The value of $a$, so that the volume of parallelepiped formed by $\hat{i}+a \hat{j}+\hat{k}, \hat{j}+a…
The value of $a$, so that the volume of parallelepiped formed by $\hat{i}+a \hat{j}+\hat{k}, \hat{j}+a \hat{k}$ and $a \hat{i}+\hat{k}$ becomes minimum is
$\frac{1}{\sqrt{3}}$
3
-3
$\sqrt{3}$
Solution
We know volume of parallelopiped whose edges are
$\begin{aligned}
& \overrightarrow{\mathbf{a}}, \overrightarrow{\mathbf{b}}, \overrightarrow{\mathbf{c}}=[\overrightarrow{\mathbf{a}} \overrightarrow{\mathbf{b}} \overrightarrow{\mathbf{c}}] \text {. } \\
& \therefore \quad[\overrightarrow{\mathbf{a}} \overrightarrow{\mathbf{b}} \overrightarrow{\mathbf{c}}]=\left|\begin{array}{lll}
1 & a & 1 \\
0 & 1 & a \\
a & 0 & 1
\end{array}\right|=1+a^3-a,
\end{aligned}$
Let $\quad f(a)=a^3-a+1 \Rightarrow f^{\prime}(a)=3 a^2-1$
$\Rightarrow \quad f^{\prime \prime}(a)=6 a$
For maximum or minimum, put $f^{\prime}(a)=0$ $\Rightarrow a= \pm \frac{1}{\sqrt{3}}$ which shows $f(a)$ is minimum at $a=\frac{1}{\sqrt{3}}$ and maximum at $a=-\frac{1}{\sqrt{3}}$.