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  • Please solve RD Sharma class 12 chapter 23 Scaler and Dot Product exercise Multiple choice question 26 maths textbook solution

Please solve RD Sharma class 12 chapter 23 Scaler and Dot Product exercise Multiple choice question 26 maths textbook solution

Answers (1)

Answer:

Option (b)  \frac{\pi }{3}

Hint:

You must know the formula to find angle between two vectors  \cos \theta=\frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}

Given:

|\vec{a}|=\sqrt{3},|\vec{b}|=4 \text { and } \vec{a} \cdot \vec{b}=2 \sqrt{3}

Solution:

\begin{aligned} &\frac{\vec{a} \cdot \vec{b}}{|\vec{a}| \cdot|\vec{b}|}=\cos \theta \\\\ &\frac{2 \sqrt{3}}{\sqrt{3} \cdot 4}=\cos \theta \end{aligned}

\cos \theta=\frac{1}{2} \quad\left[\because \cos \frac{\pi}{3}=\frac{1}{2}\right]

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

Hence, option (b) is correct

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