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

In summary, the homework statement is trying to find a formula for A x (B x C). convert B x C into A B and C and solve for A.
  • #1
thercias
62
0

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.
 
Physics news on Phys.org
  • #2
hi thercias! :wink:

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 :smile:
 
  • #3
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...
 
  • #4
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.
 
  • #5
hi thercia! :smile:
thercias said:
(A'x(B'xC'))/(A'.B'xC')

what are you doing?? :confused:

you need to convert B'xC' into A B and C​
 
  • #6
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?
 
Last edited:
  • #7
(CxA) x (AxB)

[put CxA = P]

= P x (Q x R) = (P.R)Q etc :wink:
 
  • #8
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.
 
  • #9
think!

CxA.A = … ? :smile:
 
  • #10
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')
 
  • #11
that's it, isn't it? :wink:
 
  • Like
Likes 1 person
  • #12
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
 

1. What is the purpose of proving A = B'xC'/(A'.B'xC')?

The purpose of proving A = B'xC'/(A'.B'xC') is to show mathematically that this equation is true and to provide a logical explanation for its validity. This proof can be used to solve problems and make further calculations using this equation.

2. What are the steps involved in proving A = B'xC'/(A'.B'xC')?

The steps involved in proving A = B'xC'/(A'.B'xC') include simplifying the equation, identifying any patterns or relationships, applying mathematical properties and rules, and using logical reasoning to connect each step to the final solution.

3. How can I use the proof of A = B'xC'/(A'.B'xC') in real-life situations?

The proof of A = B'xC'/(A'.B'xC') can be used in various real-life situations, such as in engineering, computer science, and data analysis. It can also be applied in problem-solving and decision-making processes that involve logical reasoning and mathematical calculations.

4. Are there any limitations to the proof of A = B'xC'/(A'.B'xC')?

Like any mathematical proof, there may be certain limitations to the proof of A = B'xC'/(A'.B'xC'). These limitations may include specific assumptions or conditions that must be met for the proof to be valid, as well as the potential for human error in the calculation and reasoning process.

5. Can this proof be applied to other similar equations?

Yes, the proof of A = B'xC'/(A'.B'xC') can be applied to other similar equations that involve logical reasoning and mathematical properties. However, the specific steps and methods used may vary depending on the equation and its unique characteristics.

Similar threads

  • Calculus and Beyond Homework Help
Replies
1
Views
2K
Replies
6
Views
7K
  • Calculus and Beyond Homework Help
Replies
3
Views
2K
  • Calculus and Beyond Homework Help
Replies
6
Views
852
  • Calculus and Beyond Homework Help
Replies
2
Views
454
  • Calculus and Beyond Homework Help
Replies
2
Views
180
  • Calculus and Beyond Homework Help
Replies
1
Views
750
Replies
0
Views
402
  • Calculus and Beyond Homework Help
Replies
16
Views
4K
  • Calculus and Beyond Homework Help
Replies
1
Views
909
Back
Top