Question

The value of so that the volume of the parallelopiped formed by the vectors $\overrightarrow{i}+\lambda \overrightarrow{j}+\overrightarrow{k},\overrightarrow{j}+\lambda \overrightarrow{k}$ and $\lambda \overrightarrow{i}+\overrightarrow{k}$ becomes minimum, is

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

Answer 1

The correct option is C $\frac{1}{\sqrt{3}}$

Volume of the parallelopipe $=\left[\begin{array}{ccc}\overrightarrow{i}+\lambda \overrightarrow{j}+\overrightarrow{k}& \overrightarrow{j}+\lambda \overrightarrow{k}& \lambda \overrightarrow{i}+\overrightarrow{k}\end{array}\right]$

$=\left|\begin{array}{ccc}1& \lambda & 1\\ 0& 1& \lambda \\ \lambda & 0& 1\end{array}\right|\left[\begin{array}{ccc}\overrightarrow{i}& \overrightarrow{j}& \overrightarrow{k}\end{array}\right]$

$=1(1-0)-\lambda (0-{\lambda }^2)+1(0-\lambda )$

$=1+{\lambda }^3-\lambda$

Minimizing the above term, we get $3{\lambda }^2=1$

$\Rightarrow \lambda =\frac{1}{\sqrt{3}}$