Does the unit vectors carry any unit information? Does the unit vectors carry any unit information ?
For example, when we define a vector in position space $\vec r =(2 m)\hat x + (3 m)\hat y$, the length of the vector has some unit, in this case meters, but when we talk about the unit vectors, say $\hat x$, does it have any unit ? If so, how does not affect unit of $\vec r$ ? If it doesn't have any unit, how can it has any meaning ? I mean, saying "1 in the direction of $\vec x$" does not have any meaning.
 A: In what, IMO, is the best way to think about this sort of thing....
Scalars are dimensionless
The important point being that the product of two scalars is a scalar, and if we are attaching units to things, being dimensionless is the only way this can work.
Types of units are one-dimensional vector spaces
For example, you might have a one-dimensional vector space of lengths. Some elements of this vector space would be "meter", "kilometer", "inch", and so forth. Of course, you have scalar multiplication, so you also have elements like "3 meters".
The tensor product lets you multiply vectors in general, and that is how you combine units. e.g. the tensor product of a space of times with a space of speeds would be the space of lengths. e.g. on individual vectors,
$$ \mathrm{second} \otimes (\text{meters per second}) = \mathrm{meter} $$
This combines well with scalar multiplication:
$$ (3 \text{ second}) \otimes (10 \text{ meters per second}) = (3 \cdot 10) \text{ meter} $$
You really don't want to combine units and unit vectors
For example, you might decide that $\left( \frac{3}{5} \text{meter}, \frac{4}{5} \text{meter} \right)$ should be a unit vector, because $\sqrt{(\frac{3}{5})^2 + (\frac{4}{5})^2} = 1$.
However, $\left( \frac{3}{5000} \text{kilometer}, \frac{4}{5000} \text{kilometer} \right)$ is the same vector, and you probably don't want to call that a unit vector.
Vectors with dimensions still make sense
If you have the usual (dimensionless) copy of $\mathbb{R}^3$, you could take the tensor product with, say, lengths, and then you'd something you could think of as a three dimensional space of vector distances. 
Because the space of lengths is one-dimensional, the tensor product acts a lot like scalar multiplication. For example, if $v \in \mathbb{R}^3$ is a vector, then $v \otimes \text{meters}$ would be a vector distance; of course, we usually just write the product with juxtaposition rather than $\otimes$: i.e. as $v \text{ meters}$.
Unlike vectors with dimension, dimensionless unit vectors are a reasonable notion to talk about (if we equip $\mathbb{R}^3$ with an inner product).
