Unit vector of the direction of v? I was going over some of my notes and I saw that I wrote the following under a Resolving the Vector heading in my Direction Angles notes:
$$\vec u = \frac {\vec v}{\|\vec v\|} = \langle \cos(\theta), \sin(\theta)\rangle$$
Can someone explain this formula to me and provide an example of how to use it?
 A: You can think of a vector as being comprised of (or associated with) three different things: $(1)$ a linear subspace, $(2)$ an orientation, and $(3)$ a magnitude.
A linear subspace means the vector is associated with one unique line that passes through the origin.  An orientation means it points toward one side or the other of that line.  A magnitude is just a scalar (e.g. a real number).
Thus we could say that a vector is an oriented, weighted linear subspace.
What does "unit vector" mean?  It means that the magnitude of that vector is $1$.  So the weight of that oriented linear subspace is "unity" (the fancy mathematical term meaning $1$).
Vectors can be acted on be scalars to change their weight.  For instance if the vector $\vec v$ has a magnitude (denoted $\|\vec v\|$) of $4$, then if it is acted on (by "scalar multiplication") by the scalar $\frac 12$, the orientation and linear subspace associated with $\vec v$ stay exactly the same.  The weight is what changes: it becomes $2$.
So if you want to make a unit vector out of some arbitrary vector with magnitude $\|\vec v\|$, you just need to multiply it by the reciprocal of that magnitude: $\dfrac 1{\|\vec v\|}\vec v$ is the unit vector in the direction of $\vec v$.

That is one general interpretation of vectors.  However, to see why in the specific case of $\Bbb R^2$ unit vectors have the form $(x,y)=(\cos(\theta),\sin(\theta))$, it's better to consider complex numbers.
Any vector in the plane $\Bbb R^2$ can be represented by a complex number.  Complex numbers can be written in two ways: Cartesian form and polar form.
The Cartesian form of a complex number $z$ is $z= x+iy$, where $x, y \in \Bbb R$ and $i$ is the imaginary unit.  $i$ is a special number which doesn't exist in the ordinary "real numbers", with the property that $i^2=-1$.  The $x$ and $y$ here are indeed the $x$ and $y$ coordinates of a vector in the plane.
The polar form of a complex number $z$ is $z=re^{i\theta}$, where $r,\theta \in \Bbb R$.  Here $r$ is the magnitude of the vector and $\theta$ is the angle the vector makes with the positive $x$-axis.  The relationship between the two forms is called Euler's formula: $re^{i\theta} = r(\cos(\theta) + i\sin(\theta))$.
So if a vector in the plane is a unit vector, then it is represented in polar form as $z=1e^{i\theta} = e^{i\theta} = \cos(\theta) + i\sin{\theta}$.  Thus the $x$ coordinate of such a vector is $\cos(\theta)$ and the $y$ coordinate is $\sin(\theta)$.
A: (Edited now that the question has ben edited/restored.)
The vector $\mathbf{u}$ is a vector that has the same direction as the original vector $\mathbf{v}$, but has length (magnitude) equal to $1$.
Lets see how that relates to direction angles and what this $\theta$ is. 
Lets consider $2$-dimensional vectors, i.e., vectors with $2$ coordinates; lets call them $x$, $y$. Let
$$
\mathbf{v} = 
\begin{bmatrix}
v_{x}\\
v_{y}
\end{bmatrix},
$$
be any vector. 
Then, 
$$
\mathbf{u} = 
\begin{bmatrix}
u_{x}\\
u_{y}
\end{bmatrix}
$$
is computed based on $\mathbf{v}$ as follows:
$$
u_{x} = \frac{v_{x}}{\sqrt{v_{x}^{2}+v_{y}^{2}}} \quad\text{and} \quad
u_{y} = \frac{v_{y}}{\sqrt{v_{x}^{2}+v_{y}^{2}}}.
$$
The denominator is the length of $\mathbf{v}$ as the definition requires. 
You can also verify that the length of $\mathbf{u}$ is equal to $1$.
We know that $\mathbf{u}$ is (by construction) a vector on the unit circle. Hence, it can be written in the form 
$$
\mathbf{u} = 
\begin{bmatrix}
\cos \theta\\
\sin \theta
\end{bmatrix},
$$
where $\theta$ is the angle of $\mathbf{u}$ with the $x$-axis. (I assume you are familiar with the trigonometric circle.)
From that we infer that $u_{x}=\cos\theta$ (and hence you can compute $\theta$ easily). 
Note that for a vector $\mathbf{u}$ on the unit circle we could also write
$$
\mathbf{u} = 
\begin{bmatrix}
\cos \theta_{x}\\
\cos \theta_{y}
\end{bmatrix},
$$
where $\theta_{x}$ is the angle of $\mathbf{u}$ with the $x$-axis ($\theta_{x}$ is the same as $\theta$ before),
and $\theta_{y}$ is the angle of $\mathbf{u}$ with the $y$-axis.
From that alternative description we can get immediately the angle with each axis by looking at the appropriate coordinate of $\mathbf{u}$.
Of course, the two descriptions are equivalent (Why?).
Now, why is this useful? Note that $\mathbf{u}$ points in exactly in the same direction as $\mathbf{v}$. We only normalized the magnitude. So $\mathbf{u}$ and $\mathbf{v}$ have the same direction angles. But from $\mathbf{u}$ we get the direction angles in a pretty straightforward way from the corresponding coordinates. Hence, you can see that as a way of computing the direction angles of $\mathbf{v}$: first normalize the length to get $\mathbf{u}$ and subsequently compute the angles $\theta_{x} = \cos^{-1}u_{x}$ and $\theta_{y} = \cos^{-1}u_{y}$.
A: To take a specific example, let $\vec v=(5,7)$, then $|v|=\sqrt {5^2+7^2}=\sqrt {74}$  A unit vector in the same direction as $\vec v$ is $\vec u=\frac {\vec v}{|\vec v|}=\frac 1{\sqrt{74}}(5,7)=\left(\frac 5{\sqrt74},\frac 7{\sqrt{74}}\right)$  If you compute $|\vec u|$ you get $1$ as promised.
