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.


  • $\begingroup$ 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 $A^{-1}$ $\endgroup$
    – shadab Ali
    Feb 12 '19 at 13:28
  • 3
    $\begingroup$ @shadabAli This is totally wrong. And no mathematician would ever write $A^{-1}$ unless $A$ is a nonsingular (square) matrix. $\endgroup$ Aug 21 '20 at 17:46

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 $$ {\bf a} = \frac{{\bf B}\times {\bf C}}{{\bf A}\cdot({\bf B}\times {\bf C})} \quad {\bf b} = \frac{{\bf C}\times {\bf A}}{{\bf B}\cdot({\bf C}\times {\bf A})} \quad {\bf c} = \frac{{\bf A}\times {\bf B}}{{\bf C}\cdot({\bf A}\times {\bf B})} $$

From Reciprocal Lattice

[..] 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 $$ \frac{{\bf B}\times{\bf C}}{{\bf A}\cdot({\bf B}\times{\bf C})} = \frac{{\bf c}\cdot({\bf a}\times {\bf b})\,{\bf a}}{{\bf c}\cdot({\bf a}\times {\bf b})} = {\bf a} $$

  • $\begingroup$ 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? $\endgroup$
    – sean read
    Apr 8 '15 at 10:42
  • $\begingroup$ The last bit above was to understand the last sentence of the cited text. Which proof and solution do you mean? $\endgroup$
    – mvw
    Apr 8 '15 at 11:26
  • $\begingroup$ 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 $$ {\bf a} = \frac{{\bf B}\times {\bf C}}{{\bf A}\cdot{\bf B}\times {\bf C}} $$ $\endgroup$
    – sean read
    Apr 8 '15 at 11:47
  • $\begingroup$ I added that bit. $\endgroup$
    – mvw
    Apr 8 '15 at 13:19
  • $\begingroup$ It's worth noting that a . b x c = b . c x a = c . a x b is, as they are each the volume of the parallepiped whose sides are a, b, and c. $\endgroup$ Mar 18 '21 at 20:09

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.

  • $\begingroup$ 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... $\endgroup$
    – sean read
    Apr 7 '15 at 12:06
  • 1
    $\begingroup$ @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. $\endgroup$ Apr 7 '15 at 12:11

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