Physics uses mathematical models to describe physical observations. You will meet this if you continue studying physics.
The mathematics is used rigorously with all its definitions. What makes it a model is a set of "assumptions", "postulates", "principles" which pick up a subset of all the space the mathematical functions can range. I will try to demonstrate with your "vectors", showing that they model observations given some assumptions.
You ask:
First, why is it that in physics, magnitude and direction of vectors are emphasized while in linear algebra components of vectors are emphasized?
Because vector algebra is used in physics to model impacts, for example, which by observation depend on magnitude and direction. In other uses of vectors the components are emphasized , but you have to continue in physics studies to encounter examples.
Also, for elementary physics (kinematics, dynamics) is R2 and R3 the vector space from which we operate and do our calculations?
We use the real numbers for end results of calculations. In special models complex numbers are important because of the simplicity of the form, but the end result of calculations has to be in real numbers.
Finally, in physics, it is emphasized that vectors are particularly useful because they are not tied to any one coordinate system. What does this mean exactly?
It means that the origin of the vectors is assumed according to the experimental set up that needs to be modeled.
In linear algebra we always write vectors emanating from the origin of the coordinate system, so it would seem as though the vectors are tied to the coordinate system.
The vectors used in physics assume as a coordinate system for the vectors the point where the vector is used to describe a force, or a velocity depending in the model. The origin could be a function of the coordinates, i.e a moving vector.
$\newcommand{\norm}[1]{\lVert #1 \rVert}\newcommand{\ip}[2]{\left<#1,#2\right>}$I want to be at first somewhat mathematically abstract and return to the physical interpretation later. In mathematics you define a vector space $V$ over a field (like real numbers etc.) $F$ to be as a triplet $(V,+, \cdot)$ where $+$ denotes the vector addition and $\cdot$ denotes the scalar multiplication, which has the following properties:
- $u + v = v + u$
- $u + (v + w) = (u + v) + w$
- There exists an element $ 0 \in V$ such that $v + 0 = v$
- For every $v ∈ V$, there exists an element $−v ∈ V$, called the additive inverse of $v$, such that $v + (−v) = 0$
- $ \alpha(\beta v) = (\alpha\beta)v $
- $ 1v = v$ where $1 \in F$
- $\alpha(u + v) = \alpha u + \alpha v$
- $ (\alpha + \beta)v = \alpha v + \beta v$
for all $u,v,w \in V$ and $\alpha, \beta \in F$. Note that the existence of the zero vector $0 \in V$ is an axiom, so without it you cannot have a vector space. Note however that the direction or the magnitude of a vector doesn't come up in the definition, so what is going on here? The thing is that we physicist like to have additional structures on vector spaces, such as a scalar product of two vectors and a magnitude of a vector. Furthermore we want to have that (in physicists notation) $ \vec v \cdot \vec v = \norm{v}^2$ where $\norm{\cdot}$ denotes the magnitude of a vector. So when we know what a scalar product is then we know what the magnitude of a vector is (in some sense). Then the question is this: What is the definition of a scalar product (I'll denote the scalar product as $\ip \cdot \cdot$ from now on.) Scalar product is a function $\ip \cdot \cdot : V\times V \to \mathbb R$, (I take a real vector space for simplicity) which obeys following axioms:
- $\ip \cdot \cdot$ is linear in both of its arguments
- $\ip vu = \ip uv$ for all $u,v \in V$
- $\ip xx \geq 0$ and $\ip xx = 0 \iff x = 0$
So the question might be: Where is the $\cos \theta$ term, with which we defined the scalar product? The troubling part is that there is no $\cos \theta$ term. As it turns out a scalar product satisfies this wonderful inequality called Schwartz-Inequality:
$$ \norm x \cdot \norm y \geq \ip xy $$
note that for $x,y \neq 0 $ we can write this as:
$$ \frac{\ip xy} {\norm x \cdot \norm y} \leq 1 $$
now we can define $\cos \theta := \frac{\ip xy} {\norm x \cdot \norm y}$ since the term on the right is always less than 1. You now see that we define the angle to be $\frac{\ip xy} {\norm x \cdot \norm y}$ not the other way around. So what does this mean for the zero vector? Well nothing because you see that we have divided by $\norm x$ which is zero for the case of the zero vector. Thus it doesn't make any mathematical sense to talk about the direction of the zero vector.
So let's talk about the physical meaning of this whole mathematical abstract explanation and for the sake of the argument let's think about vectors as arrows. Take two vectors, one of which is pointing up and the other pointing left and let's make the length of these vectors to go to zero. Obviously in both cases we get the zero vector because remember $\norm x = 0 \iff x =0$. However there is an ambiguity about the direction of the zero vector. Since both of the vectors were showing in different directions and a vector (in this a the zero vector) should show in only one direction. At this point we say that the direction of the zero vector is ill-defined. The question is this: Does it matter in which direction it points to? Of course not! Think of it this way: If I push you from your back or from your side it makes a difference. You would certainly be able to tell the difference (Corresponding to a force with nonzero magnitude). However if I don't push you from your back or your side, I bet you cannot tell the difference. (This corresponds to applying a force with magnitude zero.) Another example is the velocity vector. It makes quite a difference when you are driving your car to the south or to the east but if you are not driving, well then you are both "not going" to east and south (which is why the direction is ill-defined) However note that in the absence of the zero vector you wouldn't be able to say that you are not moving. Another great analogy is just the number line. There are positive numbers and there are negativer numbers but what if I asked you what is the sign of zero. Well it is (vaguely speaking) both and neither positive and negative so we can say that the sign of zero is ill-defined or you can arbitrarily choose a sign however it makes you happy.
The moral of the story is that you need the zero vector to describe (vaguely speaking) the absence of something. Concretely you need the zero vector in order to say that there is an inverse to a vector (see additive inverse in the way beginning). More like how you need the number zero.
Best Answer
This is one of those things that (intentionally) gets conflated, though it may be better if we were more consistent about keeping them separate.
So, points don't form a vector space. It makes no sense to ask "what's the location of New York plus the location of DC". However, given two points we can subtract them and get a displacement, and we can add that displacement to points to get new points. The mathematical structure for this is called (among other things) a torsor.
Your text is, however, accurate. If we choose a particular point to be our origin, call it $O$, then we can make a vector $r = P - O$ and call it a position vector for $P$. Now the difference between a position vector and a "regular" vector is that it changes when we change what we consider the origin. When we perform a translation on a system, position vectors change, regular vectors do not.