Question

The value of $a$ for which the volume of parallelepiped formed by the vectors $\hat {i}+a\hat{j} + \hat{k} \,\, \hat{j}+a\hat{k}$ and $a\hat{i}+\hat{k}$ is minimum, is

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

Answer 1

Answer: (A)

$\frac{1}{\sqrt{3}}$

Answer 2

The correct answer is A:$\frac{1}{\sqrt{3}}$
Given that;
The volume of parallelopiped is given by$[\vec{a}\space\vec{b}\space\vec{c}]$;
Where, $\vec{a}=\hat{i}+a\hat{j}+\hat{k}$
$\vec{b}=\hat{j}+a\hat{k}$
$\vec{c}=a\hat{i}+\hat{k}$
$\therefore$ Putting values we can make $\begin{vmatrix}1&a&1\\\0&1&a\\\a&0&1\end{vmatrix}$
$=1(1-0)-a(0-a^2)+1(0-a)$
$=1+a^3-a$
$\therefore$ Volume (v)=$1+a^3-a$
The question asks for the minimum value of ‘a’
So, $\frac{dv}{da}=\frac{d}{da}(1+a^3-a)$
⇒$3a^2-1=0$
⇒$3a^2=1$
⇒$a=\frac{1}{\sqrt{3}}$