Why aren't all vectors coinitial? In physics, we are learning about vectors, but not in much detail. I am hazy about many concepts. We have been taught that vectors can be parallelly shifted, provided that they maintain their length (magnitude) and their direction.

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*Does this not mean that all vectors are co-initial since I can always just shift them?

There is an answer to a similar question on this site; the answerer says that on a manifold, two vectors may not be coinitial. But I'm confused because I don't see why his explanation (using example of a particle moving along the unit circle in the $x$-$y$ plane) is specific to manifolds and not just any 2D plane (I don't know what manifolds are; I'm just going by what the answerer said).


*Also, is it really true in the first place that I can parallelly shift any vector? It seems confusing to me that the starting point of a vector has no significance.

 A: Try this: take a spherical object, or just imagine a sphere in front of you if you can't find one. Now place your index finger pointing north (up) right in the centre of the sphere, along the equator.
If you slide your finger forward to the north pole, your finger will be pointing straight ahead. But if you were to instead rotate 90 degrees along the equator and then slide your finger to the north pole, your finger is no longer pointing straight ahead, it's pointing 90 degrees away.
The name for this idea is Holonomy. Basically if you slide a vector around a loop on a curved surface, the vector can end up rotated. This does not happen when you slide vectors around on flat surfaces, however.
So you see that on a sphere, where your vector starts is important because if you try to slide your vector to a new starting point, it will depend on which path you take to the new point. There is no well-defined (i.e. independent of any choices) way to compare vectors  on the equator of a sphere versus the north pole.
Part of the mathematical formalism of a manifold, is that you don't have just one set of vectors, each point in the manifold has its own set of vectors and we call that the tangent space at that point. For two tangent vectors to be considered equal, they need to share the same starting point. We also have ways of moving vectors from one tangent space (starting at p) to another (starting at q). But, as I said, that depends on the path from p to q.
A: For simplicity I'll give this answer in $\mathbb{R}^3$, which is likely to be your context for introductory physics, but really the same thing applies in any dimension.
There is a conceptual difference between vectors and points.  I find it helpful to think of vectors as instructions for how to get from one place to another: Given two points $A = (a_1, a_2, a_3)$ and $B = (b_1, b_2, b_3)$, we can construct the vector $\vec{AB} = \begin{bmatrix}b_1-a_1 \\ b_2-a_2 \\ b_3-a_3\end{bmatrix}$.  The vector $\vec{AB}$ describes how to get from the point $A$ to the point $B$, in the sense that if you start at the point $A$ and travel $b_1-a_1$ units in the $x$-direction, $b_2-a_2$ units in the $y$-direction, and $b_3-a_3$ units in the $z$-direction, then you'll end up at $B$.  Notice, though, that $\vec{AB}$ remembers neither $A$ nor $B$.  If $A = (1, 1, 1)$ and $B = (1, 0, 0)$ then $\vec{AB} = \begin{bmatrix}0\\-1\\-1\end{bmatrix}$, but if $A = (3, 2, 1)$ and $B = (3, 1, 0)$ then we also have $\vec{AB} = \begin{bmatrix}0\\-1\\-1\end{bmatrix}$.
Now suppose that I say $\vec{v} = \begin{bmatrix}1\\2\\3\end{bmatrix}$.  This gives me instructions of how to move: $1$ unit in the $x$-direction, $2$ in the $y$-direction, and $3$ in the $z$-direction.  This information does not tell me where to start.  If I start at $(0, 0, 0)$ I'll end up at $(1, 2, 3)$, but if I start at $(5, 3, 2)$ I'll end up at $(6, 5, 5)$.  The vector itself hasn't changed, so it makes equally good sense to draw the vector starting at $(0, 0, 0)$ or at $(5, 3, 2)$ or anywhere else you want.  That's why we can shift around any vector we want - we're not changing the instructions given by the vector, we're changing where we've decided to start that process, and the vector has no opinion about the starting point.
Despite this, in physics you will often care not just about vectors but also about where they are placed.  After all, a force applied at one end of a see-saw will do something very different than the same force applied at the centre of the see-saw!  The shifting does, however, let you shift the entire system around as you please, provided you move everything around in the same way. You might do this to create points and vectors with entries that are nicer numbers to use.
