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

The Correct Option is A

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, \(\frac{dv}{da}=\frac{d}{da}(1+a^3-a)\)
⇒\(3a^2-1=0\)
⇒\(3a^2=1\)
⇒\(a=\frac{1}{\sqrt{3}}\)