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) \[0\le \theta \le \frac{3\pi }{4}\]
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 \[m=n=\cos \theta \] \[\Rightarrow \left| {\vec{c}} \right|=\left| \cos \theta \,\vec{a}+\cos \theta \,\vec{b}+p(\vec{a}\times \vec{b}) \right|=1\] 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\[\theta \]) \[\therefore \frac{\pi }{4}\le \cos \theta \le \frac{3\pi }{4}\]You need to login to perform this action.
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