Solving Algebraic Relations of Reciprocal Lattice Vectors

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The discussion revolves around proving algebraic relations involving reciprocal lattice vectors a', b', and c'. The user initially struggles with how to approach the proof, considering methods like direct component expansion and the Levi-Civita symbol. Key insights include recognizing that the vectors are orthogonal and using scalar relationships to express a, b, and c in terms of the primed vectors. The user successfully completes the proof by transforming the expressions and confirming the identities a=a, b=b, and c=c. The conversation highlights the importance of understanding vector relationships in the context of reciprocal lattices.
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Homework Statement



Basically I have reciprocal lattice vectors:

a'=\frac{b\times c}{a\cdot(b\times c)}
b'=\frac{c\times a}{a\cdot(b\times c)}
c'=\frac{a\times b}{a\cdot(b\times c)}

And I have to prove that these relations hold:

a=\frac{b'\times c'}{a'\cdot(b'\times c')}
b=\frac{c'\times a'}{a'\cdot(b'\times c')}
c=\frac{a'\times b'}{a'\cdot(b'\times c')}

The Attempt at a Solution



I really dk where to start :\

Do I try with the direct component expansion or can I do it with Levi-Civita symobol:

a\cdot(b\times c)=\varepsilon_{ijk}a_ib_jc_k and b\times c=\varepsilon_{ijk}b_jc_ke_i

And then it would be:

a'=\frac{\varepsilon_{ijk}b_jc_ke_i}{\varepsilon_{ijk}a_ib_jc_k}

but what can I do with it?
 
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from given data u can find thata`xb`=b`xc`=c`xa`=1so they are a set of orthogonal vectors

now use this to solve the required thing
 
Well I found in Arfken Weber that this is used in reciprocal lattice. Where it's also the information you gave.

In my problem I only have the three primed vectors a',\ b',\ c' with the given forms and I have to proove the following. There is no mentioning of that information, not even a hint :\
 
Note that

a\cdot b' = a \cdot c' =0,

so we can write

a = \alpha ( b'\times c')

for some scalar \alpha. Find similar expressions for b and c, using the symmetry to relate the scalars. You can compute the proportionality by computing

a\cdot b\times c.

You will need the identity

(A\times B)\times (C\times D) = (A\cdot B\times C)D - (A\cdot B\times D)C.
 
Thanks for the hint :)
 
Am I doing this right?

a\cdot b'=0\Rightarrow\frac{b'\times c'}{a'\cdot(b'\times c')}\cdot b'=\frac{b'\cdot(b'\times c')}{a'\cdot(b'\times c')}=\frac{c'\codt(b'\times b')}{c'\cdot(a'\times b')}=0

I used scalar triple product for the numerator - either b'\times b'=0 or I use the determinant and see that I have two same rows - therefore determinant is 0.

EDIT:

I've proved it! Yay for me XD

Basically you transform everything and in the end just show that a=a, b=b and c=c ^^ Thank you all for help ^^
 
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