
As we know, the volume of tetrahedron formed by the coterminous edges $\mathbf{a}, \mathbf{b}$ and $\mathbf{c}$,
$
\begin{aligned}
& V=\frac{1}{6}\left[\begin{array}{lll}
\mathbf{a} & \mathbf{b} & \mathbf{c}
\end{array}\right] \\
& 4=\frac{1}{6}\left[\begin{array}{lll}
\mathbf{a} & \mathbf{b} & \mathbf{c}
\end{array}\right]
\end{aligned}
$
Now, the volume of parallelopiped formed, by the coterminous edges $\mathbf{a} \times \mathbf{b}, \mathbf{b} \times \mathbf{c}$ and $\mathbf{c} \times \mathbf{a}$.
$
\begin{aligned}
\mathrm{V} & =\left[\begin{array}{lll}
\mathbf{a} \times \mathbf{b} & \mathbf{b} \times \mathbf{c} & \mathbf{c} \times \mathbf{a}
\end{array}\right] \\
\mathrm{V} & =\left[\begin{array}{lll}
\mathbf{a} & \mathbf{b} & \mathbf{c}
\end{array}\right]^2 \\
\mathrm{~V} & =(24)^2=576 \quad(\text { from Eq. (i)) }
\end{aligned}
$
