What is the definition of unit vector? In many physics textbooks it is given the following definition of unit vector: "A unit vector is every vector whose magnitude is 1 unit". I don't like this definition.
On one hand, it is quite common to use a notation for unit vectors (for instance a hat, $\hat{u}$) different to the one used for vectors in general (usually, an arrow, $\vec{A}$). We could have a vector $\vec{A}$ in an exercise or problem and find at the end that it has a magnitude of 1 unit. ¿Would $\vec{A}$ be a unit vector? I don't think so.
I think that unit vectors arise as a consecuence of normalizing another vector $\vec{A}$, that is, by dividing it by its magnitude,
$$ \hat{u} = \frac{\vec{A}}{|\vec{A}|} $$
so that we get a vector $\hat{u}$ with just vector $\vec{A}$'s direction information.
According to units, at this point we can consider two paths:

*

*Suppose that $|\vec{A}|$ has the same units as $\vec{A}$. Then $\hat{u}$ is a dimensionless quantity.


*Suppose that $|\vec{A}|$ is dimensionless. Then $\hat{u}$ has the same units as $\vec{A}$.
I think the first option is the one that is usually used.
This last thing would also be a reason to not consider every vector with magnitude 1 unit a unit vector. In order to be a unit vector it must be dimensioness.
Is correct my definition of unit vector?
 A: $(1,0,0)={(2,0,0)\over2}$.  According to your naming scheme, the left hand side is not a unit vector but the right hand side is.  This seems to be a problem.
For that matter, whenever $(x,y,z)$ has length $1$, then $(x,y,z)={(x,y,z)\over 1}$ both is and is not a unit vector.  Oops!
A: A unit vector has magnitude $1$ - as in, the dimensionless number $1$.  Not $1\ \mathrm{cm}$ or $1\ \mathrm{kg}$ or $1\ \mathrm{N}$ or $1\ \mathrm{J}$. It's also not hard to show that for any vector $\vec A$, the dimensions of $\vec A$ and $\vert \vec A \vert$ are the same.
A: 
In many physics textbooks it is given the following definition of unit vector: "A unit vector is every vector whose magnitude is 1 unit".

Yes, that is the definition of unit vector. I cannot remember seeing any other definitions.

Is correct my definition of unit vector?

You did not actually give a definition, unless you are referring to this formula:

$$ \hat{u} = \frac{\vec{A}}{|\vec{A}|} $$

But this formula is equivalent to the definition above, as shown below.

On one hand, it is quite common to use a notation for unit vectors (for instance a hat, $\hat{u}$) different to the one used for vectors in general (usually, an arrow, $\vec{A}$).

Yes, if we know that a vector is a unit vector, we can use a different notation to convey this extra information. But we do not have to.

We could have a vector $\vec{A}$ in an exercise or problem and find at the end that it has a magnitude of 1 unit. ¿Would $\vec{A}$ be a unit vector?

Yes, it would be a unit vector. We might not say that it is a unit vector; this does not matter.

I think that unit vectors arise as a consecuence of normalizing another vector $\vec{A}$, that is, by dividing it by its magnitude,
$$ \hat{u} = \frac{\vec{A}}{|\vec{A}|} $$
so that we get a vector $\hat{u}$ with just vector $\vec{A}$'s direction information.
According to units, at this point we can consider two paths:

*

*Suppose that $|\vec{A}|$ has the same units as $\vec{A}$. Then $\hat{u}$ is a dimensionless quantity.


*Suppose that $|\vec{A}|$ is dimensionless. Then $\hat{u}$ has the same units as $\vec{A}$.
I think the first option is the one that is usually used.

It is not a choice. $|\vec{A}|$ does have the same dimension as $\vec{A}$, by definition. $|\vec{A}|$ can only be dimensionless if $\vec{A}$ itself is dimensionless.
You actually say this yourself: $\hat{u}$ has “just vector $\vec{A}$'s direction information”. This means that $\hat{u}$ is dimensionless.
A: I think that your confusion lies in the word "unit". In the definition of unit vector, "A unit vector is every vector whose magnitude is 1 unit," the word does not really refer to units like meter and second. It rather means just '1' and could be skipped.
A true unit vector has no physical dimension (like force) but only a direction.
If $\vec{F}$ is a physical vector (e.g. a force), then $|\vec{F}|$ is its size and has the same physical dimension (e.g. force), and $\hat{u} = \vec{F}/|\vec{F}|$ is a unit vector without physical dimension.
A: Just to expand on J. Murrays answer: Every unit vector is a vector, but not every vector is a unit vector. Hence, if you use the "hat" notation to indicate unit vectors its fine to write
$
\hat u = \vec u = \vec e_u = \hat e_u = \ldots
$. So, if we consider a force of magnitude 1N pointing in the $x$-direction, we  write
$\vec F = F \cdot \hat x = 1N \;\hat x = 1N \;\vec e_x$, where $\vec e_u$ is the unit vector, not $\vec F$.
A: Your problem is basically interpretation.
Definition of unit vector:
$$\hat{u}=\dfrac{\vec{A}}{|\vec{A}|}$$
So a unit vector is a mathematical object that is different from a vector of length 1 [UNIT], be it meters, Newtons, whatever. This is because by definition the unit vector has no physical units, since its given by the ratio
of the vector by its norm (and vector and norm always have the same physical units).
Thus, you can always write any vector in terms of any unit vector without any trouble having to figure out the physical units of the quantities they represent.
For instance, say you have a force $\vec{F}=[3\hat{x}+4\hat{y}]$ Newtons. And you want a particle to displace with a velocity of 10 m/s in the same direction as $\vec{F}$. You could represent it in terms of the base unit vectors, $\hat{x}$ and $\hat{y}$, by doing the scalar product, BUT YOU DON'T NEED TO DO IT.
Just write,
$$ \vec{v}=10 \dfrac{\vec{F}}{|\vec{F}|} [{m/s}]$$
This is possible, again, because by definition the unit vector has no physical unit attached to it. So it is very important to separate it from vectors of length 1 [unit].
The only remark here is that if you want to derive $\vec{v}$ written in terms of the unit vector $\vec{F}/|\vec{F}|$, and $\vec{F}$ is not constant, you have to derive $\vec{F}$ as well. This is particularly important when representing your problem in a rotating frame of reference.
