Question

If the vectors $\overrightarrow{b}=(\tan \alpha ,-1,\sqrt{4\sin \alpha /2})\overrightarrow{c}=(\tan \alpha ,\tan \alpha ,\frac{-3}{\sqrt{\sin \alpha /2}})$ are orthogonal and the vector $\overrightarrow{a}=(1,3,\sin 2\alpha )$ makes an obtuse angle with $z$-axis, then $\alpha$ is

• A. $(2n+1)\pi +{\tan }^{-1}2$
• B. $(2n+1)\pi -{\tan }^{-1}2$
• C. $2(n+1)\pi -{\tan }^{-1}2$
• D. $2(n+1)\pi +{\tan }^{-1}2$

Answer 1

Answer: (B)

$(2n+1)\pi -{\tan }^{-1}2$

$\overrightarrow{b}\cdot \overrightarrow{c}=0\Rightarrow {\tan }^2\alpha -tan\alpha -6=0\Rightarrow \tan \alpha =3,-2$

Also $\overrightarrow{a}$ makes an obtuse angle with $z$-axis therefore $\overline{a}\cdot \hat{k}<0\Rightarrow \sin \alpha 2\alpha <0$ If $\tan \alpha =3$ then
$\sin 2\alpha =\frac{2\tan \alpha }{1+{\tan }^2\alpha }=\frac{4}{3}>0$

Now, $\tan 2\alpha >0,\sin 2\alpha <0\Rightarrow \alpha \in$ third quadrant now $\tan \alpha =-2$

Now, $\tan \alpha =-2$
$\Rightarrow \tan (\pi -\alpha )=2$
$\therefore \alpha =(2n+1)\pi -{\tan }^{-1}2,n\in I$