angle of A with +ve x axis = tan^-1(3/2) = 56.31 deg suppose B makes angle @ with +ve x-axis. and x component of B is +ve so @ will be less than 90 deg so @ can be either 56.31 +45 = 101.31 or 56.31 - 45 = 11.31. so @ will be 11.31 deg. suppose magnitude of B is b then vector B = bcos11.31i+bsin11.31j and scalar p[roduct=A.B = (2i+3j). (bcos11.31i+bsin11.31j) = 3 (2bcos11.31) + (3bsin11.31) = 1.96b+0.588b = 3 b = 1.18 so vector B = 1.18cos11.31 + 1.18sin11.31 = 1.15i + 0.231 j The angle between vector A=2.00 +3.00j and vector B is 45.0. The scalar product of vectors A and B is 3.00. If the x component of vector B is positive, what is vector B.