Origin of vectors Background 
I was reviewing notes of physics, and i realized that something about the mathematics of vectors was wrong in my head.
Example-problem 
Suppose a vector is $A=5\textbf{i} + 3\textbf{j}$, and other $B=7\textbf{i}+3\textbf{j}$. Then $A-B=C=-2\textbf{i}$.
The question is: Why C is not placed at the origin, from $x=0$ to $x=-2$? Is not it what $C=-2\textbf{i}$ indicates? 
The general question is: does a vector have an origin?

Notice the calculation for $B-A$ is not correct in the question.
$$B-A = (7i -3j) - (5i + 3j) = (7-5)i + (-3 -3)j = 2i - 6j$$
So $i$ represents a unit vector in $x$ direction, and $j$ represents a unit vector in $y$ direction.
So what $B-A$ means is that it has component in $x$ with $2$ of unit vectors, and component in $y$ with $-6$ unit vectors.
EDIT
Now $B$ has been edited to $B=7i+3j$, so we have $A-B=C=-2i$, so how to draw that?
So a vector is a line-segment with a direction, and any translation to it won't change it. e.g. a line segment starting from $(0,0)$, ending at $(3,4)$ is the same vector as line segment starting from $(2,1)$, ending at $(5,5)$; and they are all referring to vector $3i + 4j$. Having discussed this, let's assume we always draw vector starting from $(0,0)$, and you could move them accordingly if you want, but all of them represents the same vector.
$C=-2i$ meaning C has component of $-2$ on $x$ axis, and component of $0$ on $y$ axis. Thus it is a vector starting from $(0,0)$ and ending at $(-2,0)$.
To extend this, let $C = ai + bj$, where $a,b \in \mathbb R$, then $C$ has component of $a$ on $x$ axis, and component of $b$ on $y$ axis, and the vector thus starts at $(0,0)$ and ends at $(a,b)$.
