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    				| 1. | 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 beA. `pi/2`B. 0C. `pi`D. `pi/4` | 
| Answer» Correct Answer - A::C `(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` | |