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

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
Answer» SOLUTION :Given :`theta = 90^(@) , r = (b)/(2)` Formula :`r^(2) = (a^(2) +b^(2) + 2ab cos PHI) "". . . (1)` `tan theta = (b sin phi)/(a + b cos phi)"" . . . (20` Since,`theta = 90^(@) , tan 90^(@) = infty` i.e., `a + b cos phi = 0 (or) a = - b cos phi "". . . (3)` (3) in (1) ` r^(2) = (( - b cos phi)^(2) + b^(2) + 2 (-b cos phi) b cos phi )` `= b^(2) - b^(2) cos^(2) phi` `= b^(2) (1 - cos^(2) phi)` `r^(2) = b^(2) sin^(2) phi` `sin phi = pm (r)/(b) = pm (b)/(2b) = pm (1)/(2)` `phi = 30^(@) (or) 150^(@)` `phi`cannot be`30^(@)` as, resultant MAKES an angle `90^(@) `with a . `:. a and b is 150^(@)` `:. Ans: (C)150^(@)`