My question is quite similar to the one I have already found at Math Stack, i.e. Why shift the result of subtracting vectors?. As one of the answers explains, a vector is uniquely determined by it's length and orientation, so we can shift the arrow that represents it any way we want and still refer to the same vector.

However, I'm still missing something here. If the representation of a vector is not unique, then how for example interpret the equation of a line $$ \vec{r}(t) = \vec{a} + t\vec{v}, $$ where $\vec{v}=(a,b,c)$ is some direction vector and $\vec{a}=\vec{OP}$ ($P=(x_0,y_0,z_0)$ is a given point). If for a fixed $t=t_0$ we have $\vec{r}(t_0) = \vec{a} + t_0\vec{v}$ I can clearly see the vector addition and that the endpoint of $\vec{r}(t_0)$ determines a point on the line. But if I will take $\vec{r}(t_0)$ and shift it to some other point in $\mathbb{R}^3$, the endpoint will be moved as well (but the vector will be the same in the sense of the previous definition). In fact, by shifting $\vec{r}(t_0)$ I can produce arbitrary point as an endpoint of it, which is clearly absurd.

I am clearly missing something here and none of my textbooks explains this (perhaps obvious) situation.

  • $\begingroup$ This is not true in spherical or curvilinear basis vectors. $\endgroup$ – marshal craft May 12 '17 at 12:22
  • $\begingroup$ Maybe its a complicated relating to linear algebra, and vector spaces, and then topology. $\endgroup$ – marshal craft May 12 '17 at 12:45

Another way to look at the situation is as follows.

We can make Cartesian coordinates for a three-dimensional Euclidean space by choosing an $x$ axis, a $y$ axis, and a $z$ axis, which define a coordinate system on this space. The point where the three axes intersect is the origin of this coordinate system and is given the name $(0,0,0)$ in the coordinate system. Each point $p$ in space, which is defined by its position in the space, has some coordinates $(p_x,p_y,p_z)$ in the coordinate system.

We can also define a three-dimensional vector space in which each vector $v$ is described by three coordinates $(v_x,v_y,v_z).$ A vector in this vector space doesn't really have a "position" in the sense that we think of the point $p$ in the Euclidean space as having a position, but the vector does have a length and a direction.

Because either a point or a vector can be described by three coordinates, it is convenient to write equations involving objects of three coordinates that are alternatively interpreted either as coordinates of points or coordinates of vectors. Hence in $$ \vec{r}(t) = \vec{a} + t\vec{v}, $$ we start with the three coordinates of a point $a$ through which the desired line passes and treat them as coordinates of a vector $\vec a.$ We take the three coordinates of a vector $\vec v$ parallel to the desired line, and add some multiple of this vector to $\vec a.$ The result is a vector $\vec{r}(t),$ whose three coordinates we interpret as the coordinates of a point $r(t)$ on the line.

In order to make sense out of this, we might use the concept of position vectors. The position vector of a point $p$ is the vector $\vec p$ equal to the distance and direction from the origin of coordinates to $p.$ Constructed this way, the vector $\vec p$ always has the same coordinates in the vector space as the point $p$ has in the Euclidean space, so it's easy to swap a vector for a point or a point for a vector in calculations.

We may even stop representing points as separate objects and talk only about their position vectors. This is often what is meant by the equation $\vec{r}(t) = \vec{a} + t\vec{v},$ where $\vec{r}(t)$ is a position vector of a point on the line and that is simply how we identify that point.


There is another way to think about this. Namely, forget the notion of "moving a vector".

All vectors start from the origin. a "vector" $\vec{p} = (x, y, z)$ is just the arrow from $(0, 0, 0)$ to $(x, y, z)$

Now, in this model, your "line" equation:

$$ r(t) = \vec{a} + t\vec{v} $$

at a particular choice of $t$ gives me the "position vector" $r(t)$ from the origin.

For vector subtraction, we always do the triangle thing about the origin, since all our vectors start form the origin! there is no "moving a vector to align them". Vectors always have the same start point (the origin) and the only thing matters is where they end: the $(x, y, z)$.

The cleanest way to think of a vector in my opinion is the way I described, since it eliminates a whole class of confusion between "position vector" and "non position vector" and all of that. Forget it - all vectors start from the origin.

  • $\begingroup$ Ok, so two things here: 1. So with your interpretation, the triangle rule for subtracting vectors is not properly pictured, because the result vector is translated and we should just perform regular addition $\vec{u}-\vec{v} = \vec{u} + (-\vec{v})$ and leave it as it is (I refer to the pictures in the link I included)? $\endgroup$ – Mat Dyl May 12 '17 at 12:34
  • $\begingroup$ 2. Doesn't representing vectors this way breaks when we want to visualize a vector field? $\endgroup$ – Mat Dyl May 12 '17 at 12:35
  • $\begingroup$ Yeah, the triangle rule is simply a visual aid, it is not the "formal" definition of vector subtraction. As the comment points out, I would argue the second diagram describes $\vec{u} - \vec{v}$, since it shows the actual vector (at the origin). The third diagram shows the sum of the two vectors. $\endgroup$ – Siddharth Bhat May 12 '17 at 12:36
  • $\begingroup$ No, it doesn't break down for a vector field, since a vector field maps each point in a space to a vector. For simplicity, if you consider a 1D vector field, it is a function $f: \mathbb{R} \rightarrow V$. So, $f(x)$ is a vector. It is not "located" at a point in $R$ (in fact, $V$ may not even be compatible with $R$). The function $f$ simply tells you how go from each point in $R$ to a vector "at the origin in $V$. We can visualise this by "placing" each vector in $V$ at a point in $R$. $\endgroup$ – Siddharth Bhat May 12 '17 at 12:41
  • $\begingroup$ What exactly do you mean by "vector at the origin in $V$"? $\endgroup$ – Mat Dyl May 12 '17 at 13:04

Ok so to start let's say we have really this Euclidean space $\Bbb R^3$ and we will call it $G$ for global or the underlying space. Now we can define a basis $\mathbf x, \mathbf y, \mathbf z$. Also we need to define coordinates $x,y,z$ so that any coordinate or position can be represented as an n-tuple $(x,y,z)$.

Now though this coordinate n-tuple kind of behaves like a vector too. In fact we can change coordinates just as we can change basis?

So could we change coordinates, and change the basis, or just one or the other?

But underneath we may or may not want to represent the same object with the new coordinates and or basis.

Now we can move $r(t)$ around but it still looks like a line right or maybe it is a adjacent line?

In one case the vector covaries and the other contravaries. The two notions are dual and you could say every vector space has a dual space.

So $x_0\mathbf x_0 + y_0\mathbf y_0+z_0\mathbf z_0$ is a vector and now consider $x_0 \mathbf x_1+y_0\mathbf y_1+z_0 \mathbf z_1$. Considering it as coordinate vector $(x_0,y_0,z_0)$ the two are the same.

Or we could take $( \mathbf x_0, \mathbf y_0, \mathbf z_0)$ and $( \mathbf x_1, \mathbf y_1, \mathbf z_1)$ to be the vectors, here they are not the same "entity".

  • $\begingroup$ See differential geometry for more thorough explanation. $\endgroup$ – marshal craft May 12 '17 at 13:09

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