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Vector proofs

  1. Jun 10, 2006 #1
    i have to prove the diagonals of a rhombus intersect at right-angles using the scalar dot product.

    i have set up a cartesian plane system where A lies on the origin. numbering the sequential points clockwise, i let B = (a,b), C = (a+c, b) and D = (c,0). I then thought if i set up vectors AC and DB and found the dot product between them, i could get a value of zero. unfortunately thats not the case so where did i go wrong?
     
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  3. Jun 10, 2006 #2

    Hootenanny

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    Perhaps you made an arithmetic error? Your method appears good.
     
  4. Jun 10, 2006 #3

    Hurkyl

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    Well, he didn't set up a rhombus; he set up a parallelogram.
     
  5. Jun 10, 2006 #4

    Hootenanny

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    So that would explain why my dot product was non zero. I fixed it for a rhombus and the proof works fine.
     
  6. Jun 10, 2006 #5
    hmm...where would my flaw lie? i think i just may be overlooking a critical property of a rhombus.
     
  7. Jun 10, 2006 #6

    Hurkyl

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    What's the definition of a rhombus?
     
  8. Jun 10, 2006 #7
    a parallelogram having equal sides. i still can't see how the way i defined the coordinates is wrong, maybe just a little obscure (can be simplified).
     
  9. Jun 10, 2006 #8

    Hurkyl

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    So did you, anywhere in your set-up, impose the condition that all the sides were equal?
     
  10. Jun 10, 2006 #9
    ahh im confused, the way i have my diagram, a traditional rhombus, it looks right. i can see how a square would be different.

    so back to basics:
    AB = BC = CD = DA

    but that doesn't necessarily mean the coordinates will be represented that way, just the magnitude.
     
  11. Jun 10, 2006 #10

    Hurkyl

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    What if you set a = 400, b = 400, and c = 1?


    In the way you set up your problem, the vector AB = (a, b), and the vector AD = (c, 0). But clearly they are not automatically equal!
     
  12. Jun 10, 2006 #11
    hmm i see your point. but then what would AB = ?

    basically a^2 + b^2 = c^2

    so AB would equal (sqrt(c^2-b^2), sqrt (c^2 - a^2)), can't be right, much too messy...
     
  13. Jun 10, 2006 #12

    Hurkyl

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    I suspect that this is all you need -- go back and redo your work, but make use of this relation.


    Incidentally, this problem is easier if you don't use coordinates at all!
     
  14. Jun 11, 2006 #13
    oh ok, i just left my coordinates as is but when i got down to the dot product,

    after i get a^2 + ac - ac - c^2 + b^2
    i stated that a^2 + b^2 = c^2
    so c^2 - c^2 = 0


    i hope thats sufficient. what other method would have been easier Hurkyl?
     
  15. Jun 11, 2006 #14

    Hurkyl

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    Do essentially the exact same thing, but never use coordinates: leave everything in terms of vectors.
     
  16. Jun 11, 2006 #15

    Hurkyl

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    All right, I'll spoil the answer if you haven't worked it out yet yourself!


    Let x and y be the vectors denoting two sides of our rhombus.

    Then, as you noted, the two diagonals are:

    x + y

    and

    x - y

    and because the sides are equal:

    x² = y²

    The dot product of the diagonals is:

    (x + y).(x - y) = x² - y² = 0

    and thus, they are perpendicular.


    (Where I've used the notation that x² means x.x)
     
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