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Can I say that vector is more like a "unique identity" of an entity in space rather than calling it an entity with magnitude and direction ?

For example a line. A vector $(10,10,0)$ is the identity of a unique line that starts from $(0,0,0)$ and ends up at $(10,10,0)$ .

Can I apply this notion everytime, everywhere in math and physics ?

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4 Answers 4

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In fact, in maths first you define vector space as a module over a field (take a look at wiki article). And only then you define a vector as an element of a vector space. Moreover, in most cases the "direction" is not a notion one can easily describe. Take, for example, Lebesgue spaces, or any banach space of infinite dimension.

In physics, mostly in analtical mechanics, it's not uncommon to see a reasoning of the type "let's take a vector from point $A$ to point $B$", which begs to define vectors as a class of equivalence (magnitude and direction). All such approaches are equivalent to a formal one, so it's up to you to chose one that makes the reasoning concise and clear.

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The description of vectors as having magnitude and direction is misleading in that it applies only to some vector spaces and not to others. I think, though, that "unique identity of an entity" is even worse, since it strikes me as an essentially meaningless phrase.

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  • $\begingroup$ what vector spaces don't use direction to specify itself ? $\endgroup$
    – Vishwas
    Commented Jul 12, 2013 at 18:10
  • $\begingroup$ @VishwasGagrani For example, the vector space of all continuous real-valued functions on $[0,1]$ doesn't involve magnitude and direction in the usual sense of those words. $\endgroup$ Commented Jul 12, 2013 at 18:40
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In 2D and 3D, I think it's quite reasonable to define a vector as a "directed arrow", or as a displacement of position, or as "a thing that has magnitude and direction (but no fixed position)". You can then show (roughly) that these things form a vector space by checking the mathematical axioms via physical reasoning.

In more abstract settings, a vector is typically defined as an element of a vector space, as the other answer said. This always seems a bit circular, to me. So, if you only care about 2D and 3D, I'd suggest using the less formal (and more physical) notions that I gave above.

Your idea of identifying a vector with a line segment is not quite correct. The line segment has a fixed position in 3D space, whereas vectors have no position.

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Well, i got my answer through another source. So, vectors are not any unique identity on a graph. They are have a magnitude and direction. For example a vector of 5 magnitude and subtending an angle of 30 degree. There are infinite possibilities of such vectors on a graph. Thus vector is not an identity of a unique entity.

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  • $\begingroup$ One of possible approaches to vectors is to introduce classes of equivalence. Let's for simplicity take $\mathbb R^2$. On the set of all segments $\vec {AB}$ joining two points on our plane, we say $\vec{CD}\equiv \vec{FG}$ if they have the same direction (angle) and magnitude. We, of course, need to define "direction", prove that it's indeed equivalence and show that these classes of equivalence form a vector space, which we conventionally call $\mathbb R^2$. $\endgroup$ Commented Jul 12, 2013 at 13:37

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