Derive Unit Vector from Rotation Vector I have a rotation vector
<45.0, 45.0, 45.0>

Up Vector:
<0.0, 1.0, 0.0>

How can i get a unit vector representation of the rotation vector? Basically i want to see which direction an object is looking at. 
I might have wrong impression of the correct naming, please feel free to correct my question. 
 A: If $\vec{z} = (45,45,45)$ then the unit direction is simply
$$ \hat{z} = \frac{(45,45,45)}{\|(45,45,45)\|} =( \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}})$$
Now with the up direction $\hat{y} = (0,1,0)$ you can find the x-axis direction with
$$ \vec{x} = \vec{y} \times \vec{z} \\ \hat{x} = \frac{\vec{x}}{\| \vec{x} \|} = ( \frac{1}{\sqrt{2}},0,\frac{1}{\sqrt{2}}) $$
 NOTE: $\times$ is the vector cross product 
A: The simple way to find a unit vector from any vector is by performing scalar division on it. Let $\vec v$ be your so-called "rotation vector", and $\vec w$ be the unit vector. To change the magnitude of a vector, we can use the scalar multiplication and division operations. The vector $a$  divided by some scalar $s$ has a magnitude $||a||/s$. Similarly, the vector $\vec v$ divided by it's magnitude $||\vec v||$  is equal to $||\vec v|| / ||\vec v||$, or $1$. Recall that a vector of length 1 is a unit vector. This means the vector $\vec v / ||\vec v||$ has a magnitude of 1.
Also remember that the scalar multiplication and division operations do not affect the direction of the vector, but only it's length. So if a vector of length 1 is heading the same direction of another vector, it is considered a unit vector of that direction.
TL;DR: The unit vector with the direction of some "rotation vector" $\mathbf{\vec v}$ is equal to the vector divided by it's magnitude, or 
$$\frac{\vec v}{||v||}$$
