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  • Explain solution RD Sharma class 12 chapter 23 Scaler and Dot Product exercise Multiple choice question 20 maths

Explain solution RD Sharma class 12 chapter 23 Scaler and Dot Product exercise Multiple choice question 20 maths

Answers (1)

Answer:

Option (c) 4

Hint:

Magnitude of a vector =\sqrt{a^{2}+b^{2}+c^{2}}

Given:

\vec{a} \text { and } \vec{b}  are unit vectors and the greatest value of  \sqrt{3}|\vec{a}+\vec{b}|+|\vec{a}-\vec{b}|=?

Solution:

\sqrt{3}|\vec{a}+\vec{b}|+|\vec{a}-\vec{b}|=\sqrt{3} \sqrt{|\vec{a}|^{2}+|\vec{b}|^{2}+2 \vec{a} \cdot \vec{b}}+\sqrt{|\vec{a}|^{2}+|\vec{b}|^{2}-2 \vec{a} \cdot \vec{b}}

 

\sqrt{3}|\vec{a}+\vec{b}|+|\vec{a}-\vec{b}|=\sqrt{3} \sqrt{1+1+2 \cos \theta}+\sqrt{1+1-2 \cos \theta} \quad[\because|\vec{a}|=|\vec{b}|=1]

 

=\sqrt{3} \sqrt{2+2 \cos \theta}+\sqrt{2-2 \cos \theta}

 

\begin{aligned} &=\sqrt{3} \sqrt{2(1+\cos \theta)}+\sqrt{2(1-\cos \theta)} \\\\ &=\sqrt{3} \sqrt{2\left(2 \cos ^{2} \frac{\theta}{2}\right)}+\sqrt{2\left(2 \sin ^{2} \frac{\theta}{2}\right)} \end{aligned}

\begin{array}{ll} =\sqrt{3}\left(2 \cos \frac{\theta}{2}\right)+2 \sin \frac{\theta}{2} & {\left[\because \cos \theta=1-\sin ^{2} \frac{\theta}{2}\right]} \\\\ =2 \times 2\left(\frac{\sqrt{3}}{2} \cos \frac{\theta}{2}+\frac{1}{2} \sin \frac{\theta}{2}\right) & {[\because \sin x \cos y+\cos x \sin y=\sin (x+y)]} \end{array}

\begin{aligned} &=4\left(\sin \frac{\pi}{3} \cos \frac{\theta}{2}+\cos \frac{\pi}{3} \sin \frac{\theta}{2}\right) \\\\ &=4\left[\sin \left(\frac{\pi}{3}+\frac{\theta}{2}\right)\right] \end{aligned}

Now,

\begin{aligned} &-1 \leq \sin \left(\frac{\pi}{3}+\frac{\theta}{2}\right) \leq 1 \\\\ &-4 \leq 4\left[\sin \left(\frac{\pi}{3}+\frac{\theta}{2}\right)\right] \leq 4 \end{aligned}

Hence, maximum value of \sqrt{3}|\vec{a}+\vec{b}|+|\vec{a}-\vec{b}| \text { is } 4

So, option (C) is correct

Posted by

infoexpert26

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