If `veca` and `vecb` are two non collinear unit vectors and `iveca+vecbi=sqrt(3)` then find the value of `(veca-vecb).(2veca+vecb)`
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`because a=b=1` so `a^(2)+b^(2)+2ab cos theta=(sqrt(3))^(2)=3rArr1+1+2cos theta=3thetacos theta=(1)/(2)` `(veca-vecb).(2veca+vecb)=2a^(2)-b^(2)-abcos theta=2-1-(1)/(2)=(1)/(2)`
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