Is A x [B x (C x D)] Always Equal to Zero?

[SAMIA]

Hello I hope you can help me in solving this proof

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


Thank you

 

[LCKurtz]

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]

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.

 

[augray]

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))=-k≠0