CGPET 2016 Solved Paper

Since, the vector a=(1,3,sin‌2‌α) makes an obtuse angle with the Z - axis. Therefore, its z - component is negative.
i.e. sin‌2‌α<0
∴ −1≤sin‌2‌α<0....(i)
Since, b and c are orthogonal.
∴ b.c=0
⇒ tan2α−tan‌α−6=0
⇒ (tan‌α−3)(tan‌α+2)=0
⇒ tan‌α=3,−2
∴ tan‌α=3
Then, sin‌2‌α=

2‌tan‌α

1+tan2α

=

6

10

>0, which is a contradiction to ......(i)
∴ tan‌α=3 is not possible.
Thus, tan‌α=−2 and for this value of tan‌α, we get
tan‌2‌α=

2‌tan‌α

1−tan2α

=

4

3

Since, sin‌2‌α<0‌and‌tan‌2‌α>0
Therefore, 2α is in third quadrant.
Also, √sin‌

α

2

is meaningful, if 0<sin‌

α

2

<1
when these conditions are satisfied, α is given by
α=(4n+1)π−tan−12.