A particle moves through an xyz coordinate system while a force acts on it. When the particle has the position vector vector r = (2.00 m)ihat - (3.00 m)jhat + (2.00 m)khat, the force is vector F = Fxihat + (7.00 N)jhat - (6.40 N)khat and the corresponding torque about the origin is t = (5.20 N·m)ihat + (2.00 N·m)jhat + (-2.20 N·m)khat. Determine Fx.