Question

If the volume of the parallelepiped formed by three non-coplanar vectors $a,b,$ and $c$ is $4$ cubic units, then $\left[a\times bb\times cc\times a\right]$ is equal to

• A. $64$
• B. $16$
• C. $4$
• D. $8$

Answer 1

The correct option is B

$16$

Explanation for the correct option :

Step 1. Find the value of $\left[a\times bb\times cc\times a\right]$.

The parallelepiped is formed by three non-coplanar vectors $a,b,$ and $c$ and its volume is $4$ cubic units. So $\left[abc\right]=4$.

Now $\left[a\times bb\times cc\times a\right]$ can be extended as:

$\left[a\times bb\times cc\times a\right]=\left(a\times b\right)·\left\{\left(b\times c\right)\times \left(c\times a\right)\right\}$

Now using the property $\left(a\times b\right)\times \left(c\times d\right)=\left[abd\right]c-\left[abc\right]d$ the expression can be now modified as:

$\left(a\times b\right)·\left\{\left(b\times c\right)\times \left(c\times a\right)\right\}=\left(a\times b\right)·\left\{\left[bca\right]c-\left[bcc\right]a\right\}$

Now using $\left[bcc\right]=0$ and $\left[bca\right]=\left[abc\right]=\left[cab\right]$ it can be further simplified as:

$\begin{array}{rcl}\left(a\times b\right)·\left\{\left[bca\right]c-\left[bcc\right]a\right\}& =& \left(a\times b\right)·\left\{\left[abc\right]c-0\right\}\\ & =& (a\times b)·c\left[abc\right]\\ & =& \left[abc\right]\left[abc\right]\\ & =& {\left[abc\right]}^2\\ & =& 4^2\\ & =& 16\end{array}$

So, the value of $\left[a\times bb\times cc\times a\right]$ is $16$.

Hence, the correct option is B.