Vector Algebra Question 207
Question: If the vectors $ \vec{a} $ and $ \vec{b} $ are linearly independent satisfying $ (\sqrt{3}tan\theta +1)\vec{a}+(\sqrt{3}sec\theta -2)\vec{b} $ =0, then the most general values of $ \theta $ are
Options:
A) $ n\pi -\frac{\pi }{6},n\in z $
B) $ 2n\pi \pm \frac{11\pi }{6},n\in z $
C) $ n\pi \pm \frac{\pi }{6},n\in z $
D) $ 2n\pi +\frac{11\pi }{6},n\in z $
Show Answer
Answer:
Correct Answer: D
Solution:
- [d] $ \sqrt{3}\tan \theta +1=0 $ and $ \sqrt{3}\sec \theta -2=0 $
$ \Rightarrow \theta =\frac{11\pi }{6} $
$ \Rightarrow \theta =2n\pi \pm \frac{11\pi }{6},n\in Z $
BETA
