Vector Algebra Question 121
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} ) $
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Answer:
Correct Answer: B
Solution:
- [b] Let $ \overset{\to }{\mathop{x}},=\sqrt{3}(\overset{\to }{\mathop{a}},\times \overset{\to }{\mathop{b}},) $ and $ \overset{\to }{\mathop{y}},=\overset{\to }{\mathop{b}},-(\overset{\to }{\mathop{a}},\cdot \overset{\to }{\mathop{b}},)\overset{\to }{\mathop{a}}, $ Clearly $ \overset{\to }{\mathop{x}},\cdot \overset{\to }{\mathop{y}},=0\Rightarrow \overset{\to }{\mathop{x}}, $ and $ \overset{\to }{\mathop{y}}, $ are perpendicular So, one angle is $ \frac{\pi }{2} $ . Also $ |\overset{\to }{\mathop{x}},|=\sqrt{3}|bsin\theta | $ , where $ \theta $ is angle between vectors $ \overset{\to }{\mathop{a}}, $ and $ \overset{\to }{\mathop{b}}, $ $ (|\vec{a}|=1) $ $ |\overset{\to }{\mathop{y}},|=\sqrt{{{{ \vec{b}-(\vec{a}.\vec{b})\vec{a} }}^{2}}}=\sqrt{b^{2}-{{(\vec{a}.\vec{b})}^{2}}} $ $ =\sqrt{b^{2}-b^{2}\cos \theta }=| b\sin \theta | $
$ \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
