Question

The value of a so that the 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 $\frac{1}{\sqrt{3}}$

$V=[\hat{i}+a\hat{j}+\hat{k}\hat{j}+a\hat{k}a\hat{i}+\hat{k}]$

$=(\hat{i}+a\hat{j}+\hat{k})\cdot \left\{\right(\hat{j}+a\hat{k})\times (a\hat{i}+\hat{k}\left)\right\}$
$=(\hat{i}+a\hat{j}+\hat{k})\cdot (\hat{i}+a^2\hat{j}-a\hat{k})=1+a^3-a$

$\frac{dV}{da}=3a^2-1,\frac{d^{2}V}{d{a}^2}=6a,\frac{dV}{da}=0$
$\Rightarrow 3a^2-1=0\Rightarrow a=\pm \frac{1}{\sqrt{3}}$
At $a=\frac{1}{\sqrt{3}}$

$\frac{d^{2}V}{d{a}^2}=\frac{6}{\sqrt{3}}>0$
$\therefore$ $V$ is minimum at $a=\frac{1}{\sqrt{3}}$.
Hence, option '$A$' is correct.