Is a unit vector really unitless and dimensionless? According to my textbooks, a unit vector has no units and no dimensions, but is only used to specify direction. It only shows the orientation of a corresponding vector in space. I think it's true, or that's what it looks like. It makes sense because a unit vector is defined as 'a vector' divided by its magnitude. Since we have the same numerical value in numerator and the denominator, a unit vector has a magnitude of 1 unit. Likewise, we have the same unit in both numerator and the denominator, that makes a unit vector 'unitless', and hence dimensionless. That's why I think a unit vector has no dimensions. Please correct me if I'm wrong.
But, another question naturally comes to our mind. Why if I say, "a force of 1 N due east" or "a displacement of 1m, 30° NOE"?
Both force and displacement are vector quantities, and both have a magnitude of 1 unit in the above two examples. 
My question is, can we call these two "unit vectors"? That's what I'm struggling to understand. There's no reason why we can't call these two unit vectors. Because both have a magnitude of 1 unit, and both are vectors. However, both have units, and hence both are not dimensionless.
 A: If $\vec{v}$ is a vector with a physical unit, then its unit vector is defined as:
$$\hat{v}=\frac{\vec{v}}{||\vec{v}||}$$
Where:
$$||\vec{v}||=\sqrt{\sum_i v_i^2}$$
where every components $v_i$ has the physical unit. This clearly means that the unit vector is dimensionless. 
A: Something to realize is that your vector of magnitude $1\ \rm N$ only has "unit" length because you chose to measure or represent your force in Newtons. If you chose some other unit, like pounds, then you would not have $1$ pound of force.
On the other hand, your actual unit vectors are indeed unitless$^*$. This is because unit vectors are defined as the ratio between two things with the same units. They will always have a (unitless) magnitude of $1$. In fact, this is true for any unitless quantity, since they do not depend on your choice of units (which is an intentionally redundant statement).

$^*$ I have always found this amusing. Unit vectors are unitless. 
A: 
Why if I say, "a force of 1 N due east" ?? Or "a displacement of 1m, 30° NOE" ?
  Both force and displacement are vector quantities, and both have a magnitude of 1 unit in the above two examples. My question is, can we call these two "unit vectors" ??

No, those are not unit vectors. Let $\textbf{F}=(1\ \text{N})\hat{\textbf{x}}$. (Some people notate $\hat{\textbf{x}}$ as $\hat{\textbf{i}}$.) Then the unit vector in the direction of $\textbf{F}$ is
$$\frac{\textbf{F}}{|\textbf{F}|} = \hat{\textbf{x}} \, ,$$
which is not the same as $\textbf{F}$. It has different units, and it is not true that $|\textbf{F}|=|\hat{\textbf{x}}|=1$, since things with incompatible units can never be equal.
A: Well, if you have a dimensionless unit vector $\mathbf{i}$, that uniquely defines a unit vector for force $\mathbf{i}_F = \mathbf{i} \cdot 1\,\mathrm{N}$, same for displacement $\mathbf{i}_D = \mathbf{i} \cdot 1\,\mathrm{m}$ and for any other units you care about. Given that all of these are related in a well-defined manner, nothing’s stopping you from expressing your quantities in terms of a force basis should you so choose. General relativity, QFT and certain other branches of physics involve basis changes which are way more complicated than the question of whether $|\mathbf{i}|$ equals 1 or 1 N. 
