If vector b→=(tan⁡α,−1,2sin⁡α/2) and c→=tan⁡α,tan⁡α,−3sin⁡α/2 are orthogonal and vector  a→=(1,3,sin⁡2α) makes an obtuse angle with the z-axis, then the value of α is

Since  a→=(1,3,sin⁡2α) makes an obtuse angle with the z-axis, its z-component is negative. Thus,     −1≤sin⁡2α<0----i But     b→⋅c→=0    tan2⁡α−tan⁡α−6=0∴    (tan⁡α−3)(tan⁡α+2)=0⇒    tan⁡α=3,−2Now, tan α= 3. Therefore, sin⁡2α=2tan⁡α1+tan2⁡α=61+9=35     (not possible as sin2α < 0)Now, if tan α= -2 ⇒ sin⁡2α=2tan⁡α1+tan2⁡α=−41+4=−45⇒ tan⁡2α>0Hence, 2α is  in the third quadrant. Also meaningful. If 0