SATHEE: Application of Derivatives 4 Question 2

Application of Derivatives 4 Question 2

####2. If the volume of parallelopiped formed by the vectors $\hat{i} + λ \hat{j} + \hat{k}, \hat{j} + λ \hat{k}$ and $λ \hat{i} + \hat{k}$ is minimum, then $λ$ is equal to

(a) $- \frac{1}{\sqrt{3}}$

(b) $\frac{1}{\sqrt{3}}$

(d) $- \sqrt{3}$

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Answer:

Correct Answer: 2. (b)

Solution:

Key Idea Volume of parallelopiped formed by the vectors $a$ , b and $c$ is $V = [a b c]$ .

Given vectors are $\hat{i} + λ \hat{j} + \hat{k}, \hat{j} + λ \hat{k}$ and $λ \hat{i} + \hat{k}$ , which forms a parallelopiped.

$∴$ Volume of the parallelopiped is

$\begin{aligned} V & = | \begin{array}{ccc} 1 & λ & 1 \\ 0 & 1 & λ \\ λ & 0 & 1 \end{array} | = 1 + λ^{3} - λ \\ \Rightarrow V & = λ^{3} - λ + 1 \end{aligned}$

On differentiating w.r.t. $λ$ , we get

$\frac{d V}{d λ} = 3 λ^{2} - 1$

For maxima or minima, $\frac{d V}{d λ} = 0$

$\Rightarrow λ = \pm \frac{1}{\sqrt{3}}$

and $\frac{d^{2} V}{d λ^{2}} = 6 λ = \begin{array}{ll} 2 \sqrt{3} > 0, & for λ = \frac{1}{\sqrt{3}} \\ 2 \sqrt{3} < 0, & for λ = - \frac{1}{\sqrt{3}} \end{array}$

$∵ \frac{d^{2} V}{d λ^{2}}$ is positive for $λ = \frac{1}{\sqrt{3}}$ , so volume ’ $V$ ’ is minimum for $λ = \frac{1}{\sqrt{3}}$

Application of Derivatives 4 Question 2

Answer:

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Application of Derivatives 4 Question 2

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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