Difference between a vector function and parametric equations What is the difference between a vector function and parametric equations? Both are capable of describing plane/space curves and both can indicate direction. The only notable difference I see is that a vector function includes a measured distance between a given point and the origin (magnitude of position vector that draws out a plane/space curve). 
 A: $\newcommand{\R}{\mathbb{R}}$
I hope this addresses the question:
So with a "Vector function" I think you mean a "vector valued function". Now, I think everything with this stuff boils down to on what your interpretation of the space is in which the function maps your input. 
To be clear let $\varphi : [0,2\pi] \to \R^2$, and $t \mapsto (cos(t),sin(t))$. So your output of the function will be in $\R^2$. Now, how do you think about this $\R^2$? If you think about it as normed space, in particular a vector space, than of course every $\varphi(t) \in \R^2$ can be viewed as a vector in the normed vector space $\R^2$. So it has a direction and a magnitude! 
Now let's look a the circle as a set of points, then you could say, the unit circle in $\R^2$, denoted by $S^1$ is equal to the set $\{(t,\varphi(t)) \in \R^2 : t\in[0,2\pi]\}$ and thus $\varphi$ is a parametrization of that set(!) and not really a vector valued function (by choice!). Of course, in $\R^n$ this is all not really that important, because it omits (canonical) so many interpretations, e.g. topological space, metric space, vector space, normed space, manifold, etc., that are all compatibile with each other. But consider a set of points that actually isn't a vector space! For example $S^1$ itself and consider the function $\psi : [0,\pi] \to S^1$ with $\varphi\big|_{[0,\pi]} = \psi$. Then $\psi$ parametrize a half circle but since $\psi(t) \in S^1$ it has neither magnitude nor direction, it is just a point in a set. But to me this is a parameteric equation, because it really captures the most important aspect of that its image is geometric objection, e.g. a curve. But that definition could be just mine alone and you always define the target space to be the $\R^n$. Which leads me to vector functions.
Now, when we talk about a vector function or vector valued function, you already implying that it will map to a vector space. And those are obviously more general than just a function from $\R \to \R^n$. You could very well have a function from $\R^m \to \R^n$ or possibly even infinite-dimensional (for example a Banach space), also, the domain as a set doesn't need to be endowed with anything on structure I think. But a vector space does not come with a norm, but it is extra structure! So saying that in general it has a magnitude is not correct I think.
In the end though, it depends all on your precise definition of these terms, you want to use! That is what I wanted to argue here!
Hope this was helpful? Did clarify things?
A: What do you mean that a vector function includes a measured distance (magnitude)?
Vector functions and parametric equations are essentially the same thing.
Sometimes, if you're working in a different coordinate system, like polar coordinates, you might see that they include the magnitude $r$ in their functions. 
For example, $(x,y)=(\cos(t),\sin(t))$ for $t\in [0,2\pi]$, versus $(r,\theta)=(1,t)$ for $t \in [0,2\pi]$. 
A vector function says the exact same thing except $\vec{v}(t)=\cos(t)\vec{i}+\sin(t)\vec{j}.
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A: I used to think they are the same, but recently I found they are different, at least for the meaning of coplanarity. For example, suppose a vector function $r(t) = <f(t), g(t), h(t)>$ and a paramatric curve $(x, y, z) = (f(t), g(t), h(t))$.
When we say $r(t)$ lies on the same plane, we mean that all vectors determined by $r(t)$ are parallel to the same plane. This means that all points $(f(t), g(t), h(t))$ determined by position vectors of $r(t)$ and the origin lie on the sample plane.
When we say the paramatric curve $(x, y, z) = (f(t), g(t), h(t))$ lie on the same plane, we just mean that all points on the paramatric curve lie on the same plane. The plane deos not need to contain the origin.
Practically, to prove the curve determined by $r(t)$ lies on the same plane with normal vector $n$, one needs to prove $n\cdot r(t)=h$ where $h$ is a constant. To prove $r(t)$ lies on the same plane, one needs to prove  $ n\cdot r(t)=0$.
