Question

If the volume of parallelepiped formed by the vectors $\hat{i}+\lambda \hat{j}+\hat{k},\hat{j}+\lambda \hat{k}$ and $\lambda \hat{i}+\hat{k}$ is minimum, then $\lambda $ is equal to

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

Answer 1

Answer: (C)

Volume of parallelepiped formed by the vectors $\hat{i}+\lambda \hat{j}+\hat{k},\hat{j}+\lambda \hat{k}$ and $\lambda \hat{i}+\hat{k}$ is given by
$V=\begin{vmatrix} 1 & \lambda & 1 \\ 0 & 1 & \lambda \\ \lambda & 0 & 1 \end{vmatrix}$
$=\left(1\right)\left\{\left(1\right) \left(1\right) - \left(0\right) \left(\lambda \right)\right\}-\left(\lambda \right)\left\{\left(0\right) \left(1\right) - \left(\lambda \right) \left(\lambda \right)\right\}+\left(1\right)\left\{\left(0\right) \left(0\right) - \left(1\right) \left(\lambda \right)\right\}$
$=1+\lambda ^{3}-\lambda $
Now, $V=1+\lambda ^{3}-\lambda $
So, $\frac{d V}{d \lambda }=3\lambda ^{2}-1$ .
For maxima or minima, $\frac{d V}{d \lambda }=0$ .
$\Rightarrow 3\lambda ^{2}-1=0$
$\Rightarrow \lambda =\pm\frac{1}{\sqrt{3}}$
Also, $\frac{d^{2} V}{d \lambda ^{2}}=6\lambda $ .
So, $\left(\frac{d^{2} V}{d \left(\lambda \right)^{2}}\right)_{\lambda = \frac{1}{\sqrt{3}}}=6\times \frac{1}{\sqrt{3}}=2\sqrt{3}>0$ and $\left(\frac{d^{2} V}{d \left(\lambda \right)^{2}}\right)_{\lambda = - \frac{1}{\sqrt{3}}}=6\times -\frac{1}{\sqrt{3}}=-2\sqrt{3} < 0$
For minimum, $\left(\frac{d^{2} V}{d \left(\lambda \right)^{2}}\right)>0$ .
So, volume is minimum at $\lambda =\frac{1}{\sqrt{3}}$ .