Question

The value of 'a' so that volume of parallelopiped formed by $\hat{i}+a\hat{j}+\hat{k},\hat{j}+a\hat{k}$ and $a\hat{i}+\hat{k}$ becomes minimum is,

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

Answer 1

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

Volume of parallelopiped $=V=\left|\begin{array}{ccc}1& a& 1\\ 0& 1& a\\ a& 0& 1\end{array}\right|$

$V=1(1-0)-a(0-a^2)+1(0-a)$

$\Rightarrow V=1+a^3-a$

$V=a^3-a=1$

$\Rightarrow \frac{dv}{da}=3a^2-1$ is minimum

when $\frac{dv}{da}=0\Rightarrow 3a^2-1=0$

$\Rightarrow a^2=\frac{1}{3}$

$\Rightarrow a=\pm \frac{1}{\sqrt{3}}$

But volume cannot be negative.

Hence volume of parallelopiped$=\frac{1}{\sqrt{3}}$units.