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What's the easiest way to understand and prove that $A \cdot B \times C = C \cdot A \times B $ ?

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Just compute $A \cdot (B \times C)$ and $C \cdot (A \times B)$ for arbitrary $A,B,C \in \mathbb{R}^{3}$ (I assume you work in three dimensional real space, otherwise what do you mean by dot and cross products?) and find out that they are equal. I don't see that there is much more to it than that. – Nils Matthes May 23 '12 at 12:53
@Nils: well. An identity involving only cross and dot products is invariant under orientation-preserving rotations, so one might hope that such a thing has a geometric interpretation that might afford a conceptually simpler proof. – Qiaochu Yuan May 23 '12 at 13:08
@NilsMatthes: although the proof is not neccesarily much simpler, the geometrical intution Qiaochu Yuan mentions is what drives the determinant-based proofs below. Of course, if one knows about wedge products, the whole thing boils down to their associativity... – Johannes Kloos May 23 '12 at 13:12
@Qiaochu: Still the easiest way to prove the above equality (for my money) is just calculating it. But I realize that my comment answers only half of the question (namely the "proof"-part) and my last sentence is false from the conceptual point of view, which you kindly pointed out. Understanding an equality is as important as seeing why it is true, something I apparently didn't have in mind while typing my comment above.;) – Nils Matthes May 23 '12 at 13:26
Thank you. Yes I am working in the three dimensional real space. – Ben May 23 '12 at 15:15
up vote 2 down vote accepted

You might make use of the fact that for $A,B,C \in \mathbb R^3$, $A \cdot (B \times C) = (A \times B) \cdot C = \det M$, where $M$ is the matrix made from the column vectors $A, B, C$. Both identities follows from the Sarrus formula for determinants of $3\times 3$ matrices.

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Show that the first is the determinant of the matrix whose rows are $A,B,C$.

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For $A = (a_1, a_2, a_3)$, $B = (b_1, b_2, b_3$, an $C= (c_1, c_2, c_3)$ you could simply just compute each side manually, so you would for example get for the left hand side:

$$\begin{align} A\cdot B\times C &= (a_1, a_2, a_3)\cdot (b_2c_3 - b_3c_2, b_3c_1 - b_1c_3, b_1c_2 - b_2c_1) \\ &= ... \end{align} $$

And then you compute the right hand side and check that you got the same thing.

Q: Is this the easiest way? A: Probably not, but it might be a good exercise in keeping track of terms.

Added: As for the understanding. The quantity $A\cdot B\times C$ is called a triple product. If the three vectors are not in the same plane, then they span a parallelepiped, and the absolute value of the triple product gives the volume of the the parallelepiped.

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The question asked for a way to "understand and prove" the claim. This might be the easiest way to prove it, but I don't think it contributes much to the understanding of it. – MJD May 23 '12 at 13:38
@MarkDominus: I added a bit more to my answer. Better? – Thomas May 23 '12 at 14:18
Thank you. Yes it is better. – Ben May 23 '12 at 15:14

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