The value of $\alpha$, so that the volume of parallelopiped formed by $\hat{i}+\alpha \hat{j}+\hat{k},…
The value of $\alpha$, so that the volume of parallelopiped formed by $\hat{i}+\alpha \hat{j}+\hat{k}, \hat{j}+\alpha \hat{k}$ and $\alpha \hat{\mathrm{i}}+\hat{\mathrm{k}}$ becomes minimum, is
$-3$
$3$
$\frac{1}{\sqrt{3}}$
$-\frac{1}{\sqrt{3}}$
Solution
Volume of parallelopiped $=\left|\begin{array}{ccc}1 & \alpha & 1 \\ 0 & 1 & \alpha \\ \alpha & 0 & 1\end{array}\right|$
$\therefore \quad \mathrm{V}=1+\alpha^3-\alpha$
For maxima or minima,
$\begin{aligned}
& \frac{\mathrm{dV}}{\mathrm{d} \alpha}=0 \\
& \Rightarrow 3 \alpha^2-1=0 \\
& \Rightarrow \alpha= \pm \frac{1}{\sqrt{3}}
\end{aligned}$
Now, $\frac{\mathrm{d}^2 \mathrm{~V}}{\mathrm{~d} \alpha^2}=6 \alpha$
For $\alpha=\frac{1}{\sqrt{3}}$,
$\frac{\mathrm{d}^2 \mathrm{~V}}{\mathrm{~d} x^2}>0$
$\therefore \quad \mathrm{V}$ is minimum at $\alpha=\frac{1}{\sqrt{3}}$.