Vector-Algebra Question 418
Question: The angles of a triangle, two of whose sides are represented by the vectors $ \sqrt{3}(\vec{a}\times \vec{b}) $ and $ \vec{b}-(\vec{a}.\vec{b})\vec{a} $ where $ \vec{b} $ is a non-zero vector and $ \vec{a} $ is a unit vector are
Options:
A) $ \tan {{,}^{-1}}( \frac{1}{\sqrt{3}} );,\tan {{,}^{-1}}( \frac{1}{2} );,\tan {{,}^{-1}}( \frac{\sqrt{3}+2}{1-2\sqrt{3}} ) $
B) $ \tan {{,}^{-1}}( \sqrt{3} );,\tan {{,}^{-1}}( \frac{1}{\sqrt{3}} );,\cot {{,}^{-1}}( 0 ) $
C) $ \tan {{,}^{-1}}( \sqrt{3} );,\tan {{,}^{-1}}( 2 );,\tan {{,}^{-1}}( \frac{\sqrt{3}+2}{2\sqrt{3}-1} ) $
D) $ \tan {{,}^{-1}}( \sqrt{3} );,\tan {{,}^{-1}}( \sqrt{2} );,\tan {{,}^{-1}}( \frac{\sqrt{2}+3}{3\sqrt{2}-1} ) $
Correct Answer: BShow Answer
Answer:
Solution:
$ \therefore \frac{|\overset{\to }{\mathop{x}},|}{|\overset{\to }{\mathop{y}},|}=\sqrt{3}=\tan \alpha \Rightarrow \alpha =\frac{\pi }{3} $ . So, $ \beta =\frac{\pi }{6} $
BETA
