Answer to Question #29445 in Vector Calculus for Akarsh Anupam

Question

A body is in equilibrium under the action of three force vectors A, B and C simultaneously. Show that A X B = B X C = C X A.

Answer 1

If the body is at equilibrium, due to the 2rd Newton'sLaw,
A + B + C = 0.
By multiplying it (in sense of vector product) by A from theleft,
we get
AxA + AxB + AxC = 0.
Some known rules:
1) Any vector product of collinear vectors is 0, so AxA =BxB = CxC = 0.
2) Changing the order of multipliers in the productchanges the sign: AxB = - BxA.
Thus, we have already AxB = - AxC = CxA. [i]
By multiplying the original equation in the same fashion byB, we get
BxA + BxC = 0;
AxB = - BxA = BxC. [ii]

Results [i] and [ii] provide enough evidence to say that A X B = B X C = C XA.