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

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  • #1
SAMIA
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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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  • #2
LCKurtz
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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.
 
  • #3
Deveno
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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.
 
  • #4
LCKurtz
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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?
 
  • #5
Deveno
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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.
 
  • #6
augray
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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
 

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