If vector ` vec b=(t a nalpha,-12sqrt(sinalpha//2))a n d vec c=(t a nalpha, t a nalpha-3/(sqrt(sinalpha//2)))` are orthogonal and vector ` vec a=(13,sin2alpha)` makes an obtuse angle with the z-axis, then the value of `alpha` is `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`

Options

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

b,d

Explanation

Since `veca = 1,3, sin 2 alpha) ` makes on abtuse angle with the z-axis its z-component is negtive, thus,
` -1 le sin 2 alpha lt 0`
But `vecb.vecc=0`
` tan^(2) alpha - tan alpha -6 =0`
`(tan alpha -3) (tan alpha + 2) =0`
` Rightarrow tan alpha 3, -2`
Now, `tan alpha =3,` therefore,
`sin2 alpha = (2 tanalpha)/(1+tan^(2)alpha)= 6/(1+9)= 3/5`
( not possible as `sin 2 alpha lt 0`)
Now , if ` tan alpha = -2`
`Rightarrow sin2 alpha= (2tan alpha)/(1+ tan^(2)alpha)= (-4)/(1+_5) = (-4)/5`
`tan 2 alpha gt0`
Hence, ` 2alpha` is the third quadrant , Also , `sqrt(sin alpha//2)` is meaningful. if `0 lt sin alpha//2 1, ` then
`alpha= (4 n + 1) pi-tan^(-1)2`
`and alpha= (4n+2) pi - tan^(-1) 2`