Question

A body is in equilibrium under the action of three force vectors vector a, vector b and vector c similtaneously. show that vector a× vector b=vector b× vector c=vector c× vector

a.

Answer 1

Start with the given information: A+B+C=0A+B+C=0. Cross with one of the vectors, say AA. Note that A×A=0A×A=0. Note that B×A=−A×BB×A=−A×B. Conclude that A×B=C×AA×B=C×A. Similarly for the other equations.

A geometric argument: The three vectors are the sides of a triangle in order. The area of the triangle is half the cross product of two of them in that order.

Answer 2

Answer:

Hence proved.

Explanation:

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.