Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
    
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
1  
@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

3 Answers 3

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.

share|improve this answer

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

share|improve this answer

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.

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.