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

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

Answers (1)

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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