# reciprocal vectors

I don't understand some of the terminology in this question. I googled reciprocal vectors and got an article on reciprocal lattices, but I'm not sure if that is what they are talking about in this question. Also, when they say that ${\bf A}$, ${\bf B}$, and ${\bf C}$ are defined by ... plus cyclic permutations, again I looked at the wikipedia article on the subject, but I still do not understand the concept. Does anyone have a link for a clear explanation?

The vectors ${\bf a}$, ${\bf b}$, and ${\bf c}$ are non-coplanar, and form a non-orthogonal vector base. The vectors ${\bf A}$, ${\bf B}$, and ${\bf C}$, defined by

$${\bf A} = \frac{{\bf b}\times {\bf c}}{{\bf a}\cdot{\bf b}\times {\bf c}},$$

plus cyclic permutations, are said to be reciprocal vectors. Show that

$${\bf a} = \frac{{\bf B}\times {\bf C}}{{\bf A}\cdot{\bf B}\times {\bf C}},$$

plus cyclic permutations.

thanks

• Reciprocal vector: the vector which has the same direction as that of vector $A$ but has magnitude reciprocal to that of vector $A$, is called the reciprocal of vector $A$ and is denoted as vector $A^{-1}$ – shadab Ali Feb 12 at 13:28

Cyclic means a cyclic permutation of the operands: $${\bf A} \to {\bf B}, {\bf B} \to {\bf C}, {\bf C} \to {\bf A} \quad a \to b, b \to c, c \to a$$ This gives $${\bf A} = \frac{{\bf b}\times {\bf c}}{{\bf a}\cdot({\bf b}\times {\bf c})} \quad {\bf B} = \frac{{\bf c}\times {\bf a}}{{\bf b}\cdot({\bf c}\times {\bf a})} \quad {\bf C} = \frac{{\bf a}\times {\bf b}}{{\bf c}\cdot({\bf a}\times {\bf b})}$$ and $${\bf a} = \frac{{\bf B}\times {\bf C}}{{\bf A}\cdot({\bf B}\times {\bf C})} \quad {\bf b} = \frac{{\bf C}\times {\bf A}}{{\bf B}\cdot({\bf C}\times {\bf A})} \quad {\bf c} = \frac{{\bf A}\times {\bf B}}{{\bf C}\cdot({\bf A}\times {\bf B})}$$

[..] the "crystallographer's" definition, comes from defining the reciprocal lattice to be $e^{2 \pi i\mathbf{K}\cdot\mathbf{R}}=1$ which changes the definitions of the reciprocal lattice vectors to be

$\mathbf{b_{1}}=\frac{\mathbf{a_{2}} \times \mathbf{a_{3}}}{\mathbf{a_{1}} \cdot (\mathbf{a_{2}} \times \mathbf{a_{3}})}$ and so on for the other vectors. The crystallographer's definition has the advantage that the definition of $\mathbf{b_{1}}$ is just the reciprocal magnitude of $\mathbf{a_{1}}$ in the direction of $\mathbf{a_{2}} \times \mathbf{a_{3}}$, dropping the factor of $2 \pi$.

Checking the magnitude:

$$\lVert{\bf A}\rVert = \frac{\lVert {\bf b} \times{\bf c}\rVert}{\lVert{\bf a}\rVert\lVert{\bf b}\times {\bf c}\rVert\cos\angle({\bf a}, {\bf b} \times{\bf c})} = \frac{1}{\lVert{\bf a}\rVert ({\bf e_a} \cdot {\bf e}_{{\bf b} \times{\bf c}})}$$

Solving the question:

Using the "bac-cab" rule ${\bf a}\times({\bf b}\times{\bf c}) = {\bf b}({\bf a}\cdot {\bf c}) - {\bf c}({\bf a}\cdot {\bf b})$ we go for the nominator of ${\bf B}\times{\bf C}$: $$({\bf c}\times {\bf a})\times({\bf a}\times{\bf b}) = {\bf a}(({\bf c}\times{\bf a})\cdot{\bf b})-{\bf b}(({\bf c}\times{\bf a})\cdot{\bf a})= {\bf a}(({\bf c}\times{\bf a})\cdot{\bf b})$$ because ${\bf c}\times{\bf a} \perp {\bf a}$. This gives $${\bf B}\times{\bf C} = \frac{{\bf a}(({\bf c}\times{\bf a})\cdot{\bf b})}{({\bf b}\cdot({\bf c}\times {\bf a}))({\bf c}\cdot({\bf a}\times {\bf b}))} = \frac{{\bf a}}{{\bf c}\cdot({\bf a}\times {\bf b})} \\ {\bf A}\cdot({\bf B}\times{\bf C}) = \frac{{\bf b}\times {\bf c}}{{\bf a}\cdot({\bf b}\times {\bf c})} \cdot \frac{{\bf a}}{{\bf c}\cdot({\bf a}\times {\bf b})} = \frac{1}{{\bf c}\cdot({\bf a}\times {\bf b})}$$ Dividing those gives the desired result $$\frac{{\bf B}\times{\bf C}}{{\bf A}\cdot({\bf B}\times{\bf C})} = \frac{{\bf c}\cdot({\bf a}\times {\bf b})\,{\bf a}}{{\bf c}\cdot({\bf a}\times {\bf b})} = {\bf a}$$

• hey thanks, do you mean to say that we can use the magnitude to show the proof? I'm a little stumped by this one, even now that I understand the terminology. do you have any suggestions on how to approach the solution? – sean read Apr 8 '15 at 10:42
• The last bit above was to understand the last sentence of the cited text. Which proof and solution do you mean? – mvw Apr 8 '15 at 11:26
• that the question is asking for, to show from $${\bf A} = \frac{{\bf b}\times {\bf c}}{{\bf a}\cdot{\bf b}\times {\bf c}}$$ that $${\bf a} = \frac{{\bf B}\times {\bf C}}{{\bf A}\cdot{\bf B}\times {\bf C}}$$ – sean read Apr 8 '15 at 11:47
• I added that bit. – mvw Apr 8 '15 at 13:19

I think they intended to write $A\cdot (B\times C)$, which is a scalar; the vectors are non-coplanar, hence this scalar is nonzero and hence we can safely divide by this scalar.

This wiki article would be a good start.

• oh I see, so then permutations would be, for example $${\bf B} = \frac{{\bf c}\times {\bf a}}{{\bf b}\cdot{\bf c}\times {\bf a}},$$ but I'm still not sure what they mean by reciprocal vectors... – sean read Apr 7 '15 at 12:06
• @seanread It's just a name. Looking at ${\bf A} = \frac{{\bf b}\times {\bf c}}{{\bf a}\cdot{\bf b}\times {\bf c}},$, if you "cancel" $\bf b\times \bf c$, then you are left with $\bf A=\frac{1}{\bf a}$, the "reciprocal" of $\bf a$. Of course, you can't legitimately carry out this cancellation but it does help to explain the name. – Kim Jong Un Apr 7 '15 at 12:11