Question

The edges of a parallelopiped are of unit length and are parallel to non-coplanar unit vectors $a,b,c$ such that $a.b=b.c=c.a=\frac{1}{2}$. Then, the volume of the parallelopiped is?

Answer 1

The volume of parallelopiped:

The volume of the parallelepiped spanned by $a,b,c$ is $\left[abc\right]=a.\left(bxc\right)$

Since, we know that ${\left[abc\right]}^2=\left[\begin{array}{ccc}a.a& a.b& a.c\\ b.a& b.b& b.c\\ c.a& c.b& c.c\end{array}\right]$

So, ${\left[abc\right]}^2=\left[\begin{array}{ccc}a.a& a.b& a.c\\ b.a& b.b& b.c\\ c.a& c.b& c.c\end{array}\right]$

$=\left[\begin{array}{ccc}1& \frac{1}{2}& \frac{1}{2}\\ \frac{1}{2}& 1& \frac{1}{2}\\ \frac{1}{2}& \frac{1}{2}& 1\end{array}\right]$

$=1\left(1-\frac{1}{4}\right)-\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}\right)+\frac{1}{2}\left(\frac{1}{4}-\frac{1}{2}\right)$

$=\frac{1}{2}$

$\Rightarrow \left[abc\right]=\frac{1}{\sqrt{2}}$

Hence, volume of the parallelopiped is $\frac{1}{\sqrt{2}}$