Question

The minimum volume of the parallelopiped formed by the vectors $i+aj+k,j+ak$ and $ai+k$ is

• A. $\frac{1}{3\sqrt{3}}$
• B. $\frac{1}{3}$
• C. $\frac{3\sqrt{3}-2}{3\sqrt{3}}$
• D. $\frac{3\sqrt{3}+2}{3\sqrt{3}}$

Answer 1

The correct option is C $\frac{3\sqrt{3}-2}{3\sqrt{3}}$

Volume of paralleopiped

$\begin{array}{c}\left|\begin{array}{ccc}1& a& 1\\ 0& 1& a\\ a& 0& 1\end{array}\right|=1-a\left(0-a^2\right)+\left(0-a\right)=a^3-a+1\\ f\left(a\right)=a^3-a+1\\ f^{\prime }\left(a\right)=3a^2-1=0\\ a=\pm \frac{1}{\sqrt{3}}\\ \underset{–}{(+)-(+)}\\ -\frac{1}{\sqrt{3}}\frac{1}{\sqrt{3}}\end{array}$

sign convention of $f^{\prime }\left(a\right)$

minimum occurs at $a=\frac{1}{\sqrt{3}}$

$\begin{array}{c}\frac{1}{3\sqrt{3}}-\frac{1}{\sqrt{3}}+1\\ =\frac{3\sqrt{3}-2}{3\sqrt{3}}\end{array}$

Then,

Option $C$ is correct answer.