Is there a uniform way to define angle bisectors using vectors? 
Look at the left figure. $x_1$ and $x_2$ are two vectors with the same length (norm). Then $x_1+x_2$ is along the bisector of the angle subtended by $x_1$ and $x_2$. But look at the upper right figure. When $x_1$ and $x_2$ are collinear and in reverse directions, $x_1+x_2=0$ and no longer represent the bisector of the angle (in this case 180 deg). The bisector should be perpendicular to $x_1$ and $x_2$. (The $x_1+x_2$ works well for the case shown in the lower right figure.)
Question: Is there a way to represent the bisector for all the three cases? I don't want to exclude the upper right case. Is it possibly helpful to introduce some infinity elements?
 A: In the plane, define $P(x_1,x_2)=\{y:y^T(x_1-x_2)=0\}.$ 
Then define the bisecting line for $x_1,x_2$ of the same length as the collection of vectors
$$\{s(x_1+x_2)+ty : s,t \in R, y\in P(x_1,x_2)\}.$$
This is a bit ungainly but it works in the three cases, since in the "bad cases" for $x_1+x_2$ the $s(x_1+x_2)$ term is the zero vector but the set $P(x_1,x_2)$ is not just the zero vector in this case, and its vectors point along the line bisecting the angle.
This answer is really only a combination of the ideas mentioned above in the comments, but I think it does give a single definition for all the cases.
A: A reasonable definition of "bisector" is the set of vectors $y$ such that $y^T(x_1-x_2) = 0$. If you are willing to accept this as the definition, then it's completely general, and does not involve any special cases.
However, this set is a plane, not a line or a vector. It's a plane perpendicular to the "paper" in your pictures. If you want your "bisector" to be a vector, you need to intersect the bisector plane with some other plane. In your pictures, the "other plane" that's being used is the plane of the two vectors. But, if the two given vectors are parallel, then this construction doesn't work.
So, in short -- "bisector plane" can be defined generally, but "bisector vector" can not (as far as I can see).
All of the above is pretty sloppy, mathematically. I have deliberately confused "affine" and vectorial concepts. Explaining in rigorous terms would take longer, and it wouldn't be any more helpful, I suspect.
A: I also would like to give a solution, which I am currently using in my work.
The key idea is to use a rotation matrix.
Suppose the angle between $x_1$ and $x_2$ is $\theta$. Let $R(\theta/2)$ be a rotation matrix, which can rotate a vector $\theta/2$. Then
$$y=R(\theta/2)x_1$$
is a unified way to express the bisector.
Of course, we need also pay attention to the details, which can be determined straightforwardly:

*

*the rotation matrix rotates a vector clockwise or counter-clockwise?


*how to define the angle $\theta$?


*the bisector should be $y=R(\theta/2)x_1$ or $y=R(\theta/2)x_2$?
EDIT: I give an example here. Consider two unit-length vectors $x_1$ and $x_2$, which will give two angles: one is in [0,pi] and the other is in (pi,2pi). We can define the angle $\theta$ such that rotating $x_1$ counterclockwise $\theta$ about the origin yields $x_2$. Here $\theta\in[0,2\pi)$. Consequently define the rotation matrix $R(\theta/2)$ rotates a vector counterclockwise $\theta/2$. (The formula of this kind of R is given here) Thus $R(\theta/2)x_1$ is a unit-length vector lying on the bisector of $\theta$.
Another thing as mentioned by coffemath is that: how to compute the angle given two vectors? Of course, it is not enough to only use $\cos \theta=x_1^Tx_2$ because $\cos \theta$ gives two angles whose sum is $2\pi$. However, if we carefully define the angle $\theta$ and $R$ we can also compute $\sin \theta$. For example, we define the angle and rotation matrix as above mentioned. Then define $x_2^{\perp}=R(\pi/2)x_2$. Then it can be calculated that $x_1^Tx_2^{\perp}=-\sin \theta$. hence from both $\cos\theta$ and $\sin\theta$, we can compute $\theta$.
