What is truly the definition of vectors in modern mathematics? About me: I've studied abstract algebra, linear algebra and mathematical analysis.
I know the definition of "vector" from linear algebra course. However, some authors, especially in precalculus, introduced or even defined a vector as a "geometric object/entity"(i.e. not an plain ordered-pair with some operations defined on it; conversely, it was essentially treated as something like segments, points, or triangles in geometry.) But from the undergraduate courses, I found that vectors are always defined as an element of a more abstract algebraic structure--namely, vector spaces. There is no such geometric meaning or any "magnitude-direction" appearance.
So, does someone define vectors as a geometric objects (or other different types)?  Or do we no longer use those ways in modern mathematics? Why? What is truly the definition of vectors in modern mathematics?
 A: To answer at least one specific question that you asked: I'm going to give a definition of a "directed segment in the plane", and then show that equivalence classes of these under a certain relation form a vector space that's isomorphic to $\mathbb R^2$, leaving out many of the details. Here goes. 
First things first: by $\mathbb R^2$, I mean the set of ordered pairs of real numbers. This (via Descartes) is in correspondence with the points of a Euclidean plane, once you choose a point $O$ and two rays $X$ and $Y$ from $O$, to serve as the origin and two coordinate axes. Even ignoring that correspondence, it also happens to have a vector space structure (over the reals as the field of scalars), with addition defined term-wise, so that $(a, b) + (c, d) = (a+c, b+d)$ and scalar multiplication defined by $c (x, y) = (cx, cy)$. Whew!
A "segment" in classical geometry is characterized by two points, so in my version, I'm going to say it consists of a pair of points. And to make a directed segment, I'm going to say this: a directed segment is an ordered pair of points in $\mathbb R^2$. 
You might object that this isn't a segment at all, but there's a pretty clear correspondence between segments and pairs of points: 2 points map to the segment between them; a segment maps to its two endpoints. The only subtlety here is "order" -- for me, the two points have the property that one is the first item in my ordered pair, and the other is the second item. 
A translation of the plane is a map of the form 
$$
T_{a,b}: \mathbb R^2 \to \mathbb R^2 : (x, y) \mapsto (x+a, y+b)
$$
where $a, b \in \mathbb R$ are fixed numbers. (We call this "translation by $(a, b)$".)
Small theorems about translations: $T_{a,b} \circ T_{c,d} = T_{a+c, b+d}$; $T_{0,0} = identity$; $T_{-a,-b} \circ T_{a,b} = identity.$ I leave the proofs to you. 
I'm now going to define an equivalence relation on directed segments: 
The directed segment $(P, Q)$ is "equivalent" to the segment $(A, B)$ if there is a translation $T_{r,s}$ such that $T_{r,s}(P) = A, T_{r,s}(Q) = B$. Informally, two directed segments are equivalent if one can be translated to the other. 
Claim: the 'equivalence' I've just defined is an equivalence relation. 
Proof: $(P,Q)$ is equivalent to $(P, Q)$ by the translation $T_{0,0}$. The other two parts follow from the other little theorems about translations. 
I'm now going to define a map $H$ that takes an equivalence class of directed segments (which is just a big set full of ordered-pairs of points in the plane) and assigns to that class a single element of $\mathbb R^2$. Here's the definition. 
For an equivalence class $C$, pick any representative $(P, Q) = ((a,b), (a', b'))$. Associate to the class $C$ the pair $H(C) = (a'-a, b'-b) \in \mathbb R^2$. 
I still have to show that $H$ is well-defined, i.e., that it doesn't depend on my choice of representative. Well, suppose that $(R, S)$ is another directed segment in the class $C$. Then there's some translation 
$$
T_{e,f}
$$
with the property that
$$
T_{e,f}(P) = R\\
T_{e,f}(Q) = S
$$
by the definition of "equivalence". if we write $(R,S) = ((c,d), (c', d'))$, this means that 
$$
a + e = c\\
b + f = d\\
a' + e = c' \\
a' + f = d'
$$
Now let's compute the point associated to $(R,S)$. According to the definition, it's just $(c' - c, d'-d)$. But we can simplify this with the rules above to get
$$
(c' - c, d'- d) = ((a'+e) - (a + e), (b' + f) - (b+f)) \\
= (a' + e - a - e,m b' + f - b - f) \\
= (a' - a, b' -b)
$$
so that we get the same value no matter what representative of the equivalence class we choose. 
But is $H$ actually a nice 1-1 correspondence between equivalence classes of directed segments and elements of $\mathbb R^2$? It is. I'll show that by writing down an inverse, $K$. 
$$
K: \mathbb R^2 \to S : (a, b) \mapsto [ ((0,0), (a, b)) ]
$$
where $S$ denotes the set of all equivalence classes of directed segments, and $[ u ]$ denotes the equivalence class of a segment $u$. 
It's easy to check that $H(K(a, b)) = (a, b)$, so this really is a nice correspondence. (You should also check that $K \circ H$ is the identity to finish up the job.) 
By the way, I haven't defined the operation of "vector addition" for these equivalence classes. One way to do it is to say "send the two classes to elements of $\mathbb R^2$ via $H$, add them there, and then use $K$ to bring the result back." If you do that, and work through the details, it turns out to be the same as this:
Pick a representative $(P,Q)$ for the first class; find a representative $(Q, R)$ for the second class that starts with the same point that the first one ended with (i.e., $Q$). THen the sum is the class that contains $(P, R)$." 
You have to prove, of course, that there IS an element of the form $(Q,R)$ in the second class --- that's a small lemma --- but once you do that, you see that this is really the old "put the tail of one vector at the head of the other and then complete the triangle" definition of vector addition. 
PS: Thanks for asking this and thereby encouraging me to think through this at last. I'd known forever that there WAS such an equivalence, but never really saw that it also implied the "parallelogram law", for instance. 
