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

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$α = (4 n + 1) π + \tan^{- 1} 2$

$α = (4 n + 1) π - \tan^{- 1} 2$

${α=(4n+2)π+tan}^{- 1} 2$

$α = (2 n + 1) \frac{π}{2} - \tan^{- 1} (3)$

answer is B.

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Detailed Solution

$a = (1, 3, \sin 2 α)$ makes an obtuse angle with the Z-axis, its Z component is negative
Thus $- 1 \leq \sin 2 α \leq 0 b u t b . c = 0 \Rightarrow \tan^{2} α - \tan α - 6 = 0 (\tan α + 2) (\tan α - 3) = 0 \Rightarrow \tan α = - 2, 3 N o w \tan α = 3 \sin 2 α = \frac{2 \tan α}{1 + \tan^{2} α} = \frac{6}{10} = \frac{3}{5} i s n o t p o s s i b l e, \sin 2 α < 0 \tan α = - 2 \sin 2 α = \frac{2 \tan α}{1 + \tan^{2} α} = \frac{- 4}{5} \tan 2 α > 0 2 α i s t h i r d q u a d r a n t 0 < \sin \frac{α}{2} < 1 α = (4 n + 1) π - \tan^{- 1} 2$