If `veca` and `vecb` are unequal unit vectors such that `(veca -vecb) xx [(vecb+veca) xx (2veca+vecb)]=veca+vecb`, then angle `theta` between `veca` and `vecb` can be

Options

A. `pi/2`
B. 0
C. `pi`
D. `pi/4`

Answer

A, C

Explanation

`(veca-vecb) xx [(vecb + veca) xx (2veca + vecb)]=vecb + veca`
`rArr {(veca -vecb).(2veca+ vecb)}(vecb+veca)-{(veca-vecb).(vecb+veca)}(2veca+vecb)=vecb+veca`
`rArr vecb+veca=vec0` or `1-veca.vecb=1`
`rArr vecb=-veca`or `veca.vecb=0`
`rArr theta=pi` or `theta=pi/2`