Q. The value of $a$ so that the volume of parallelopiped formed by $\widehat{\dot{\mathbf{i}}} + a\widehat{\dot{\mathbf{j}}} + \widehat{\mathbf{k}},\ \widehat{\dot{\mathbf{j}}} + a\widehat{\mathbf{k}}$ and $a\widehat{\dot{\mathbf{i}}} + \widehat{\mathbf{k}}$ becomes minimum is
A) $- 3$
B) 3
C) $1/\sqrt{3}$
D) $\sqrt{3}$
Explanation
(c)
Volume of parallelopiped,
$f(a) = \left| \begin{matrix}
1 & a & 1 \\
0 & 1 & a \\
a & 0 & 1
\end{matrix} \right| = 1 + a^{3} - a$
$Now,\ \ \ f^{'}(a) = 3a^{2} - 1$
$\Longrightarrow f^{''}(a) = 6a$
Put $f^{'}(a) = 0$
$\Longrightarrow a \neq \pm \frac{1}{\sqrt{3}}$
Which shows $f(a)$ is maximum at
$a = \frac{1}{\sqrt{3}}\ and\ maximum\ at\$
$a = - \frac{1}{\sqrt{3}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.