If the force and couple of an equivalent forcecouple system are perpendicular
to one another, we have the equation
F_{R} • M_{RP}
= 0
The system can be reduced to a single force F_{R} located
at a point Q. The moments of the two systems can be equated
about the point P giving,
M_{RP} = r_{PQ}
× F_{R}
(4)
By substituting Eqs. 1, 2, and 3 into Eq. 4; performing the cross products;
and equating the i, j, and k components; the three scalar equation are,
Σ(r_{PFy}
F_{z}  r_{PFz} F_{y} ) = r_{PQy} ΣF_{z}
 r_{PQz} ΣF_{y}
Σ(r_{PFz}
F_{x}  r_{PFx} F_{z} ) = r_{PQz} ΣF_{x}
 r_{PQx} ΣF_{z}
Σ(r_{PFx}
F_{y}  r_{PFy} F_{x} ) = r_{PQx} ΣF_{y}
 r_{PQy} ΣF_{x}
Solving this system of three equations for the three unknowns r_{PQx},
r_{PQy}, and r_{PQz} gives the position vector r_{PQ}
of the force F_{R},
r_{PQ} = r_{PQx}i
+ r_{PQy}j + r_{PQz}k
F_{R} = ΣF
