Kinematics Question 585
Question: If $ \overrightarrow{A}=\overrightarrow{B}-\overrightarrow{C} $ , then, the angle between $ \overrightarrow{A} $ and $ \overrightarrow{B} $ it’s
Options:
A) $ \text{ta}{{\text{n}}^{-1}}\frac{B^{2}+A^{2}-C^{2}}{2AB} $
B) $ {{\sin }^{-1}}\frac{B^{2}+A^{2}-C^{2}}{2AB} $
C) $ {{\cos }^{-1}}\frac{A^{2}+B^{2}-C^{2}}{2AB} $
D) $ {{\sec }^{-1}}\frac{A^{2}+B^{2}-C^{2}}{2AB} $
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Answer:
Correct Answer: C
Solution:
[c] Given, $ \vec{A}=\vec{B}-\vec{C} $ $ \therefore $ $ \vec{C}=\vec{B}-\vec{A} $
If $ \theta $ it’s the angle between $ \vec{A} $ and $ \vec{B} $ ,
then $ C^{2}=B^{2}+A^{2}-2ABcos\theta $
$ \therefore \cos \theta =\frac{B^{2}+A^{2}-C^{2}}{2AB} $
