Analytic Geometry / Vectors - Finding an Angle Bissector

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SUMMARY

The discussion focuses on deriving the vector, parametric, and symmetric equations of the angle bisector of angle ∠ABC, with points A=(1,2,3), B=(3,4,5), and C=(6,7,0). The user defines point D=(d_1,d_2,d_3) on the bisector and establishes two lines, r and s, based on the coordinates of points A, B, and C. The key insight is that the distances from point D to lines r and s must be equal, leading to the conclusion that finding points D and E on lines BA and BC, respectively, allows for the calculation of the bisector's equation through their average. The discussion emphasizes the importance of normalizing direction vectors for accurate results.

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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:
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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.
 
Oh, of course. Thank you very much Dick.
 

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