I am going through a derivation of a physics theorem (particulars are not important). It involves a sphere centered at the origin. The observation is made that $\hat{n} = \hat{r}$ (unit normal to surface of sphere vector and unit radial vector, respectively).

My understanding is that a vector (in physics contexts) is defined by magnitude and direction. The vectors are obviously parallel and the same magnitude. However, it bothers me that no reference to the "position" of the vectors is made. And, that it is conceivable to have $\hat{r}$ located with tail at the origin and $\hat{n}$ located at the surface of the sphere even if they are parallel and same magnitude.

If we consider vectors in a linear algebra context, it seems like this problem doesn't even exist since they are more abstract objects not necessarily existing in cartesian space, for example (at least to my understanding).

  1. Is there a rigorous way of understanding if/why two vectors can be equal even if they have different locations? Maybe more generally, what is the role of location when considering vectors?

  2. How can I connect these two "interpretations" of vectors? Am I missing some assumptions that we make when deciding to use vectors in physics? Or, plain wrong.

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    $\begingroup$ You are in a math forum. My understanding of math vectors is that there isn't any position. "Parallel" vectors are equal/indistinguishable. You might want to ask on a physics platform for force vectors. $\endgroup$ Jun 14, 2022 at 0:32
  • $\begingroup$ I would like a math answer, which is why I posted here :). $\endgroup$ Jun 14, 2022 at 0:33
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    $\begingroup$ @ChristopherQuinnLaFondJr. Given the nature of the question, I'd bet that fiber bundles/normal bundles are well beyond the OP's scope. $\endgroup$ Jun 14, 2022 at 0:42
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    $\begingroup$ It's also possible to have vectors that in a diagram are drawn with tails at different points, for example several forces applied to a rigid body, that lie in the same vector space. These vectors do not intrinsically have different locations in my view. You can add the vectors just like abstract vectors in order to find the net linear force on the object. You can also find the torque, but you need two vectors at each point where the forces were applied, one vector for the force and the other vector for displacement from the point where you want to find torque. $\endgroup$
    – David K
    Jun 14, 2022 at 0:42
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    $\begingroup$ Would it not be smart to put answers to the questions in the answer box and reserve the comment section for comments? $\endgroup$ Jun 14, 2022 at 0:45

4 Answers 4


Here is a mathematical approach. Let us stick with $\mathbb{R}^{2}$.

We can identify a line segment in $\mathbb{R}^{2}$ as a pair of ordered pairs $(a,b)\in\mathbb{R}^{2}$ and $(c,d)\in\mathbb{R}^{2}$ where $(a,b)$ denotes the origin of the line segment and $(c,d)$ indicates its end.

Based on such definition, we establish the following relation: the line segments $((a,b),(c,d))$ and $((e,f),(g,h))$ are related iff the following relation holds: \begin{align*} (c,d) - (a,b) = (g,h) - (e,f) \end{align*}

It can be proven that such relation is an equivalence relation, that is to say, reflexive, symmetric and transitive. We are now able to define the concept of a vector. Given an ordered pair $v = ((a,b),(c,d))$, the corresponding vector associated to it is the equivalence class: \begin{align*} [v] = \{((e,f),(g,h))\in\mathbb{R}^{2}\times\mathbb{R}^{2} : v \sim ((e,f),(g,h))\} \end{align*}

Thus two vectors are equal if they represent the same equivalence class.

Hopefully this helps!

  • $\begingroup$ Thank you! This was most helpful. $\endgroup$ Jun 15, 2022 at 15:32

In mathematics, you don't understand things. You just get used to them.

Dramatic quotes aside, I think the question of whether or not vectors are 'fixed' somewhere in space is pretty much just a matter of definition. Since it can be a source of confusion for students, some textbooks are beginning (at early stages of the book) to clearly make the distinction.

For instance, in Vector Calculus by Marsden and Tromba (one of the main multivariable calculus books used in western universities), they define vectors as follows (pg. 6 in 6th edition):

'Vectors (also called free vectors) are directed line segments in [the plane or] space represented by directed line segments with a beginning (tail) and an end (head). Directed line segments obtained from each other by parallel translation (but not rotation) represent the same vector. When we insist on vectors beginning at the origin, we will say that we have bound vectors.'

I suppose it follows that when you want to consider a vector fixed at a point other than the origin, you would just say something like 'the vector with tail at the point $(1,0)$' or something like that, but it isn't specified.


In the usual formulation of general vectors mathematically, i.e. as elements of a vector space, vectors do not have such a thing as "positions", any more than ordinary numbers do. You don't need a "position" to say that the temperatures taken at two different points in a room are equal, so a pair of vectors that apply at two different points don't likewise need "positions", either, to call them "equal".

The relevant property of vectors is that you can "add" them to "positions" or better, points. In particular, if $P$ is a point and $\mathbf{v}$ is a vector, then $P + \mathbf{v}$ is another point that is $|\mathbf{v}|$ distant from it, and away from $P$ toward the direction encoded in $\mathbf{v}$. Coordinatewise, if $P = (P_x, P_y)$ and $\mathbf{v} = \langle v_x, v_y \rangle$ then $P + \mathbf{v} = (P_x + v_x, P_y + v_y)$.

In the physics context, vectors like force and velocity that do not directly have units (or physical dimension) of position cannot, of course, be added to positions, but ultimately they relate to them in some other more indirect way that is still consistent with this, often involving combinations with other vectors, and constants that, when multiplied by, change their physical dimension. Of course, in your case, both vectors do have such units, so it makes sense to talk about them that way, but it's important to keep in mind regarding things like force, velocity, and so forth.

In the geometric context, to say the "vector is normal" is to actually say that the segment created by applying that vector is normal to the surface at the given point, i.e. the segment $\overline{PQ}$ with $Q = P + \mathbf{v}$ when $P$ is the desired point on the surface. Another, equivalent way, that is more directly "vectorish", is that a vector is normal to a plane if the vector is normal to any vector which translates a point in the plane to any other point in the plane, and for curved surfaces of course, the natural extension of both of these is via the tangent plane.

So how are two vectors equal? Generally, any two mathematical objects are equal if all characterizing data are identical between them. For vectors, the characteristics that are relevant are magnitude and direction, though the latter is kind of hard to formulate, so it is often simpler to instead characterize them by their Cartesian components from the start, viz. for any vector $\mathbf{v}$, $\mathbf{v}$ is defined formally as the ordered pair $\langle v_x, v_y \rangle$, where we've used the angle brackets to indicate the semantic difference versus a coordinate point. That is, the vector is this object, for the purposes of mathematical manipulation. Since these data completely characterize the vector, if we have another $\mathbf{u}$ with $\mathbf{u} = \langle u_x, u_y \rangle$, then $\mathbf{u} = \mathbf{v}$ if and only if $u_x = v_x$ and $u_y = v_y$.


I do agree with you that this isn't covered very well in physics textbooks.

The rigorous way of understanding how vectors can be equal even if they are in different locations is that equality of located vectors is an equivalence relation.

A reference I found useful was "Calculus of several variables" by Serge Lang. It starts off describing located vectors as pairs of points, then goes on to define a vector as being an $n$-tuple of reals whose magnitude and direction can be compared without specifying the location.


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