Question

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

• A. $12$
• B. $6$
• C. $2$
• D. $1$

Answer 1

Answer: (C)

Given, $\left[\overset{ \rightarrow }{a} \times \overset{ \rightarrow }{b} \overset{ \rightarrow }{b} \times \overset{ \rightarrow }{c} \overset{ \rightarrow }{c} \times \overset{ \rightarrow }{a}\right]=36$
$\Rightarrow \left[\overset{ \rightarrow }{a} \overset{ \rightarrow }{b} \overset{ \rightarrow }{c}\right]^{2}=36$
$\Rightarrow \left[\overset{ \rightarrow }{a} \overset{ \rightarrow }{b} \overset{ \rightarrow }{c} \right]=\pm6$
Volume of the tetrahedron $=\left|\frac{1}{6} \left[\overset{ \rightarrow }{a} + \overset{ \rightarrow }{b} \overset{ \rightarrow }{b} + \overset{ \rightarrow }{c} \overset{ \rightarrow }{c} + \overset{ \rightarrow }{a}\right]\right|$
$=\left|\frac{1}{3} \left[\overset{ \rightarrow }{a} \overset{ \rightarrow }{b} \overset{ \rightarrow }{c}\right]\right|=2$ cubic units