Question

If the volume of the parallelepiped formed by the vectors $\overrightarrow{a}\times \overrightarrow{b},\overrightarrow{b}\times \overrightarrow{c}$ and $\overrightarrow{c}\times \overrightarrow{a}$ is $36$ cubic units, then the volume (in cubic units) of the tetrahedron formed by the vectors $\overrightarrow{a}+\overrightarrow{b},\overrightarrow{b}+\overrightarrow{c}$ and $\overrightarrow{c}+\overrightarrow{a}$ is equal to

• A. $12$
• B. $6$
• C. $2$
• D. $1$

Answer 1

Answer: (C)

Given, $\left[\overrightarrow{a}\times \overrightarrow{b}\overrightarrow{b}\times \overrightarrow{c}\overrightarrow{c}\times \overrightarrow{a}\right]=36$
$\Rightarrow {\left[\overrightarrow{a}\overrightarrow{b}\overrightarrow{c}\right]}^2=36$
$\Rightarrow \left[\overrightarrow{a}\overrightarrow{b}\overrightarrow{c}\right]=\pm 6$
Volume of the tetrahedron $=\left|\frac{1}{6}\left[\overrightarrow{a}+\overrightarrow{b}\overrightarrow{b}+\overrightarrow{c}\overrightarrow{c}+\overrightarrow{a}\right]\right|$
$=\left|\frac{1}{3}\left[\overrightarrow{a}\overrightarrow{b}\overrightarrow{c}\right]\right|=2$ cubic units