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Vector proof

  1. Oct 29, 2006 #1
    Prove that [tex]\frac{|B|A+|A|B}{|A|+|B|}[/tex] is the bisector of the angle formed by A and B. where i have used normal text for vector and abs value bars to represent magnitude of vector.

    i have no clue how to get started on this. i have tried many approaches such as constructing a triangle with a, b, and b-a, but i cannot seem to make any progress. a couple of hints on getting started would be appreciated
  2. jcsd
  3. Oct 29, 2006 #2
    Write two equations:

    [tex] ax+by+c = 0[/tex]
    [tex] cx+dy+e = 0 [/tex]
  4. Oct 29, 2006 #3
    and what do these equations represent?
  5. Oct 29, 2006 #4


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    I don't know off the top of my head (i'm not a vector geometry expert), but if you call the bisector vector C, taking [tex]A \cdot C[/tex] and [tex] B \cdot C[/tex] and knowing the cosine half angle formula should be a decent way to start
  6. Oct 29, 2006 #5


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    You want to know that the vector you constructed (I'll call it C) is the angle bisector of A and B. Therefore, you want to know:

    (1) The angle between A and B
    (2) The angle between A and C
    (3) The angle between B and C

    don't you?
  7. Oct 29, 2006 #6
    Suppose that A and B intersect at some point Q, and R is some point on A , and S is some point on B . Write the vector equations of the individual lines, and then of the bisector.
    Last edited: Oct 29, 2006
  8. Oct 29, 2006 #7
    so i wrote A=Q+tQR
    and B=Q+tQS,
    as my two vector equations...how can i write the bisector
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