JEE Main 12 Jan 2019 Evening Question 16
Question: Let $ \vec{a},\vec{b} $ and $ \vec{c} $ be three unit vectors, out of which vectors $ \vec{b} $ and $ \vec{c} $ are non-parallel. If $ \alpha $ and $ \beta $ are the angles which vector a makes with $ \vec{a} $ vectors $ \vec{b} $ and $ \vec{c} $ respectively and $ \vec{a}\times (\vec{b}\times \vec{c})=\frac{1}{2}\vec{b}, $ then $ |\alpha -\beta | $ is equal to [JEE Main Online Paper Held On 12-Jan-2019 Evening]
Options:
A) $ 45{}^\circ $
B) $ 60{}^\circ $
C) $ 90{}^\circ $
D) $ 30{}^\circ $
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Answer:
Correct Answer: D
Solution:
- Here, $ \vec{a}\times (\vec{b}\times \vec{c})=\frac{1}{2}\vec{b} $
$ \Rightarrow $ $ (\vec{a}.\vec{c})\vec{b}-(\vec{a}.\vec{b})\vec{c}=\frac{1}{2}\vec{b} $ Since, $ \vec{b} $ and $ \vec{c} $ are non-parallel vectors.
$ \therefore $ $ \vec{a}.\vec{c}=\frac{1}{2} $ and $ \vec{a}.\vec{b}=0 $ Since $ \vec{a},\vec{b} $ and $ \vec{c} $ are unit vectors.
$ \therefore $ $ \cos \beta =\frac{1}{2} $ and $ \cos \alpha =0 $
$ \Rightarrow $ $ \beta =60^{o} $ and $ \alpha =90^{o} $
$ \therefore $ $ |\alpha -\beta |=30^{o} $
