# Very basic question about vector orientation

by cytochrome
Tags: basic, orientation, vector
 P: 145 I was thinking about something that really confused me... Say you have a vector in R^2 that is defined by (0,2) so that it is pointing upward on the y-axis. What would the coordinates of the vector with the opposite orientation be (so that the tail of the vector is at the point (0,2) and the head is at (0,0))??? How do you define vector orientation with just coordinates?
 P: 352
 P: 749 What is the vector (a,b) such that (a,b) + (0,2) = (0,0)?
P: 810

## Very basic question about vector orientation

The thing you're describing isn't a vector.

A vector in R^2 is just a point in R^2. When we think of vectors as "arrows" they are always arrows from the origin to the specified point. So (0, 2) "as an arrow" would be the arrow starting at (0, 0) and leading up to (0, 2).

If you wanted to define a separate entity: "an arrow on a plane", you could do so. Let's say an arrow is a PAIR of vectors in R^2, (u, v). The vector u is the butt of the arrow and v is the head (where it points to).

Arrows have some similarities to vectors. Arrows form a vectorspace: you can them together and you can multiply by scalars: (u, v) + (u', v') = (u + u', v + v'), and a * (u, v) = (a * u, a * v).

We can define the "flip" operation you are curious about: flip a = flip (u ,v) = (v, u). (It happens to be a linear operator).

In a sense (that can be made technical), arrows have "more information" than vectors in R^2. Without too much difficulty, you can actually show that the vectorspace of arrows is isomorphic to R^4.

tl;dr, vectors aren't just "arrows". They are arrows based at the origin.
P: 145
 Quote by Tac-Tics The thing you're describing isn't a vector. A vector in R^2 is just a point in R^2. When we think of vectors as "arrows" they are always arrows from the origin to the specified point. So (0, 2) "as an arrow" would be the arrow starting at (0, 0) and leading up to (0, 2). If you wanted to define a separate entity: "an arrow on a plane", you could do so. Let's say an arrow is a PAIR of vectors in R^2, (u, v). The vector u is the butt of the arrow and v is the head (where it points to). Arrows have some similarities to vectors. Arrows form a vectorspace: you can them together and you can multiply by scalars: (u, v) + (u', v') = (u + u', v + v'), and a * (u, v) = (a * u, a * v). We can define the "flip" operation you are curious about: flip a = flip (u ,v) = (v, u). (It happens to be a linear operator). In a sense (that can be made technical), arrows have "more information" than vectors in R^2. Without too much difficulty, you can actually show that the vectorspace of arrows is isomorphic to R^4. tl;dr, vectors aren't just "arrows". They are arrows based at the origin.
Thank you! For some reason I wasn't thinking that vectors have to be based at the origin.

To the other replies, the vector (0,-2) is just a reflection about the x-axis. I was trying to find what the "arrow" from (0,2) to (0,0) would be defined as.
Sci Advisor
P: 5,435
 Quote by cytochrome Thank you! For some reason I wasn't thinking that vectors have to be based at the origin.
They don't. Google for the difference between "free vectors" and "bound vectors". Unless the distinction is important, they are both just called "vectors".

Nether free vectors or bound vectos have to be based at the origin, but bound vectors are only meanigful if they are "attached to" some definite point in space. Free vectors just specify a magnitude and a direction, not "attached to" any particular position.
 P: 1,065 A vector that "starts at 0.2" is not a meaningful statement. It is also incorrect to say that vectors are based at the origin. I could move a vector 2000 units to the left and it is the exact same vector, and it's not based at the origin. This is also why the law of transformation for 1st order tensors (and indeed any order tensor) only contains terms relating the angles of the axes of the two coordinate systems, and says nothing of where the origin is in the two coordinate systems. So in other words, if you have two coordinate systems A and B with the only difference being the location of the origin, using the law of transformation to describe vector v in coordinate system B given its description in coordinate system A would show that this vector is described in the exact same manner in both coordinate systems.
P: 810
 Quote by AlephZero They don't.
 Quote by 1MileCrash A vector that "starts at 0.2" is not a meaningful statement. It is also incorrect to say that vectors are based at the origin.
In the most technical sense, vectors don't start anywhere. They aren't arrows. They also don't necessarily have a length and you can't necessarily take angles between them.

However, it's just really darned convenient to image vectors in R^n as arrows starting at the origin. It's a perfectly valid representation of them. Scaling is done through scaling the length, and addition is done through constructing a standard parallelogram.

There are other interpretations as well, of course.

Although...

 Google for the difference between "free vectors" and "bound vectors". Unless the distinction is important, they are both just called "vectors".
I've never heard these terms before. I don't feel like they are widely used. (At least not in mathematics).
P: 1,065
 Quote by Tac-Tics However, it's just really darned convenient to image vectors in R^n as arrows starting at the origin. It's a perfectly valid representation of them. Scaling is done through scaling the length, and addition is done through constructing a standard parallelogram.
I didn't say it was wrong, I said it was not meaningful. It is not meaningful when describing or defining a vector. Obviously, it is just the natural thing to do when drawing one, but the geometrical idea of a vector is just that, a geometrical description of a mathematical object. A vector is truly nothing more than a group of 3 numbers that obey the laws of transformation for a 1st order tensor.

 I've never heard these terms before. I don't feel like they are widely used. (At least not in mathematics).
Have you heard of "local tensors?" It's really the same idea.
Sci Advisor
P: 2,895
 Quote by cytochrome Say you have a vector in R^2 that is defined by (0,2) so that it is pointing upward on the y-axis. What would the coordinates of the vector with the opposite orientation be (so that the tail of the vector is at the point (0,2) and the head is at (0,0))??? How do you define vector orientation with just coordinates?
As you can see from the answers, there is more than one mathematical definition for the term "vector". A "vector" in the sense of a element of a mathematical "vector space" does not necessarily have a representation with a finite number of coordinates, nor does it have a defined orientation and magnitude. (This completely contradicts the saying "A vector is anything that has magnitude and direction", which I've seen in secondary school physics textbooks.) I'll leave the definition of a "vector" in the sense of a specialized tensor to be explained by those who know such things.

Your question is fairly specific, because it asks about a "vector" that either "is" a line segment or represents some properties of a line segment in 2-D Euclidean space. This narrows the meaning of "vector", but it still leaves some ambiguity. You have to decide if you want to talk about "free vectors" or "bound vectors", as a previous post suggests. The usual method in physics is to think in terms of "free vectors", but only compute with "bound vectors".

You aren't using the term "orientation" correctly. Two line segments of unequal lengths can have the same or opposite orientations. You are asking for more than a vector with "opposite orientation". Decide whether you want to talk about "free vectors" or "bound vectors" and then rephrase you question in one of those contexts.

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