Vector Algebra Question 254
Question: Let $ \vec{a}\cdot \vec{b}=0 $ where $ \vec{a} $ and $ \vec{b} $ are unit vectors and the unit vector $ \vec{c} $ is inclined at an angle $ \theta $ to both $ \vec{a} $ and $ \vec{b} $ . If $ \vec{c}=m\vec{a}+n\vec{b}+p(\vec{a}\times \vec{b}),(m,n,p\in R) $ , then
Options:
A) $ -\frac{\pi }{4}\le \theta \le \frac{\pi }{4} $
B) $ \frac{\pi }{4}\le \theta \le \frac{3\pi }{4} $
C) $ 0\le \theta \le \frac{\pi }{4} $
D) $ 0\le \theta \le \frac{3\pi }{4} $
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Answer:
Correct Answer: B
Solution:
- [b] $ \vec{c}=m\vec{a}+n\vec{b}+P(\vec{a}\times \vec{b}) $ Taking dot product with $ \vec{a} $ and $ \vec{b} $ , we have $ m=n=\cos \theta $
$ \Rightarrow | {\vec{c}} |=| \cos \theta ,\vec{a}+\cos \theta ,\vec{b}+p(\vec{a}\times \vec{b}) |=1 $ Squaring both sides, we get $ {{\cos }^{2}}\theta +{{\cos }^{2}}\theta +p^{2}=1 $ Or $ \cos \theta =\pm \frac{\sqrt{1-p^{2}}}{\sqrt{2}} $ Now $ -\frac{1}{\sqrt{2}}\le \cos \theta \le \frac{1}{\sqrt{2}} $ (for real value of $ \theta $ )
$ \therefore \frac{\pi }{4}\le \cos \theta \le \frac{3\pi }{4} $
BETA
