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Moving Vectors Around

  1. Jun 21, 2012 #1
    Why is it that I am allowed to move vectors around in a coordinate system?
  2. jcsd
  3. Jun 21, 2012 #2


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    Why not? Thgere is no law against it.
  4. Jun 21, 2012 #3
    There might be in some of the southern states.
  5. Jun 21, 2012 #4
    You can move anything around in a co-ordinate system. Vectors in the mathematical sense are geometric objects that are co-ordinate independent. This is particularly useful in Physics as these objects represent physical properties that are invariant i.e. are measured to be the same by different observers.

    I think this is what you are asking.
  6. Jun 21, 2012 #5
    If you move a vector without changing its magnitude or direction, then you have not changed the value of the vector. However, you may have changed its effects. For example, a force has a point of application --it makes a difference where you push something. But, as long as the magnitude and direction are the same, the value of the vector is the same.

    This is usually mentioned in school soon before the class covers the topic of graphical addition of vectors. The teacher places the tail of one vector to begin at the head of another vector, and makes the comment, "Do you remember that rule that I haven't changed the value of a vector if I move it, as long as the magnitude and direction haven't been changed? Well, I'm using that rule now."
  7. Jun 22, 2012 #6


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    It is part of the DEFINITION of "vector". I suggest you review the definition.
  8. Jun 22, 2012 #7
    Because an arrow is not a vector. A vector is the class of ALL arrows with the same direction and magnitude. An individual arrow is a representative of the vector.

    In other words, the arrow from the origin to (1,1) is a representative of the same vector as the arrow from (1,1) to (2,2).

    Does that make sense? The vector is the entire collection of all arrows with the same length in the same direction.

    It's a little like being on vacation and withdrawing money from a branch of your bank near your vacation spot. The branch is not the bank; the branch is a representative of the bank. All branches of the bank are equivalent; you can do your transactions at any branch. Each branch represents the bank. The branch is not the bank. The bank is the collection of ALL its branches. When you need to do business, you can choose the branch that's the most convenient.

    And when we have a vector problem, we are free to choose the representative of the vector that's most convenient.

    A vector is an abstract entity. Arrows are concrete. An arrow starts at one particular point and goes in a particular direction for a particular length. A vector is what we get when we conceptually group all of the arrows together that have the same magnitude and direction, ignoring their starting point.

    Hope something in there helped.
    Last edited: Jun 22, 2012
  9. Jun 22, 2012 #8
    Let me try to explain it formally. I'll be using the 2-dimensional plane [itex]\mathbf R^2[/itex], but you can easily generalize this with [itex]\mathbf R^n[/itex].

    In affine geometry, we can see [itex]\mathbf R^2[/itex] as a set :
    • of points (let's note simply [itex]\mathbf R^2[/itex] the set [itex]\mathbf R^2[/itex] seen as a set of points)
    • or a vector space, in other words, a set of "vectors" (let's note [itex]\overrightarrow{\mathbf R^2}[/itex] the set [itex]\mathbf R^2[/itex] seen as a set of vectors).

    Note that [itex]\mathbf R^2[/itex] and [itex]\overrightarrow{\mathbf R^2}[/itex] are the same set (the set of couples of reals), but perceived differently.

    You can make linear combinations (sum, difference, multiplication by a scalar) in [itex]\overrightarrow{\mathbf R^2}[/itex], using the usual operations on vectors. More formally, we define [itex]+[/itex] as the binary operation [itex](a,b) + (c, d) = (a+c, b+d)[/itex], and so on.

    Note that you can only do this with [itex]\overrightarrow{\mathbf R^2}[/itex] : as, for instance, adding two vectors seems pretty intuitive (you just put the start of the second arrow at the end of the first one), it's counterintuitive to visualize the same operation done with points (if you want to visualize it with points, you'd actually "imagine" vectors between the origin and the points).

    If we take two points [itex]A[/itex] and [itex]B[/itex] in [itex]\mathbf R^2[/itex] (notice there's no arrow here, A and B are points) we define the vector [itex]\overrightarrow{AB}[/itex] as the difference [itex]B - A[/itex]. You might ask yourself: "but I thought we couldn't make sums, differences, etc. with points as we could with vectors". Well yes, but here we'll just use the regular definition of the substraction of two couples of reals, i.e. if [itex]A, B[/itex] have [itex](x_A, y_A), (x_B, y_B)[/itex] for standard coordinates, we'll define [itex]B-A = (x_B - x_A, y_B - y_A)[/itex].

    Now, if you take four points, [itex]A[/itex], [itex]B[/itex], [itex]A'[/itex], [itex]B'[/itex], such that "the arrow that links [itex]A[/itex] and [itex]B[/itex] is equal to the arrow that links [itex]A'[/itex] and [itex]B'[/itex]", it is easy to show that the difference [itex]B-A[/itex] will be equal to the difference [itex]B'-A'[/itex], in other words, that [itex]\overrightarrow{AB} = \overrightarrow{A'B'}[/itex].
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