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  • Need Solution for R.D.Sharma Maths Class 12 Chapter 23 Scalar or Dot Products Exercise 23.1 Question 33 Sub Question 1 Maths Textbook Solution.

Need Solution for R.D.Sharma Maths Class 12 Chapter 23 Scalar or Dot  Products  Exercise 23.1 Question 33 Sub Question 1 Maths Textbook Solution.

Answers (1)

Answer:  \frac{\pi }{4}

Hint: You must know the rules of solving vectors.

Given:   Find the angle between two vectors \vec{a} and \vec{b} if  \mid \vec{a}\mid =\sqrt{3},  \mid \vec{b}\mid =2 & \vec{a} . \vec{b} =\sqrt{6}

Solution: Let \theta be the angle between \vec{a} and \vec{b}

                                                        \mid \vec{a}\mid =\sqrt{3},  \mid \vec{b}\mid =\sqrt{2} & \vec{a} . \vec{b} =\sqrt{6}

We know that,

\begin{aligned} &\vec{a} \cdot \vec{b}=|\vec{a}||\vec{b}| \cos \theta \\\\ &\sqrt{6}=(\sqrt{3})(2) \cos \theta \\\\ &\cos \theta=\frac{\sqrt{6}}{2 \sqrt{3}} \\\\ &\cos \theta=\frac{1}{\sqrt{2}} \\\\ &\theta=\cos ^{-1}\left(\frac{1}{\sqrt{2}}\right) \\\\ &\theta=\frac{\pi}{4} \end{aligned}

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