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Explain solution RD Sharma class 12 chapter 23 Scaler and Dot Product exercise Multiple choice question 29 maths

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Answer:

Option (a)  \frac{\pi }{6}

Hint:

You must know the identity (a-b)^{2}=a^{2}+b^{2}-2 a b

Given:

|\vec{a}|=|\vec{b}|=1 \text { and } \sqrt{3} \vec{a}-\vec{b} to be a unit vector

Solution:

\frac{\vec{a} \cdot \vec{b}}{|\vec{a}| \cdot|\vec{b}|}=\cos \theta                                                                            [\because|\vec{a}|=|\vec{b}|=1]

\vec{a} \cdot \vec{b}=\cos \theta                                                                                                                                              According to the given question,

\begin{aligned} &|\sqrt{3} \vec{a}-\vec{b}|=1 \Rightarrow|\sqrt{3} \vec{a}-\vec{b}|^{2}=1 \\\\ &3|\vec{a}|^{2}+|\vec{b}|^{2}-2 \sqrt{3}(\vec{a} \cdot \vec{b})=1 \\\\ &3+1-2 \sqrt{3} \cos \theta=1 \end{aligned}

\begin{aligned} &-2 \sqrt{3} \cos \theta=-3 \\\\ &2 \sqrt{3} \cos \theta=3 \\\\ &\cos \theta=\frac{3}{2 \sqrt{3}} \end{aligned}                    \left[\because \frac{\sqrt{3}}{2}=\cos \frac{\pi}{6}\right]

\begin{aligned} &\cos \theta=\frac{\sqrt{3}}{2} \\\\ &\theta=\frac{\pi}{6} \end{aligned}

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