Vague understanding of vectors in coordinate plane Albeit having done a lot of problems of vectors,i still struggle in visualizing displacement vectors in the coordinate plane. Let's stay in cartesian plane as of now for simplicity. A vector is generally expressed as $a\hat{i}+b\hat{j}$. This shows how much to walk in the $x$ and $y$ axis measuring from the origin to reach the tip. But a vector being expressed in this form inadvertently deals with the origin? If we had to shift this vector so that none of the head or tail is at the origin, what will be the expression of that vector? If two points $A(x, y)$, $B(m, n)$ are joined with a vector from $A$ to $B$, i saw that the displacement vector is given by $(m-x)\hat{i}+(n-y)\hat{j}$. But $(m-x)\hat{i}+(n-y)\hat{j}$ generally means going $m-x$ units in the x axis from the origin and then $n-y$ units in the y axis.
I look forward to having the kind attention of physics lovers in this regard.
 A: I guess what you are referring to is the difference between the position and displacement vector.

*

*A position vector is a vector from the origin to a point in the coordinate system. Its tail is always at the origin and the head at any point in coordinate system.

*A displacement vector is vector with neither the tail nor the head at the origin of the coordinate system.

*It takes two position vectors to point the tail and head of a displacement vector, respectively.

*If $A$ and $B$ position are vectors pointing to the tail and head of a displacement vector, $C$, then the displacement vector $C=B-A$.

A: Do a Google search for "free, bound, and sliding" vectors. I remember suffering from similar confusions. In mathematics, vectors can be slid all over the coordinate plane and still represent the same object: a directed line segment.
In classical mechanics, position vectors represent displacements of some particle from an origin, which is identified with some physical point in space, so that you can make sense of the components of the position vector. The tail of the vector is identified with the origin. If you slid the vector around, it would retain its mathematical meaning but lose its physical meaning. Therefore, the position vector is a vector bound or fixed to the origin. You can think of it as a relative position vector as ${\bf r} - {\bf r}_O = a \hat{\bf i} + b\hat{\bf j}$, where ${\bf r}_O = {\bf 0}$ is identified with the origin.
An example of a sliding vector is the application of a force on a rigid body. The tip of the vector represents the material point where the force is applied. The physics don't change if you slide the force vector along its line of action.
Free vectors include objects like pure couples or angular velocities. They are not associated with any particular point. It doesn't matter where you draw them in your diagrams, so long as you have the magnitude and direction correct.
A: Points in a plane are part of affine space, so a vector doesn't mean anything without an origin. It's confusing because a point in space isn't a vector, it a point.
When you subtract two points, you get an abstract vector that connects them, the origin then becomes irrelevant. The vector connecting two points doesn't depend on the location (or existence) of the origin.
The problem arrises when you implicitly make the origin one of the points and then forget about it, so the vector represents the point minus the origin, but you treat the vector as representing the point.
Affine spaces have origins, vector spaces do not, and points aren't vectors.
Edit: I think the confusion is resulting from the naive picture of a vector as something with a head and tail, or, represented by an arrow. That's a nice way to introduce vectors, but it's not why we use them. We use them because they're the fundamental geometric objects that represent rotations.
If you look at the simplest functions on the unit sphere that are eigenfunction of rotations, you will get $Y_1^m(\theta, \phi)$ for $m\in(-1,0,1)$. They have the following linear combinations:
$$ Y_1^1(\theta,\phi) +Y_1^{-1}(\theta,\phi)\propto \sin{\theta}\cos{\phi}= x$$
$$ -i(Y_1^1(\theta,\phi) -Y_1^{-1}(\theta,\phi))\propto \sin{\theta}\sin{\phi}= y$$
$$ Y_1^0(\theta,\phi) \propto \cos{\theta}= z$$
What emerges is $(\hat x, \hat y,\hat z)$
which closed under rotations, and transform accordingly.
The associated gradients $-i\partial_ x, -i\partial_ y,-i\partial_z$ are the generators of translations in Euclidean space. This is why vectors are special.
Points? Some random point $P$ in space....not so much. You can't rotate it by itself. If you want to describe that point mathematically, you need to define an origin $O$. From there you use the vector:
$$\vec P \equiv P-O$$
where the subtraction on the affine space of points yields a vector. Meanwhile, the vector $\vec P$, along with an affine  point called the origin $O$, can be used to define the point $P$:
$$P = O + \vec P$$
$P$ is a point, and does not care about $O$. The origin is completely arbitrary. Now in some physical problems, there may be a natural choice of $O$, like the center of the earth in defining ECEF coordinates in cartography, but if you are working with an rando infinite plane, it is totally arbitrary.
If you start to think of the point $P$ as being the same as the vector $\vec P$, and assign it vector behavior, you're going to have problems. Points aren't vectors.
