Answer to Question #29445 in Vector Calculus for Akarsh Anupam

Question #29445
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.
1
Expert's answer
2013-05-02T10:50:19-0400
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.

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Comments

Assignment Expert
26.06.17, 18:29

The sequence of multiplications is not important, because A x B = - B x A.

vishwa
26.06.17, 10:44

why r we multiplying the "original equation" first by A and then by B

Assignment Expert
29.05.17, 16:01

Dear Treasa, You're welcome. We are glad to be helpful. If you liked our service please press like-button beside answer field. Thank you!

Treasa
27.05.17, 22:48

Thank you.....

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