The resultant of two vectors A and B is perpendicular to vector A and its magnitude is equal to half of the magnitude of vector B . Then the angle between A and B is :(a)30^(@)(b)45^(@)(c) 150^(@)(d)120^(@)

The resultant of two vectors A and B is perpendicular to vector A and

1.	The resultant of two vectors A and B is perpendicular to vector A and its magnitude is equal to half of the magnitude of vector B . Then the angle between A and B is :(a)30^(@)(b)45^(@)(c) 150^(@)(d)120^(@)
Answer» `30^(@)` `45^(@)` `150^(@)` `120^(@)` Solution :(i) MAGNITUDE of RESULTANT: `C^(2)=A^(2)+B^(2)+2ABcostheta` `(B^(2))/(4)=A^(2)+B^(2)+2AB((A)/(B))` `-(3)/(4)B^(2)=3A^(2)impliesB^(2)=4A^(2)` i.e., B=2A (ii) Directionof resultant `tanalpha=(Bsintheta)/(A+Bcostheta)` `oo=(Bsintheta)/(A+Bcostheta)` `:.A+Bcostheta=0` `COSTHETA=-(A)/(B)=-(A)/(2A)=-(1)/(2)` `costheta=-(1)/(2)""theta=cos^(-1)(-(1)/(2))` `theta=120^(@)`