Canonical choice for simple roots? The Dynkin diagram of a simple Lie algebra tells me about the lengths of all simple roots and about the angles of adjacent simple roots. Since the angle between non-adjacent simple roots is not specified it seems there is some freedom in choosing the simple roots.
Take for example $\mathfrak{su}(4)$. We can choose
$$
\begin{align}
\alpha_1 &= (1,0,0) \\
\alpha_2 &=\frac{1}{2}(-1,\sqrt{2},-1) \\
\alpha_3 &= (0,0,1) \\
\text{or}& \\
\alpha_3' &= (0, -\frac{2\sqrt{2}}{3}, -\frac{1}{3})
\end{align}
$$
I was wondering if there is any reason to prefer $\alpha_3$ over $\alpha_3'$ other than that $\alpha_3$ looks simpler. Is there any canonical way of choosing the root vectors?
Also in what sense does this choice not matter and how can one see this? I notice that $\alpha_3'$ cannot be written down as a linear combination of $\alpha_1, \alpha_3, \alpha_3$ with integer coefficients, so it seems the root systems constructed with $\alpha_3$ and $\alpha_3'$ will be different.
 A: First, I want to stress that there is an incorrect statement in your question as per Hagen von Eitzen's comment.
Now, what you are considering here are not different root systems, but different embeddings of the same root system into a Euclidean space, in your case of $A_3$ into $\mathbb{R}^3$ with the usual scalar product. Quite generally, there are uncountably many such embeddings, since if you start with one, you can let any orthogonal map of the Euclidean space (here, $O(3)$) act on it and get another one; and conversely, every other embedding will be the image of a fixed one under an orthogonal map. (That is more or less the precise version of the theorem that the root system is determined by the Dynkin diagram.) In very few (finitely many) cases, that orthogonal mapping will actually restrict to an automorphism of the root system. (And then has a chance to be in the Weyl group, and then you would be in the situation of Tobias Kildetoft's comment).
Now you have an extra condition, namely, you fixed choices for $\alpha_1$ and $\alpha_2$. (Note that you had uncountably many choices in both steps there.) Now with these fixed, there are only two choices for $\alpha_3$ left. Why? Because the only orthogonal maps $\in O(3)$ that leave both $\alpha_1, \alpha_2$ fixed are: 1) the identity and 2) the reflection at the plane spanned by $\alpha_1, \alpha_2$. Indeed it is easily checked that your $\alpha_3'$ is the reflection of your $\alpha_3$ at the plane spanned by $\alpha_1, \alpha_2$.
More generally, if you want to embed ("realise") a root system of rank $n$, or a base thereof, in the Euclidean space $\mathbb{R}^n$, you have uncountably many choices for each of the first $n-1$ simple roots, and once you have fixed those, you have only two choices for the last one, one being the reflection of the other at the hyperplane spanned by your first $n-1$ choices.
In very rare cases, these two choices give the same root system, that is, that reflection is not just an isomorphism of one embedding to a different one, but an automorphism of one embedding. This is obviously the case for the two-dimensional root systems $A_2$, $B_2$ and $G_2$ (no matter which of the two basis roots you fix first). Another instance is $A_3$ if you fix the outer nodes of the Dynkin diagram, $\alpha_1$ and $\alpha_3$ first. (If you look at this picture, $\alpha_1$ and $\alpha_3$ are two orthogonal vectors in the horizontal plane: after fixing those, you have two choices for $\alpha_2$, pointing away from the quadrant enclosed by your $\alpha_1$ and $\alpha_3$, but they are each other's reflections at the horizontal plane.) 
Your example seems to be the most elementary case where the reflection is not an automorphism (as you checked, $\alpha_3'$ is not in the root system with basis $\alpha_1, \alpha_2, \alpha_3$). It seems to me that in no other case the reflection at the hyperplane spanned by $n-1$ of the $n$ basis roots is an automorphism of the root system. I checked that by hand for type $A_n$. Maybe somebody has a general argument for the other cases (or a counterexample).
