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
$V=\left|\begin{array}{ccc}1& \lambda & 1\\ 0& 1& \lambda \\ \lambda & 0& 1\end{array}\right|$
$=\left(1\right)\left\{\left(1\right)\left(1\right)-\left(0\right)\left(\lambda \right)\right\}-\left(\lambda \right)\left\{\left(0\right)\left(1\right)-\left(\lambda \right)\left(\lambda \right)\right\}+\left(1\right)\left\{\left(0\right)\left(0\right)-\left(1\right)\left(\lambda \right)\right\}$
$=1+{\lambda }^3-\lambda$
Now, $V=1+{\lambda }^3-\lambda$
So, $\frac{dV}{d\lambda }=3{\lambda }^2-1$ .
For maxima or minima, $\frac{dV}{d\lambda }=0$ .
$\Rightarrow 3{\lambda }^2-1=0$
$\Rightarrow \lambda =\pm \frac{1}{\sqrt{3}}$
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 $\lambda =\frac{1}{\sqrt{3}}$ .