From the analogy of couples to translations which was pointed out in § 7, we may infer that a couple is sufficiently represented by a "free" (or non-localized) vector perpendicular to its plane. The length of the vector must be proportional to the moment of the couple, and its sense must be such that the sum of the moments of the two forces of the couple about it is positive. In particular, we infer that couples of the same moment in parallel planes are equivalent; and that couples in any two planes may be compounded by geometrical addition of the corresponding vectors. Independent statical proofs are of course easily given. Thus, let the plane of the paper be perpendicular to the planes of two couples, and therefore perpendicular to the line of intersection of these planes. By § 4, each couple can be replaced by two forces ± P (fig. 43) perpendicular to the plane of the paper, and so that one force of each couple is in the line of intersection (B); the arms (AB, BC) will then be proportional to the respective moments. The two forces at B will cancel, and we are left with a couple of moment P · AC in the plane AC. If we draw three vectors to represent these three couples, they will be perpendicular and proportional to the respective sides of the triangle ABC; hence the third vector is the geometric sum of the other two. Since, in this proof the magnitude of P is arbitrary, It follows incidentally that couples of the same moment in parallel planes, e.g. planes parallel to AC, are equivalent. [Illustration: FIG. 43.] [Illustration: FIG. 44.] Hence a couple of moment G, whose axis has the direction (l, m, n) relative to a right-handed system of rectangular axes, is equivalent to three couples lG, mG, nG in the co-ordinate planes. The analytical reduction of a three-dimensional system can now be conducted as follows. Let (x1, y1, z1) be the co-ordinates of a point P1 on the line of action of one of the forces, whose components are (say) X1, Y1, Z1. Draw P1H normal to the plane zOx, and HK perpendicular to Oz. In KH introduce two equal and opposite forces ± X1. The force X1 at P1 with -X1 in KH forms a couple about Oz, of moment -y1X1. Next, introduce along Ox two equal and opposite forces ±X1. The force X1 in KH with -X1 in Ox forms a couple about Oy, of moment z1X1. Hence the force X1 can be transferred from P1 to O, provided we introduce couples of moments z1X1 about Oy and -y1X1, about Oz. Dealing in the same way with the forces Y1, Z1 at P1, we find that all three components of the force at P1 can be transferred to O, provided we introduce three couples L1, M1, N1 about Ox, Oy, Oz respectively, viz. L1 = y1Z1 - z1Y1, M1 = z1X1 - x1Z1, N1 = x1Y1 - y1X1. (5) It is seen that L1, M1, N1 are the moments of the original force at P1 about the co-ordinate axes. Summing up for all the forces of the given system, we obtain a force R at O, whose components are X = [Sigma](X_r), Y = [Sigma](Y_r), Z = [Sigma](Z_r), (6)