Proving vectors are rotations about some axis

Question

My question arises from having trouble understanding this question and answer:

I have trouble understanding the reasoning. Can anyone explain?

Answer 1

The last sentence of the proof is not really related to the rest of the proof. Fully written it can look as follows (disclaimer! I got a bit of a bruteforce approach):

Imagine a linear map $R$ with a matrix $\begin{bmatrix}0&0&1\\1&0&0\\0&1&0\end{bmatrix}$. It sends $i$ to $j$, $j$ to $k$ and $k$ to $i$. Also, this map is essentially our rotation, but we kinda do not know it yet.

We know though that this map is a rotation in $\mathbb R^3$ since it is an orthogonal matrix with positive determinant.

$a = (1,1,1)^T$ is invariant under the map (this is the moment which has something to do with symmetry), since it is the rotation axis.

All we need to prove that the rotation angle is indeed $120°$.

To find the angle we can take a vector $v$ that is orthogonal to $a$ and calculate angle between $v$ and its image under the rotation.

Take for example $v = (1,-1,0)^T$, $R(v) = (0,1,-1)^T$, the angle is $120°$.

The last thing we should do is to show that rotation is in right direction, to do that we need to show that $\det(v,R(v),a)$ is positive.

As a result we get that $R$ is indeed the described rotation and, as stated earlier, it sends $i$ to $j$, $j$ to $k$ and $k$ to $i$. q.e.d.

Answer 2

About symmetry and stuff:

Imagine you look at vectors $i$,$j$,$k$ along the vector $a = (1,1,1)^T$. Then they will look in different directions and will be of the same size, as if they are pointing in 3 vertices of an equilateral triangle (with $a$ pointing in the center of that triangle, downwards).

So, from that perspective it indeed looks like $i$ maps to $j$ maps to $k$ maps to $i$ again.

The only problem is that maybe their height/depth (the only thing which we do not see) is different, but the first part of the proof says that $i-j\perp a$, so the height is the same (same with $j$ and $k$).

I don't think this is a formal proof, but it is easier to visualize, I guess.