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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?

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Solution

## 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, ${\left[abc\right]}^{2}=\left[\begin{array}{ccc}a.a& a.b& a.c\\ b.a& b.b& b.c\\ c.a& c.b& c.c\end{array}\right]$ $=\left[\begin{array}{ccc}1& \frac{1}{2}& \frac{1}{2}\\ \frac{1}{2}& 1& \frac{1}{2}\\ \frac{1}{2}& \frac{1}{2}& 1\end{array}\right]$ $=1\left(1-\frac{1}{4}\right)-\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}\right)+\frac{1}{2}\left(\frac{1}{4}-\frac{1}{2}\right)$ $=\frac{1}{2}$ $â‡’\left[abc\right]=\frac{1}{\sqrt{2}}$Hence, volume of the parallelopiped is $\frac{1}{\sqrt{2}}$

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