# Shouldn't a 3D vector be 6 numbers instead of 3?

1. Feb 8, 2010

### LucasGB

A 3D vector is completely defined by its three Cartesian coordinates, which specify the position of the terminal point of the vector. But shouldn't there be 3 other coordinates to specify its starting point?

I understand vectors have no location, and therefore 3 numbers is enough. But is this a convention (mathematicians said, "let's say 2 vector are equivalent even if they're in different places") or is this a requirement from the mathematics?

2. Feb 8, 2010

### DavidSnider

If you specified a starting point it would be an absolute quantity rather than a relative one.

3. Feb 8, 2010

### LucasGB

I don't understand.

4. Feb 8, 2010

### D H

Staff Emeritus
That is the answer to your question in a nutshell (but only for vectors in R3). The mathematical concept of vector is a very generic one. It is not limited to vectors in three space.

5. Feb 8, 2010

### LucasGB

I can see that, but I believe the question really is:

Do we define vectors to have no location, or is this required for logical consistency?
Do we define vectors to be specified only by the coordinates of their terminal point, or does this follow from the mathematics?

6. Feb 8, 2010

### D H

Staff Emeritus
You are thinking only of displacement vectors, and only in three space. The concept of vector is an extremely generic idea. Vectors in general do not have a location. They don't even have to have a "length". They most certainly are not limited to displacements. There are velocity vectors, force vectors, ... Broadening things a bit, one can think of functions as being vectors in an infinite-dimensional space.

7. Feb 8, 2010

### Hurkyl

Staff Emeritus
We define "vector space" to mean a certain kind of algebraic structure -- it has addition, subtraction, multiplication by real numbers, which satisfy a list of axioms which ensures arithmetic behaves nicely.

We call something a vector when we view it as an element of a vector space.

If we specify an origin in Euclidean space, then we can create an algebraic structure whose elements are the points, addition is given by the parallelogram law, and so forth. This algebraic structure happens to be a vector space.

There is a 2-algebraic structure called an "affine space". It has points and arrows*. An arrow has an source and a target, which are objects. There are a variety of algebraic operations; e.g. one can subtract objects to give an arrow. (one cannot add objects)

The main property is that the set of arrows whose source is a fixed object constitute a vector space.

We can create a 2-algebraic structure from Euclidean space by setting:
• Objects are points
• Directed line segments are arrows
• P-Q is the arrow from Q to P
• Arrow addition is the parallelogram law, and so forth

This structure is an affine space.

P.S. the phrase "affine space" is used in several different ways in mathematics. So take care if you go googling around.

*: Arrow is just a name for certain elements of a category. It's only relationship with the English word is that it is fairly common to write them using arrows instead of variables.

8. Feb 8, 2010

### mathman

Vectors do not have a location. They are directions with magnitude. If you specify two triplets then you a defining a line segment, not a vector. The difference is a vector which describes the length of the segment and its direction.

9. Feb 8, 2010

### Hurkyl

Staff Emeritus

In an affine space, there is a standard way to translate an arrow based at point P to become an arrow based on point Q.

If we so desired, we could think of an arrow as representing some 'virtual' vector, with all translates of a given arrow naming the same vector. Then, the process of fixing an origin to create a vector space is really just coming up with a concrete representation of these virtual vectors.

I think this points-and-vectors approach is the more common way to describe an affine space -- that is, Q-P just gives you a vector, rather than an arrow from P to Q.

("virtual" is not a mathematical term, just a descriptive one I've chosen)

10. Feb 9, 2010

### Tac-Tics

You can think of it like this: a vector's "starting point" is always at the origin.

A vector isn't a "pointed line segment". In R^n, it's a direction-length pair. When visualizing a vector, you can move a vector around, and as long as you don't rotate or stretch it, it's the same vector.

For example, the vector from (0, 0, 0) to (1, 0, 0) is the same vector as the one from (1, 1 ,1) to (2, 1, 1).

11. Feb 9, 2010

### Landau

@Tac-Tics: LucasGB already knows this, his question is 'why?'.

The key concept here is affine space, as Hurkyl mentioned. From wikipedia:

12. Feb 9, 2010

### LucasGB

Alright, thanks for the replies. So let me see if I got this straight (please excuse my ignorance on this subject):

Vectors exist in an affine space, and one of the properties of an affine space is that it has no origin. Therefore, vectors can be equivalent even if they are in different locations.

Since we define that vectors exist in an affine space, this means we define vectors to have no location, it is not a necessary consequence of mathematics. Is this correct?

If you were writing a book on vectors and had to choose between one of these alternatives, which one would you pick?

1 - We define two vectors to be equivalent when they have respectively equal components, regardless of where their initial point is.

2 - It is a necessary consequence of the definition of vectors that they must be considered equivalent if they have respectively equal components, regardless of where their initial point is.

13. Feb 9, 2010

### Werg22

A 3d vector is defined as an arrow emanating from the origin, rather than an arrow rooted at an arbitrary point. That's why we only need 3 coordinates to describe it (i.e., we give the coordinates of its ending point).

14. Feb 9, 2010

### Hurkyl

Staff Emeritus
Vectors exist in a vector space. The term "vector" has no meaning except to indicate that we are treating a particular mathematical object as being an element of a particular vector space, which may be left implicit, or sometimes even explicitly specified.

Given any vector space, there is a canonical way to create an affine space out of it.

To each affine space, we can construct lots of different, but isomorphic, vector spaces:
• Each choice of a point to be an "origin" gives us a different vector space
• We can form a vector space out of displacements modulo translation

Okay, that's not quite true. In matrix algebra, we use the term "row vector" and "column vector". But that is a different (albeit closely related) usage of the word.

Neither. "Components" and "initial points" have nothing to do with vectors. There are certain vector spaces for which those terms might have meaning -- but it is a mistake to treat them as if they were part of the concept of "vector space".

Last edited: Feb 9, 2010
15. Feb 9, 2010

### lugita15

I think you have it backwards. The definition of vector came before the definition of vector space. The idea of vectors as quantities with magnitude and direction which can be represented as directed line segments in 2D and 3D space is quite old. It was only much later that people started looking at the properties of operations between vectors, and thus arrived at a definition of vector spaces as sets endowed with operations analagous in certain respects to vector addition and scalar multiplication. So I think it's a red herring to
talk about vector spaces when explaining why vectors don't carry information about their initial points.

Here's my explanation: a vector is defined to be a quantity with magnitude and direction. When representing it as a directed line segment, we represent the direction as the direction of the line segment, and the magnitude as the distance between the initial and final point. Since we only care about the distance between the points and the direction of the line segment, the actual absolute locations of the points doesn't matter. So by specifying the location, or coordinates, of the terminal point relative to the initial point, we know enough to completely specify the magnitude and direction, which is after the only information that a vector carries.

In short, to answer the OP's question, no, it's not just a matter of convention, it's actually a fact that follows from the definition.

I hope that helps.

Last edited: Feb 9, 2010
16. Feb 10, 2010

### HallsofIvy

Historically, that's true. But the definition of "vector space" is logically prior to the definition of vectors themselves. That was what Hurkyl was saying.

You are thinking in terms of vector definitions, Hurkyl is thinking in terms of mathematics definitions.

17. Feb 10, 2010

### lugita15

Yes, it's true that you can "derive," for example, the fact that vector addition is commutative for ordinary 2D and 3D vectors from the vector space axiom of commutativity. But I think that such a "derivation" would not really shed any more insight into why vector addition for ordinary 2D and 3D vectors is commutative. This is because the only reason that commutativity was chosen an axiom for vector spaces is because it is a property of ordinary 2D and 3D vectors. A better explanation would be to draw a parallelogram and to show that the resultant vector is the same regardless of the order in which two vectors are added.

It's analogous to the following issue in geometry: Suppose someone were to ask you why Euclidean space is locally flat. If you answered that it is because Euclidean space is a manifold and all manifolds are locally flat, that would achieve absolutely nothing, because Euclidean space and its local properties formed the prototype from which the general definition of manifolds was constructed.

18. Feb 10, 2010

### LucasGB

That's why sometimes it's so hard to learn math, people don't seem to understand what you just wrote.

19. Feb 10, 2010

### jgm340

Try this:

To understand why a 3D vector has 3 numbers, consider why a 1-dimensional vector only has one number. Examine the statement: "7 + -5 = 2". I could also just as well say "8 + -6 = 2" or "3.45 + -1.45 = 2". This is because we are equating any 1-dimensional line segment from any point on the number line to any other to a vector from the origin to a single point. So, the numbers we use normally can be thought of as 1-dimensional vectors.

You can see here vector addition in 1 dimension with the (3) vector and the (-6) vector:

Now suppose instead we used two numbers to represent a 1-dimensional 'vector' in the form (a : b). Then we'd end up with 'vectors' like (6 : 8) which is not equal to (7 : 5)
Here's what it would look like if we used your proposed vectors (with both start and end points):

It's not clear how we might add these. Maybe like this?

But then we have a 2-dimensional resultant 'vector'. You can try to think of an alternate scheme that fits with the normal properties of addition. We would probably want the sum of two 'vectors' translated by the same amount to be the translation of their sum, so perhaps we could take the midpoint of the start-points of the two vectors to find the start point of the new vector, and then the typical vector sum to find the magnitude and direction. (in one dimension, the direction is -/+). This would look like this:

But what exactly would this get us? It comes down to usefulness. We only have one number for 1-dimensional vectors/numbers, and this makes addition work in the way we know it.

Then you just need to realize that vector addition in 3 dimensions is just a generalization of addition in 1 dimensions. So all the nice properties of 1-dimensional vectors (numbers) translate into 3-dimensional vectors (numbers).

SHORT VERSION:

Why don't we write the vector (5,3,6) as something like [(0,0,0);(5,3,6)]???

The same reason we write the number 12 as "12" instead of "[0;12]"

20. Feb 10, 2010

### LucasGB

@jgm340: thank you very much, that was unbelievably clear. The reason we don't define vectors to have starting points is because by not doing so we can move them around, and moving them around we can define operations such as sum and subtraction! If they had origins, then we wouldn't be able to move them around, or they would become different vectors, and therefore, there would be no way to define these operations!

@Hurkyl, lugita15, David Snider, D_H, Mathman, Tac-Tics, Landau, Werg22, HallsofIvy: thank you all very much for your help!