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Why are two parallel vectors with the same magnitude equivalent?

Why is their start point irrelevant?

How can a vector starting at $\,(0, -10)\,$ going to $\,(10, 0)\,$ be the same as a vector starting at $\,(10, 10)\,$ and going to $\,(20, 20)\,$?

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The start point certainly can be relevant. Imagine I choose to drop a rock either onto your foot or two feet away onto the floor. I suspect the start point for that force vector (both pointing down, with the same magnitude) will be important to you. – KCd Aug 27 '12 at 1:39
What is a vector? The answer to the question that’s posed must depend on your definition of “vector”. For me, a vector is not an arrow but something rather more abstract that is only described by an arrow. Others have different conceptions. – Lubin Aug 27 '12 at 3:56
@Lubin can you elaborate on that? That's kind of what my question was asking in a weird way. What is a vector? – mr real lyfe Aug 27 '12 at 3:58

This is because both these vectors have the same magnitude and direction. Both may have different frames of reference, but for all computational purposes, they are identical. We have centered each vector by taking the starting point as the origin and retaining only the displacement with respect to that point.

However, if for examples, these vectors signify forces acting on a rigid (non-point mass) body, then they could result in different torques and hence have different meaning.

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Vectors denote a direction and length, rather than start and end points. Sometimes they are called free vectors. You can translate the start and end points of a vector by the same amount, and still get the same vector. To see why observe that for each coordinate we have $x_1-x_2 = (x_1+\Delta)-(x_2+\Delta)$. So two vectors with same direction & length are indistinguishable in that sense.

So two parallel vectors have the same direction. The direction of a vector $v = (x ,y)$ is given by the unit vector $u = (\dfrac{x}{\sqrt{x^2 + y^2}}, \dfrac{y}{\sqrt{x^2 + y^2}})$ of length $1$. The example is 2D for simplicity.

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