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Analytic Geometry / Vectors - Finding an Angle Bissector

  1. Jun 10, 2013 #1
    Question:

    What are the vector, parametric and symmetric equations of the angle bissector of angle ∠ABC, given that A=(1,2,3), B=(3,4,5) and C=(6,7,0).

    Attempt at resolution:

    Well, I defined some D=(d_1,d_2,d_3) to be a point in the angle bissector, and two lines r:X=(3+2a,4+2a,5+2a) and s:Y=(3+3b,4+3b,5-5b) for any a,b, which are the lines defined by points A and B and B and C respectively.

    So, I realize that the distance between D and r must be the same as the distance between D and s, and that D=(d_1,d_2,d_3)=(3-2m+3n,4-2m+3n,5-2m-5n) for some m and n. The problem is that I do not seem to be able to find some efficient way to solve for these restrictions.
     
    Last edited: Jun 10, 2013
  2. jcsd
  3. Jun 10, 2013 #2

    Dick

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    Normalize the direction vectors for the two lines. In other words, if you can find a point D on BA that's one unit away from B and a point E on BC that's one unit away from B, then F=(D+E)/2 will be on the bisector, right? Once you have have a single point on the bisector that's not B, you can find its equation.
     
  4. Jun 10, 2013 #3
    Oh, of course. Thank you very much Dick.
     
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