# Proved A x [B x (Cx D) ] = 0

Hello I hope you can help me in solving this proof

proved A x [B x (Cx D) ] = 0

Thank you

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LCKurtz
Homework Helper
Gold Member
Hello Samia, welcome to PF. Since this is your first post you may not have read the rules, one of which is you must show what you have tried. Nevertheless I will give you a Hint: Try the calculation with 4 different vectors.

Deveno
Hello Samia, welcome to PF. Since this is your first post you may not have read the rules, one of which is you must show what you have tried. Nevertheless I will give you a Hint: Try the calculation with 4 different vectors.
that's just...cruel.

presumably A,B,C and D lie in R3?

if so, A must be a linear combination of B,C, and D.

the cross-product is distributive, is it not?

from here it get easy....because of a certain elementary property of cross-products.

LCKurtz
Homework Helper
Gold Member
that's just...cruel.

presumably A,B,C and D lie in R3?

if so, A must be a linear combination of B,C, and D.

the cross-product is distributive, is it not?

from here it get easy....because of a certain elementary property of cross-products.
How does it get easy when it is false? And what is cruel about suggesting that most anything will give a counterexample?

Deveno
How does it get easy when it is false? And what is cruel about suggesting that most anything will give a counterexample?
except, obviously, for the ones i tried. please excuse me while i shave my facial omelette.

This theorem can't be proved, because (as noted by LCKurtz) it is simply false. A simple counter can be shown by letting A=B=C=i, D=j where i,j, and k are the unit vectors in the x,y, and z directions (respectively). Then

A x (B x (C x D))=i x (i x (i x j))
=i x (i x (k))
=i x (-j)
A x (B x (C x D))=-k0