Question

The volume of a tetrahedron determined by the vectors $\overset{ \rightarrow }{a},\overset{ \rightarrow }{b},\overset{ \rightarrow }{c}$ is $\frac{3}{4}$ cubic units. The volume (in cubic units) of a tetrahedron determined by the vectors $3\left(\overset{ \rightarrow }{a} \times \overset{ \rightarrow }{b}\right),4\left(\overset{ \rightarrow }{b} \times \overset{ \rightarrow }{c}\right)$ and $5\left(\overset{ \rightarrow }{c} \times \overset{ \rightarrow }{a}\right)$ will be

Answer 1

$\frac{1}{6}\left[\overset{ \rightarrow }{a} \overset{ \rightarrow }{b} \overset{ \rightarrow }{c}\right]=\frac{3}{4}\Rightarrow \left[\overset{ \rightarrow }{a} \overset{ \rightarrow }{b} \overset{ \rightarrow }{c}\right]=\frac{9}{2}\ldots \left(1\right)$
Volume of tetrahedron $=\frac{1}{6}\left[3 \left(\overset{ \rightarrow }{a} \times \overset{ \rightarrow }{b}\right) 4 \left(\overset{ \rightarrow }{b} \times \overset{ \rightarrow }{c}\right) 5 \left(\overset{ \rightarrow }{c} \times \overset{ \rightarrow }{a}\right)\right]$
$=\frac{1}{6}\times 3\times 4\times 5\left[\overset{ \rightarrow }{a} \overset{ \rightarrow }{b} \overset{ \rightarrow }{c}\right]^{2}$
$=10\times \frac{81}{4}=202.5$ cubic units