1
$\begingroup$

If $A, B, C, D$ are points in the plane $R^3$, let $\overrightarrow{u}=\overrightarrow{AB}$ and $\overrightarrow{v}= \overrightarrow{CD}$ (Note: indeed, we can consider any euclidean space, but I am setting the $R^3$ to get intuitive ideas better).

Does it then make sense to add $\overrightarrow{u}+\overrightarrow{v}$?

For me, it doesn't make sense since to add two vectors they should start from the same point. On one hand, from the point of view of Manifold Theory (which I guess is the real theory behind this), the reason why we cannot add them is because they belong to two different tangent spaces (actually two different vector spaces), where the tangent spaces are based on $A$ and $C$ respectively.

On the other hand, I have heard some people say that we are able to add them because we can move one of the vectors, say $\overrightarrow{v}$, in such a way to make his starting point coincide with $A$ (the starting point of $\overrightarrow{u}$). This new vector, call it $\overrightarrow{v\;'}$, is just a translation of $v$ and thus they are thought to be equivalent and because of this we can do the addition. Another way that some people justify the addition of $\overrightarrow{u}$ and $\overrightarrow{v}$ is because (they say) $\overrightarrow{u}=B-A$ and $\overrightarrow{v}=D-C$ and then we are treating them as vectors starting from the origin and of course having vectors like those we can add them.

What I think about the meaning of $\overrightarrow{u}=B-A$ (and respectively for $\overrightarrow{v}=D-C$) is because $\overrightarrow{u}$ has coordinates with respect to the canonical basis in the tangent space $T_A R^3$ given by $B-A$. Also, the coordinates of $\overrightarrow{v\;'}$ in the tangent space $T_A R^3$ are $D-C$ and these is the reason why we consider $\overrightarrow{v}$ and $\overrightarrow{v\;'}$ equivalent.

Where does this equivalence make sense? I think just for geometrical purposes. For instance, when we talk about a normal vector to some plane (contained in $R^3$), we represent this by drawing a vector with its starting point belonging to the given plane. When we draw it we set the normal vector starting at ANY point of the plane, but if the plane has equation $ax+by+cz+d=0$, we say that the normal vector is $(a,b,c)$, which is a vector with its starting being the origin!

What I think here that makes sense to all of these is because when we draw the normal vector starting at any point of the plane, what we mean is that no matter what vector we can pick with the property of being perpendicular to the plane, then its coordinates in the tangent space of its starting point are exactly $(a,b,c)$.

Now, I have another way to justify why it only makes sense when the vectors start at the same point. From Physics, if we have two forces acting on some object, we can add them to get the total force, but this makes sense because we have the two forces acting at the some point, however, if we have one force ($\overrightarrow{u}$) acting on one point ($A$) and another force ($\overrightarrow{v}$) acting on another point ($C)$, what sense does it have to add them? No one, I guess.

I hope my concern is clear and the reasons explained above also.

Thanks in advance.

$\endgroup$
  • 2
    $\begingroup$ A vector and a directed line segment are not the same thing. A vector doesn't 'start' anywhere; it is merely a direction and a magnitude. Also directed line segments have these, but in addition they have endpoints. You're just talking about two different types of objects here. $\endgroup$ – dbx Aug 25 '18 at 2:01
  • $\begingroup$ As a directed line segment is defined by an ordered pair of points, the set of line segments in $\Bbb{R}^3$ is naturally in bijection with $\Bbb{R}^6$, where you have addition of points (vectors). Of course, this comes down to simply adding the endpoints of the vectors. $\endgroup$ – Servaes Aug 25 '18 at 15:06
  • $\begingroup$ There is no reason to invoke tangent spaces when talking about vectors ! $\endgroup$ – Yves Daoust Aug 25 '18 at 15:09
1
$\begingroup$

it doesn't make sense since to add two vectors they should start from the same point ... I have heard some people say that we are able to add them because we can move one of the vectors

Let's make this concrete. Let $A = (2, 0, 3)$ and $B = (-1,2,1).$ Then the vector from $A$ to $B$ is $$ v = \langle -3, 2, -2 \rangle.$$

Look at that again: $\langle -3, 2, -2 \rangle.$ You can write it differently (sometimes people use different brackets, sometimes we write the numbers in a column instead of a list), but the vector is still completely defined by three numbers. Where in $\langle -3, 2, -2 \rangle$ do you see a starting point?

The thing is, you don't even need two points to define a three-dimensional vector. The three coordinates of the vector itself are sufficient. The fact that we can then link this to the displacement between two points is something you can derive, not a definition of "vector."

On one hand, from the point of view of Manifold Theory (which I guess is the real theory behind this), the reason why we cannot add them is because they belong to two different tangent spaces (actually two different vector spaces), where the tangent spaces are based on A and C respectively.

You started with four points $A,$ $B,$ $C,$ and $D$ in $\mathbb R^3.$ If you want to embed a two-dimensional manifold in $\mathbb R^3$ that passes through $A$ and $C$ in such a way that the tangent spaces at $A$ and $C$ respectively contain the vectors $B-A$ and $D-C,$ you can do that, but we still have $\mathbb R^3$ and vectors over $\mathbb R^3$ such as $C - A$ do not magically become undefined.

The tangent spaces at $A$ and $C$ then are two-dimensional subspaces of $\mathbb R^3.$ Addition of vectors from those two subspaces follows the same rules as addition of vectors from any subspaces of $\mathbb R^3$: if you add two vectors from the same subspace, the sum will be in that subspace too; if you add two vectors from two distinct subspaces, the sum is not necessarily in either subspace but is still a vector in $\mathbb R^3.$

From Physics, if we have two forces acting on some object, we can add them to get the total force, but this makes sense because we have the two forces acting at the some point

That's not the only way it can work. We add forces acting on different points all the time.

Consider a box of mass $m$ resting on a flat table. You will frequently see it written that the force the table exerts on the box is a vector of magnitude $mg$ directed straight upward. But on what point is that force exerted? Actually it is a sum of forces exerted at every point of contact between the box and the table.

More generally, if several forces act on different points of a rigid object, the object experiences an acceleration proportional to the vector sum of those forces. The object may be subject to torque as well, due to the different moment arms of the points on which the forces are acting; that is the effect of distributing the forces over multiple points.

$\endgroup$
  • $\begingroup$ 1- what do you understand by saying $<-3,2,-2>$ is the vector from $A$ to $B$? For me, it means that by looking now the origin at $A$, the coordinates of vector $AB$ are $<-3,2,-2>$ (in other words, you are looking at the tangent space of $R^3$ at $A$ and this vector has coordinates $B-A$). 2- Your reason why we can add $AB$ and $CD$ ( because they belong to two dimensional spaces), I think it is just to force them to be added just because they belong to two (different) dimensional spaces! $\endgroup$ – Math Guy Aug 26 '18 at 15:03
  • $\begingroup$ 1. Again, a vector is just a vector. It has no fixed start point. All these extra interpretations you are adding to it (manifolds, tangent spaces, etc.) are just that: extra interpretations. 2. Two vectors defined in the same three dimensional space (as your vectors $u$ and $v$ most certainly are) can be added in $\mathbb R^3.$ And that is what we do when we add $u + v.$ $\endgroup$ – David K Aug 26 '18 at 15:41
  • $\begingroup$ Now, if you have some particular reason to do some vector calculations in the tangent space of some manifold at $A,$ then of course the vector $C - D$ is irrelevant to that calculation; it does not make sense even to mention it in the context of the tangent space at $A$, let alone add it to a vector in that tangent space. But that is not the question you asked: you specified two vectors in $\mathbb R^3,$ not in some disassociated tangent spaces. Not every application of vectors is about tangent spaces. $\endgroup$ – David K Aug 26 '18 at 15:44
  • $\begingroup$ but you still dont explain why the vector (or whatever you can call it, that's not my point) from $A$ to $B$ is $<-3,2,-2>$. I guess if you can clarify that then I can follow your explanations. I explained what $<-3,2,-2>$ means to me in terms of points $A$ and $B$. This is important to be understood because if I can follow your reasoning then I can understand why you say that we can certainly add $AB$ and $BC$. $\endgroup$ – Math Guy Aug 26 '18 at 21:29
  • $\begingroup$ The reason it's that particular vector is because those are the differences in coordinates between $A$ and $B.$ For the $x$ coordinate, $(-2)-1=-3$; for the other coordinates, $2-0=2$ and $1-3=-2.$ Now a question: what is the vector from $E=(4,-1,3)$ to $F=(1,1,1)$? If your answer is not $\langle -3,2,-2 \rangle,$ explain why. Then explain how the exact same notation, $\langle -3,2,-2 \rangle,$ can mean the same thing in two places. $\endgroup$ – David K Aug 27 '18 at 0:40
0
$\begingroup$

Vectors translate freely, they don't have specific endpoints, just a length and a direction.

The sum of two vectors is obtained by constructing a parallelogram such that one side, say $AB$ has the same length and direction as the first vector, and the other side $BC$ has the length and direction of the second vector. Then the sum vector has the length and direction of $AC$.


In various problems, it is customary to speak of the position vector of a point, the length of which is the distance from the origin to the point and the direction is the one from the origin to the point. Though it indicates the absolute position of the point (as the origin is fixed), it remains a plain vector.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.