# Cross product invariant under SO(3)-matrices?

1. Apr 30, 2010

### Marin

Hi there!

I'm trying to prove the following obvious statement, but am somehow stuck :(

Let $$\vec a,\ \vec b\in\mathbb{R^3}$$ , let M be in SO(3) and x be the cross product

prove: $$M(\vec a\times\vec b)=M\vec a\times M\vec b$$

I tried using the epsilon tensor, as in physics, but it doesn't really produce an opportunity, as you can convince yourself:

$$(M\vec a\times M\vec b)_i=\varepsilon_{ijk}(M\vec a)_j(M\vec b)_k=\varepsilon_{ijk}M_{jl} a_l M_{km}b_m...$$

where as usual summation is over repeated indices

Now I want to use the fact that M is orthogonal, i.e.
$$M_{ij}M_{jk}^t=\delta_{ik}$$,
and preserves the orientation of the basis but I don't know where exactly this has to come into the proof...

What I want to end up with is

$$...=M_{ij}\varepsilon_{jkl}a_kb_l=(M(\vec a\times\vec b))_i$$

The statement seems to me obvious and can be envisioned very quickly by the right-hand-rule; I don't know why establishing it makes real problems

If you have any idea, I'd be glad to see it :)

With regards,
marin

2. Apr 30, 2010

### Simon_Tyler

Hi Marlin,

I'll give you a hint (mainly 'cause I'm too lazy to latex up the answer)
The 'preserving the orientation of the basis' bit is equivalent to saying det(M)=1
Rewrite the determinant using the epsilon tensor then a little algebraic rearrangement should get you the result.
(The formula I'm thinking of can be found http://en.wikipedia.org/wiki/Cross_product#Algebraic_properties"...)

Last edited by a moderator: Apr 25, 2017
3. May 1, 2010

### Live2Learn

Last edited by a moderator: Apr 25, 2017
4. May 1, 2010

### Marin

Thanks a lot :)

5. May 5, 2010

### Live2Learn

More Generally
Where A is any invertable 3x3 matrix and M(A) is the matrix of the minors of A

$$A\vec a\times A\vec b=M(A)(\vec a\times\vec b)$$

I got this result by playing with the algebra. Hope I didn't make a mistake.