Is length/distance a vector?

I have heard that area is a vector quantity in 3 dimensions, e.g. this Phys.SE post, what about the length/distance? Since area is the product of two lengths, does this mean that length is also a vector quantity, and why?

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Would be fine if you could give us what you mean by area, length and vector. The answer to your question depend on the definition you use. –  FraSchelle Jun 8 '13 at 6:02

Length and distance are not vector quantities (they are scalar quantities), but position and displacement are vector quantities (at least according to common terminological conventions). Here is how all of these are defined. Note that I am restricting the discussion here to vectors in three-dimensional Euclidean space $\mathbb R^3$.

Every point in three-dimensional space can be specified by a triple of real numbers $\mathbf x = (x,y,z)$ given its coordinates with respect to three axes. This triple is called the position of the point and is clearly a vector.

The length of any vector, such as a position vector, is defined as $$|\mathbf x| = \sqrt{x^2 + y^2 + z^2}$$ Notice that, by definition, length is a positive real number. Given two points $\mathbf x_1=(x_1, y_1, z_1)$ and $\mathbf x_2=(x_2, y_2, z_2)$, the displacement vector pointing from point $1$ to point $2$ is defined as $$\mathbf x_{21} = \mathbf x_2 - \mathbf x_1 = (x_2-x_1, y_2-y_1, z_2-z_1)$$ The length of the displacement vector is called the distance between the two points and is therefore given by $$d(\mathbf x_1, \mathbf x_2) = |\mathbf x_{21}| = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}$$ Note. I have heard some using the terms distance and displacement interchangeably, or using the term displacement for what I have called distance, and using distance to refer to the total length of a path along which an object travels.

Having said all of this, there is in fact a product that allows one to construct area vectors given two position vectors. It's called the cross product. If you take the cross product $$\mathbf x_1\times\mathbf x_2$$ of two position vectors, then you get a vector whose length is the area of the parallelogram spanned by these vectors, and whose direction is perpendicular to this parallelogram.

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Vector is just defined as an element of a vector space (see en.wikipedia.org/wiki/Vector_space#Definition for instance). As such, a length could be seen as a vector if you allow negative length to exist. Otherwise inverse does not exist in general. It is clear that the definition of length given by joshphysics is not a vector for this reason. This is because the length he defined (the most common definition in everyday life) first requires the notion of scalar product to be defined. This is not necessary to define a vector. –  FraSchelle Jun 7 '13 at 17:57
I disagree that position is a vector, position is specified by coordinates, but it's not a vector; it's not a quantity with a magnitude or direction...the displacement vector is a different thing and is a vector. Just thinking in terms of differential geometry, coordinates and vectors are not the same thing. @Oaoa What you are saying would be right in a different context, but not in this one. –  JLA Jun 8 '13 at 4:44
@JLA Huh? In the classical mechanics of systems of particles, positions are represented by elements of $\mathbb R^3$ which forms a vector space under the standard operations of scalar multiplication and component-wise addition. As a result, there is absolutely nothing wrong with using the (completely standard) terminology "position vector." Certainly positions, like those of points in spacetime as in GR, can in some contexts be represented by points on manifolds which aren't necessarily equipped with a vector space structure, but that's a context which I doubt the OP was referring to. –  joshphysics Jun 8 '13 at 5:09
I didn't interpret the OP's question in the formal sense of a vector space, that isn't usually what physics people mean when they use the term vector, otherwise anything could be a vector. They usually mean vectors as in velocity vectors, formally tangents to curves or whatever. It makes sense to write the position of a particle at $(0,1)$ as $\hat{y}$ in cartesian coordinates, but how would you do so in polar coordinates? In polar coordinates the same position is written $(1,\frac{\pi}{2})$, but I don't think identifying that with $\hat{r}+\frac{\pi}{2}\hat{\theta}$ works. –  JLA Jun 8 '13 at 5:17
I'm not saying it's not standard terminology, I just don't think they are the same kind of vector, and it can be confusing to bag them together...It is standard to call angular momentum a vector, but it's clearly not the same kind as a velocity vector. If you change the place of the origin, the position vector changes, a velocity vector won't. –  JLA Jun 8 '13 at 5:48

The length is a scalar. The length has no direction. Whether, a vector is $$\left[\begin{gathered} 5\\ 3 \\ \end{gathered}\right]$$

or $$\left[\begin{gathered} 3\\ 5 \\ \end{gathered}\right]$$

The length is still the same and the vectors only have different directions, i.e. are rotations of each other. So the length has NO direction.

I don't know why you think that area is a vector but it is a scalar, it has no direction. The area of a bivector is actually its magnitude. In case, you don't know what a bivector is, it is something with a magnitude (its area) and 2 directions.

But the area IS a scalar. Also note that the bivector actually lives in the exterior space, not $\mathbb R^n$ or $\mathbb I^n$ or $\mathbb C^n$

Or maybe you are confused that the cross product of two vectors (which is a vector by itself) has a magnitude equal to the area of the bivector spanned by them? Keyword here is "MAGNITUDE". It (edit: I realised this could be confusing, I mean that the cross product of the two vectors itthemselves, which is a vector not their magnitude) is NOT EQUAL to the area of the bivector. Rather than that, the magnitude of this vector cross product is equal to the area of the bivector.

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@dimension0 Down vote. Everything is wrong. I can define a length to be $\int_{\alpha}^{\beta}dx$ and it will not be a scalar, but a vector (an element in a vector space). The opposite is simply $\int_{\beta}^{\alpha}dx$. It is a basic fact that an area is orientable. As such, it is a vector $V$ even in the simple meaning of a collection of point $V=\left(\alpha, \beta, ... \right)$. It seems you define area and bivector the way you want and you change the definition with each sentence. –  FraSchelle Jun 8 '13 at 6:08
Moreover, in the question, there is a direct mention of this post physics.stackexchange.com/questions/14165/… when the relation between vector and area is well explained. So down-vote again. –  FraSchelle Jun 8 '13 at 6:11
@Olaoa: I am talking about the standard euclidean definition of vector's magnitude, which does not have to be the only one. It is the bivector which is oriented while the area only measures the magnitude of a bivector. Could you please explain how I change the definition? In my head, I'm using the same definition again and again. From your next post, the reason why you think my post is wrong is clear. I'm not defining area as the vector pointing from a vertex of a parallelogram to the vertex on the other side of the diagonal. I'm defining it (if you want that rigorous math) to be the (contd.) –  Dimensio1n0 Jun 8 '13 at 6:20
(contd.) determinant of the matrix with its columns as the 2 vectors spanning the parallelogram. –  Dimensio1n0 Jun 8 '13 at 6:20
Then you have to define what you call an area. I can define an area as $\int dx \wedge dy$ which is certainly not a scalar. I continue to believe that this question is ill-defined. Since there is no definition of either length, area or vector, one can answer whatever we want. As soon as the OP will have defined these three notion, (s)he will answer the question by him/herself. –  FraSchelle Jun 8 '13 at 6:26