# Why do position vectors have an arrow on top of them?

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According to the vector definition, the vectors have both the direction and magnitude such as displacement vectors which should possess arrows on the top of them because they have displacement so they express a direction. On the other hand, position vectors are stationary, they do not have any displacement so why do they possess arrows on the top of them? Could you explain, please?

fresh_42
Mentor
They are the displacement of the origin of the coordinate system, resp. of the point ##\{\,0\,\}## in the vector space.

ZapperZ
Staff Emeritus
According to the vector definition, the vectors have both the direction and magnitude such as displacement vectors which should possess arrows on the top of them because they have displacement so they express a direction. On the other hand, position vectors are stationary, they do not have any displacement so why do they possess arrows on the top of them? Could you explain, please?
If you ask me "where is the ball?", and I reply "Oh, it's 3 meters away", do you think you have sufficient information to locate the ball immediately?

No, because if it is 3 meters away from me, it could be 3 meters away in any direction! The only thing you can do is draw a circle of radius r = 3m, and the ball is someone on that circle.

If I say, it is 3 meters way in THAT direction (and I point), then you look at the direction that I'm pointing, and the intersection of that and the circle is the location of the particle. I've just given you the location in plane-polar coordinates. I could have easily given it to you in cartesian coordinates. And by doing that, I've defined a position vector! Each of those locations are defined in terms of unit vectors r and θ, or i and j.

It has nothing to do with whether something is stationary or moving.

Zz.

Demystifier, sophiecentaur and FactChecker
If you ask me "where is the ball?", and I reply "Oh, it's 3 meters away", do you think you have sufficient information to locate the ball immediately?

No, because if it is 3 meters away from me, it could be 3 meters away in any direction! The only thing you can do is draw a circle of radius r = 3m, and the ball is someone on that circle.

If I say, it is 3 meters way in THAT direction (and I point), then you look at the direction that I'm pointing, and the intersection of that and the circle is the location of the particle. I've just given you the location in plane-polar coordinates. I could have easily given it to you in cartesian coordinates. And by doing that, I've defined a position vector! Each of those locations are defined in terms of unit vectors r and θ, or i and j.

It has nothing to do with whether something is stationary or moving.

Zz.
I understand a bit but not completely what you mean because English is not my first language and I am learning Physics here very easily. So I request you to use lucid language so that I can understand easily and clear my all concepts. So, please could you get your point a little bit easier here in this context above?

Chestermiller
Mentor
We represent all vectors with arrows over them to distinguish them from scalars. Another way of doing this is to represent vectors using boldface, rather than with arrows over them.

nasu
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If you ask me "where is the ball?", and I reply "Oh, it's 3 meters away", do you think you have sufficient information to locate the ball immediately?

No, because if it is 3 meters away from me, it could be 3 meters away in any direction! The only thing you can do is draw a circle of radius r = 3m, and the ball is someone on that circle.

If I say, it is 3 meters way in THAT direction (and I point), then you look at the direction that I'm pointing, and the intersection of that and the circle is the location of the particle. I've just given you the location in plane-polar coordinates. I could have easily given it to you in cartesian coordinates. And by doing that, I've defined a position vector! Each of those locations are defined in terms of unit vectors r and θ, or i and j.

It has nothing to do with whether something is stationary or moving.

Zz.
I don't understand the point here in the context above 'Each of those locations are defined in terms of unit vectors r and θ, or i and j.' could you simplify it, please?

fresh_42
Mentor
'
I don't understand the point here in the context above 'Each of those locations are defined in terms of unit vectors r and θ, or i and j.' could you simplify it, please?
##(a,b)## is a position, a location with coordinates ##a## and ##b##.

##\stackrel{\longrightarrow}{(a,b)}## is a vector, sometimes called position vector. It is the vector starting at the origin ##(0,0)## and pointing to the location ##(a,b)##. It has length and direction, which is why it is denoted as a vector. The fact that both are basically written as ##(a,b)## only means, that such a notation depends on the context, i.e. whether it stands for a point or for a vector. An arrow above it resolves this ambiguity.

pbuk
Mark44
Mentor
##(a,b)## is a position, a location with coordinates ##a## and ##b##.

##\stackrel{\longrightarrow}{(a,b)}## is a vector, sometimes called position vector. It is the vector starting at the origin ##(0,0)## and pointing to the location ##(a,b)##. It has length and direction, which is why it is denoted as a vector. The fact that both are basically written as ##(a,b)## only means, that such a notation depends on the context, i.e. whether it stands for a point or for a vector. An arrow above it resolves this ambiguity.
To help distinguish them I favor writing vectors with coordinates using angle brackets -- <a, b>, and points using paretheses -- (a, b).

… position vectors are stationary, they do not have any displacement so why do they possess arrows on the top of them?
Because that's the order in which the vector is read; from origin to head. The vector doesn't signify anything moving in the real world. In some cases the arrowhead has nothin to do with (or is in a different direction than) the motion. For example, a force on a static object, where there is no motion, or the direction of a moment or angular velocity vector.

Stephen Tashi
On the other hand, position vectors are stationary, they do not have any displacement
You probably don't mean that "position vectors are stationary" You mean that an object that is stationary has a constant position vector. (Likewise, an object that has a constant velocity has a constant velocity vector.)
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"Displacement" is a word often used to suggest movement - something that would change as function of time. However, a "displacement" in the mathematics of vectors need not indicate an actual physical process. If an object has coordinates (3,2) we can imagine that the object arrived at (3,2) by making a journey from (0,0) even though it didn't really do that. That's a simple interpretation of what @fresh_42 is telling you about position vector ##\overrightarrow{(3,2)}##.

sophiecentaur
Gold Member
2020 Award
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I don't understand the point here in the context above 'Each of those locations are defined in terms of unit vectors r and θ, or i and j.' could you simplify it, please?
The displacement from me too you has a different sign from the displacement from you to me. Direction counts just as much for position as for velocity and all the other vectors. (Displacement is not distance.)

Hi,
I am new to this forum but I wanted to share with a draft of a short paper that I wrote to clarify some issues I had with the understanding of a position vector.
My experience with the position vector was bad, especially when using cylindrical and spherical coordinates. I was not able to find a logical explanation why the position vectors in these coordinate systems were different than the position vector in Cartesian coordinates. The answer may be out there somewhere but I was not able to find it. I have performed an incomplete research in this topic and I think I finally got an answer.
I am attaching my first draft of the paper and be warned that I am not an expert in math - just a math fan. Still, there is some work to be done to complete it. For instance, I need to practice and do problems with general oriented vectors in the above mentioned coordinates. Anyway, I wanted to post it hoping that I can get some comments and find out if I am just wasting my time.

I thought this thread was a good starting point. Thank you for your help.

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