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In my book it defines position vectors as vectors that area tied to the origin, ie they can't be moved like free vectors.

Granted that this is a correct definition, it got me thinking: are they even vectors at all? This is because they seem to be the same as coordinates, ie simple points on the plane. After all, coordinates also have a magnitude and a direction, don't they? They are simply not allowed to be moved, that is all. So coordinates and position vectors seem to be the same thing (except when we draw position vectors, we draw the arrow, whereas with coordinates we just draw the point). The fact that we can't move a position vector adds to my doubts regarding why they are called vectors, since they just seem to be points.

Even from the point of view of defining them as vectors because of their algebraic form, ie $x\hat i + y\hat j$, then surely the fact that we can't move them would mean that this is a contradition? Because we are defining a position vector in terms of things that can be moved, ie $\hat i$ and $\hat j$, so it would appear from this definition that $x\hat i+y\hat j$ can also be moved, meaning it isn't a position vector.

What then is a position vector?

I'm probably over thinking this though, so any hints in the right direction would be appreciated.

Edit: tl;dr: Why can't we just use coordinate notation for position vectors and agree to write position vectors and coordinates together. E.g. things like $a+\vec b$ where $a=(5,3)$ and $\vec b$ is a vector.

Raghib
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  • "They have magnitude and direction"? Then they are vectors. Period. As vectors, though, both $;(1,1);$ and $;(3,4)-(2,3);$ are exactly the same (and that's the reason, I guess, they're called "direction vectors"), though the second one is "tied" to the point $;(2,3);$ . And yes: in $;\Bbb R^n;$ it is pretty customary to "mix" between point $;(x_1,...,x_n);$ and the vector $;(x_1,...,x_n);$ tied to the origin. It makes no harm if everybody understands what's going on. – DonAntonio Jan 03 '18 at 09:51
  • @DonAntonio: Please read my edit. Also, I don't know university maths so I am slightly confused about what you mean. Are you basically saying that saying something is a position vector is based on context? – Raghib Jan 03 '18 at 09:54
  • What does it mean by moving a vector? – edm Jan 03 '18 at 09:57
  • @edm: It doesn't actually use that word but that's how I interpreted their definition: "The position vector of a particle is a vector drawn from the origin of a coordinate system to the position of the particle. For a particle at the point (x,y) the position vector $\vec r$ is $x\hat i+y\hat j$." What is the difference between this and the "free" vector $x\hat i+y\hat j$? There doesn't seem to be one, hence I am confused, except if position vectors are merely contextual, in which case it would make sense. – Raghib Jan 03 '18 at 10:01
  • Raghib - I think it depends on context, position vector are defined so (your defn.) for the purpose of taking advantage of vector properties in defining coordinates. If you want all vectors to be freely translated in any place, then that defeats our purpose! So in that sense you can say its different form vector. – jonsno Jan 03 '18 at 10:02
  • @samjoe: So are position vectors basically the overlap between vectors and coordinates? I thought that was already achieved by using cartesian unit vectors, ie $\hat i$ and $\hat j$ (which allow us to easily convert between results in vectors and coordinates). I think I see what you're saying though. – Raghib Jan 03 '18 at 10:05
  • Raghib - This definition which you gave for position vectors is very useful for various purposes, e.g. defining kinematics of a particle on some coordinates, or equation of line or plane in three dimensions etc. Unit vectors are utilised for finding position vector of some coordinate, eg, for $(2,3)$ position vector is $(2\hat{i} + 3 \hat{j})$. – jonsno Jan 03 '18 at 10:13

2 Answers2

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The position vector of the point $P$, with respect to some chosen origin $O$, is the vector $\vec{OP}$. That's all there is to it. Since it's a vector, it can of course be moved around as you like (that is, any equally directed arrow of the same length represents the same vector, regardless of where you draw it). But it's only when you “attach” it to the point $O$ that it indicates the location of the point $P$ (which is its “job” as a position vector).

Regarding addition of points and vectors, see for example this answer.

Hans Lundmark
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  • Thanks, makes sense now. When can it be useful to consider a vector as a position vector in the sense you described? For example, when defining a line in terms of vectors? – Raghib Jan 03 '18 at 10:49
  • Hello, I have posted a related question regarding when this could be useful: https://math.stackexchange.com/questions/2590026/when-can-it-be-useful-to-consider-a-vector-a-position-vector – Raghib Jan 03 '18 at 11:03
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    I'd say it's only useful when you need to perform some vector operations, like dot product and cross product. Like in mechanics, where the angular momentum of a point (with respect to some origin) is the cross product of the point's position vector (with respect to that origin) and the point's momentum vector. – Hans Lundmark Jan 03 '18 at 12:11
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    Regarding writing a parametric equation for a line or a plane, you can do without position vectors if you are willing to define the operation “point + vector = point”. Like this: $P(t) = P_0 + t \mathbf{v}$ describes all points on the line through the point $P_0$ with direction vector $\mathbf{v}$. In traditional notation, that would be $\mathbf{r}(t)=\mathbf{r}_0 + t \mathbf{v}$, where $\mathbf{r}(t)$ and $\mathbf{r}_0$ are position vectors with respect to some origin $O$ (which is really irrelevant here, since the line has nothing to do with $O$!). – Hans Lundmark Jan 03 '18 at 12:15
  • Hans- can we do that, I mean define point+ vector as point?? Thanks! – jonsno Jan 04 '18 at 03:17
  • In line equation we may not need position vector but in plane we do need right, and yes it can be done without vectors using Cartesian form, just vector form is easier sometimes? – jonsno Jan 04 '18 at 03:19
  • Yes, it's a very natural operation: $Q=P+\mathbf{v}$ is the point where you arrive if you start at the point $P$ and move as the vector $\mathbf{v}$ tells you. And no, you don't need to use position vectors to write the equation of a plane; it's just $P(s,t)=P_0 + s \mathbf{u} + t \mathbf{v}$. – Hans Lundmark Jan 04 '18 at 07:46
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At each point in the space corresponds a position vector and viceversa:

$$P(x,y)\equiv OP=x \vec i+y\vec j$$

When you consider two points

$$P(a,b)\equiv OP=a \vec i+b\vec j \quad Q(c,d)\equiv O Q=c \vec i+d\vec j$$

the vector from $P$ to $Q$ is given by

$$PQ=P-Q=(a-c,b-d)=(a-c) \vec i+(b-d)\vec j$$

From a mathematical point of view vector $PQ$ is also a vector form the origin (in a vector space all vectors are form the origin)

$$PQ=OR=(a-c,b-d)=(a-c)\vec i+(b-d)\vec j$$

such that $$OP+OR=OQ$$

Anyway you could also though to $PQ$ as a vector applied in $P$ such that:

$$OP+PQ=OQ$$

and in this sense they can be moved like free vectors.

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user
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