Proving A = B'xC'/(A'.B'xC'): A Mathematical Solution

[thercias]

Homework Statement

if A' = BxC/(A.BxC)
B' = CxA/(A.BxC)
C' = AxB/(A.BxC)

prove that A = B'xC'/(A'.B'xC')
B = C'xA'/(A'.B'xC')
C=(A'xB')/(A'.B'xC')

Homework Equations

The Attempt at a Solution

I was sort of lost on how to do this. First I tried to simplify a', b', and c' and plug those into a, b, c but that didn't work out.
then i took the dot product of A' with A
so A'.A = A.(BxC)/(A.BxC) = 1
doing the same for B' and C' makes the answer 1 too. then i did the dot product of A with A' and got 1, and same result for B and C. I'm not sure if that means anything, but now I'm kind of stuck on how to solve this.

 

[tiny-tim]

hi thercias!

do you know the formula for A x (B x C) ?

(if not, look it up!)

apply that to the numerator (top) of B' x C' …

show us what you get ​

 

[hilbert2]

The vectors in that problem are defined exactly the same way as the reciprocal lattice vectors in crystallography and solid state physics. In the problem you're basically being asked to show that the reciprocal of the reciprocal lattice is the original lattice...

 

[thercias]

Hi, yes I do know the formula for AxBxC. So if we use that for a I get
(A'x(B'xC'))/(A'.B'xC')
=(B'(A'.C')-C'(A'.B'))/(A'.B'xC')
=B'(A'.C')/(A'.B'xC') - C'(A'.B')/(A'.B'xC')

but this looks kind of messy doesn't it? I'm not quite sure how this will lead to a solution.

 

[tiny-tim]

hi thercia!

(A'x(B'xC'))/(A'.B'xC')

what are you doing??

you need to convert B'xC' into A B and C​

 

[thercias]

Truthfully, I'm not really sure. So If I'm following correctly, I convert B'xC' into
CxA/(A.BxC) x AxB/(A.BxC)
but now I'm stuck on how you're supposed to apply Ax(BxC) on this part. I'm not sure if I'm doing something wrong again.
am I supposed to find CxA x AxB?

 

[tiny-tim]

(CxA) x (AxB)

[put CxA = P]

= P x (Q x R) = (P.R)Q etc

 

[thercias]

P x (A x B)
= A(P.B) - B(P.A)

so b'xc' = (A(CxA.B)-B(CxA.A))/(A.BxC)
= A(CxA.B)/(A.BxC) - B(CxA.A)/(A.BxC)
CxA.B and A.BxC are the same so they cancel
= A-B(CxA.A)/(A.BxC) I'm not sure what relationship CxA.A and A.BxC have but I probably have to simplify that somehow.

 

[tiny-tim]

think!

CxA.A = … ? ​

 

[thercias]

hm, isn't it just zero? now that i think about it it can be represented by a determinant, and it will have repeated rows making it 0.

so b'xc' = A

so a = A/(A'.B'xC')
=A/(A'.A)

and a' = A'/(A.A')

 

[tiny-tim]

that's it, isn't it?

 

[thercias]

I guess it is! Thanks for the help.
I noticed that if you sub a' into
a= a/(a.a') we get
a=a(a.bxc)/(a.bxc)
a=a