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Vectors that constitute the diagonals of this parallelogram

  1. Sep 17, 2007 #1
    this is for my mechanics course, but its just dealing with vectors. sorry if this is in the wrong place.

    1. The problem statement, all variables and given/known data

    The problem is:
    You are given two vectors (4,0) and (0,4) that form the sides of a parallelogram. Determine the vectors that constitute the diagonals of this parallelogram and show by means of vector properties that they intersect each other at right angles.

    2. Relevant equations

    3. The attempt at a solution

    I'm assuming that you start at the origin for the vectors. Then you would add the vectors, so the diagonals would be (4,4). I'm not sure if both diagonals would be (4,4).
    Once I have the diagonals, I think that you would take the dot product and if it equals 0, then that would prove that they intersect at right angles.
  2. jcsd
  3. Sep 17, 2007 #2


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    In fact, your parallelogram is a square. You are correct both in saying that one diagonal is < 4,4 > and that *both* are not.

    Think about the diagonals (or even draw a picture) and consider what the components for that other diagonal would have to be.
  4. Sep 18, 2007 #3
    the other diagonal... one end is at (0,4) and the other end is as (4,0) so then would the diagonal be <4,-4>?
  5. Sep 18, 2007 #4


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    That would be one way of writing it; I was thinking of <-4,4>, which is really just the vector of the same length in the opposite direction.

    The point is: if we call one of the vectors defining one set of parallel sides of the parallelogram a and the vector defining the other set of parallel sides b, how could we calculate the vectors representing the two diagonals?
  6. Sep 18, 2007 #5
    the vectors representing the two diagonals would be a-b, right?
  7. Sep 18, 2007 #6


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    Well, the second one you came up with is, and the first one was a+b. Notice that the order doesn't really matter. Reversing the order of which vectors you called a and b does nothing to a+b and only reverses the direction of a-b, which just "puts the arrowhead on the other end of the diagonal".

    To sum up, our vectors were <0,4> and <4,0>, so the diagonals of the parallelogram they define are <4,4> *and* <-4,4> or <4,-4>. There is absolutely nothing special about this method that limits it to the square we were given; it applies to any vector-defined parallelogram.
    Last edited: Sep 18, 2007
  8. Sep 18, 2007 #7
    yah, it makes sense. I got the problem done, thanks for the help!
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