Application Of Derivatives Ques 72
72. If the volume of parallelopiped formed by the vectors $\hat{\mathbf{i}}+\lambda \hat{\mathbf{j}}+\hat{\mathbf{k}}, \hat{\mathbf{j}}+\lambda \hat{\mathbf{k}}$ and $\lambda \hat{\mathbf{i}}+\hat{\mathbf{k}}$ is minimum, then $\lambda$ is equal to
(a) $-\frac{1}{\sqrt{3}}$
(b) $\frac{1}{\sqrt{3}}$
(c) $\sqrt{3}$
(d) $-\sqrt{3}$
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Answer:
Correct Answer: 72.(b)
Solution:
Formula:
Maxima and Minima of functions of one variable :
Key Idea Volume of parallelopiped formed by the vectors $\mathbf{a}$, b and $\mathbf{c}$ is $V=[\mathbf{a} \mathbf{b} \mathbf{c}]$.
Given vectors are $\hat{\mathbf{i}}+\lambda \hat{\mathbf{j}}+\hat{\mathbf{k}}, \hat{\mathbf{j}}+\lambda \hat{\mathbf{k}}$ and $\lambda \hat{\mathbf{i}}+\hat{\mathbf{k}}$, which forms a parallelopiped.
$\therefore$ Volume of the parallelopiped is
$ \begin{aligned} V & =\left|\begin{array}{ccc} 1 & \lambda & 1 \\ 0 & 1 & \lambda \\ \lambda & 0 & 1 \end{array}\right|=1+\lambda^{3}-\lambda \\ \Rightarrow \quad V & =\lambda^{3}-\lambda+1 \end{aligned} $
On differentiating w.r.t. $\lambda$, we get
$ \frac{d V}{d \lambda}=3 \lambda^{2}-1 $
For maxima or minima, $\frac{d V}{d \lambda}=0$
$\Rightarrow \quad \lambda= \pm \frac{1}{\sqrt{3}}$
and $\quad \frac{d^{2} V}{d \lambda^{2}}=6 \lambda=\begin{array}{ll}2 \sqrt{3}>0, & \text { for } \quad \lambda=\frac{1}{\sqrt{3}} \\ 2 \sqrt{3}<0, & \text { for } \lambda=-\frac{1}{\sqrt{3}}\end{array}$
$\because \frac{d^{2} V}{d \lambda^{2}}$ is positive for $\lambda=\frac{1}{\sqrt{3}}$, so volume ’ $V$ ’ is minimum for $\lambda=\frac{1}{\sqrt{3}}$
BETA
