Kinematics Question 283

Question: The angle between two vectors given by $ 6\bar{i}+6\bar{j}-3\bar{k} $ and $ 7\overline{i}+4\overline{j}+4\overline{k} $ it’s [EAMCET (Engg.) 1999]

Options:

A) $ {{\cos }^{-1}}( \frac{1}{\sqrt{3}} ) $

B) $ {{\cos }^{-1}}( \frac{5}{\sqrt{3}} ) $

C) $ {{\sin }^{-1}}( \frac{2}{\sqrt{3}} ) $

D) $ {{\sin }^{-1}}( \frac{\sqrt{5}}{3} ) $

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Answer:

Correct Answer: D

Solution:

$ \cos \theta =\frac{\vec{A}\vec{B}}{AB}=\frac{42+24-12}{\sqrt{36+36+9}\sqrt{49+16+16}} $

$ =\frac{56}{9\sqrt{71}} $

$ \cos \theta =\frac{56}{9\sqrt{71}} $ \ $ \sin \theta =\frac{\sqrt{5}}{3} $ or $ \theta ={{\sin }^{-1}}( \frac{\sqrt{5}}{3} ) $



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