**What's Gone Before and What Will Ensue**: Yesterday, he first step of an exercise regarding the rotation of cross products was worked out. Today, the identity proved yesterday will be used to show that when the two vectors in a cross product are rotated by the same rotation matrix, the resulting vector of the cross product is rotated by the same rotation matrix. In the end, yet another property will be proven using tensor index notation.

The identity from yesterday is:

$\epsilon_{ijk}W_{iq}W_{jl}W_{km} = det\left(W\right)\epsilon_{qlm}$

We'll also need the definition of the cross product in index notation

$\vec{A} \times \vec{B} = \epsilon_{ijk}A_jB_k$

and the rotation matrix transpose identity

$M^TM = 1$ also known as $M_{iq}M_{in} = \delta_{qn}$

We want to prove that if $\vec{A}$ and $\vec{B}$ are both rotated by the same rotation matrix, $M_{in}$, then so is the result of the cross product, $\vec{V}$

First, rotate the two input vectors

$\vec{A^{\prime}} \times \vec{B^{\prime}} = \epsilon_{ijk}M_{jl}A_l M_{km}B_m = V^{\prime}$

We'd like to show that $\vec{V}^{\prime} = M_{in}V_n$

Go ahead and substitute the desired result in for the right hand side now. We'll see if we can make the two sides equal each other, completing the proof.

$\epsilon_{ijk}M_{jl}A_l M_{km}B_m = M_{in}V_n$

Now, rotate each side one more time using the same rotation matrix to get

$\epsilon_{ijk}M_{iq}M_{jl} M_{km} A_lB_m = M_{iq} M_{in}V_n$ .

Using the identities stated above, this gives

$\epsilon_{qlm} A_l B_m = \delta{qn}V_n$ or

$\epsilon_{qlm}A_l B_m = V_q$

The left hand side is the cross product of the unrotated input vectors and the right hand side is the result of that cross product. That wraps up the proof.

These are kind of dry posts, so let's re-re-re-introduce the

**Picture of the Day**!

This one is from Young Parisians In Lub!

## Comments