Question

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

• A. $\alpha =(4n+1)\pi +{\tan }^{-1}2$
• B. $\alpha =(4n+1)\pi -{\tan }^{-1}2$
• C. $\alpha =(4n+2)\pi +{\tan }^{-1}2$
• D. $\alpha =(4n+2)\pi -{\tan }^{-1}2$

Answer 1

Answer: (B), (D)

The correct options are
B $\alpha =(4n+1)\pi -{\tan }^{-1}2$
D $\alpha =(4n+2)\pi -{\tan }^{-1}2$

Since, $\overrightarrow{a}=(1,3,\sin 2\alpha )$ makes an obtuse angle with $z$-axis, so its $z$-component is negative.
Thus,
$-1\le \sin 2\alpha <0...\left(1\right)$
and $\overrightarrow{b}\cdot \overrightarrow{c}=0(\because \text{orthogonal})$
$\Rightarrow {\tan }^2\alpha -\tan \alpha -6=0$
$\Rightarrow (\tan \alpha -3)(\tan \alpha +2)=0$
$\Rightarrow \tan \alpha =3,-2$

Now, $\tan \alpha =3$
$\therefore \sin 2\alpha =\frac{2\tan \alpha }{1+{\tan }^2\alpha }=\frac{3}{5}$
which is not possible as $\sin 2\alpha <0$

Now, if $\tan \alpha =-2$
$\Rightarrow \sin 2\alpha =\frac{2\tan \alpha }{1+{\tan }^2\alpha }=-\frac{4}{5}$
$\Rightarrow \tan 2\alpha >0$

Hence, $2\alpha$ is the third quadrant. Also, $\sqrt{\sin \left(\alpha /2\right)}$ is defined.
If $0<\sin \left(\alpha /2\right)<1,$ then
$\alpha =(4n+1)\pi -{\tan }^{-1}2$ and
$\alpha =(4n+2)\pi -{\tan }^{-1}2$