Question

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

• A. $-3$
• B. $3$
• C. $1/\sqrt{3}$
• D. $-\sqrt{3}$

Answer 1

The correct option is C $1/\sqrt{3}$

Volume of the Parallelopiped formed by $i+aj+k,j+ak,ai+k$ is
$V=\left|\begin{array}{ccc}1& a& 1\\ 0& 1& a\\ a& 0& 1\end{array}\right|=1+a^3-a$
$\Rightarrow \frac{dV}{da}=3a^2-1$
For minimum value of $V$ we have $\frac{dv}{da}=0$
$\Rightarrow a=\pm 1/\sqrt{3}$ Also $\frac{d^{2}V}{d{a}^2}=6a>0$ for $a=1/\sqrt{3}$
Thus $V$ is minimum when $a=1/\sqrt{3}$