Solve 3-D Geometry Problem: Find Intersection Line of Bisector Planes

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Homework Help Overview

The problem involves finding the line of intersection of bisector planes associated with a triangle formed by the intersection of a plane with the coordinate axes in three-dimensional geometry. The original poster seeks guidance on how to approach the problem, particularly in determining the bisector planes and the intersection line.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants discuss the necessity of finding the bisector planes versus directly identifying points on the intersection line. There is an exploration of the significance of the origin as a point on the line and the potential role of the incenter of triangle ABC.

Discussion Status

Some participants have suggested that identifying two points on the intersection line may suffice, while others express uncertainty about the overall approach. There is an ongoing exploration of the geometric relationships involved, particularly regarding the origin's position and the incenter's relevance.

Contextual Notes

Participants note the challenge of applying two-dimensional concepts to a three-dimensional context and the potential complexity of finding the incenter in this scenario. There is also mention of the original poster's struggle with recalling relevant geometric principles.

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Homework Statement


The planes ax+by+cz=1 meets the axes OX, OY, OZ in A,B,C. A plane through the x-axis bisects the angle A of the triangle ABC. Similarly, planes through the other two axes bisect the angles B and C. Find the equation of the line of intersection of these planes.


Homework Equations





The Attempt at a Solution


Its been quite some time I have done any problems on 3-D geometry. I can find the points where the given plane intersect the axes but how do I find the bisector planes? I need a few hints to begin with.

Any help is appreciated. Thanks!
 
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You don't need to find the planes, all you need is the line of intersection. All you need is two points on that line. The origin is one point on that line, can you find another point?

I myself don't know what the formula will be, it's a strange question.
 
Last edited:
verty said:
You don't need to find the planes, all you need is the line of intersection. All you need is two points on that line. The origin is one point on that line, can you find another point?

I myself don't know what the formula will be, it's a strange question.

The following are the coordinates of A,B and C:
##A(1/a,0,0)##, ##B(0,1/b,0)## and ##C(0,0,1/c)##

Since the planes bisect the angles, I guess the other point would be the incentre of triangle ABC. Correct? How did you find that origin is a point satisfying the line.

The given answer is:
$$\frac{x}{\sqrt{b^2+c^2}}=\frac{y}{\sqrt{a^2+c^2}}=\frac{z}{\sqrt{b^2+a^2}}$$
Is this what you get?
 
I didn't get an answer, I knew the problem would reduce to finding the incenter, which I didn't try to find. I want you to figure out why the origin is on that intersection, keep thinking about that, draw a picture if necessary.

I don't know how I would find the incenter. I would try to solve it in two dimensions first. There may be no easy way to answer it.
 
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verty said:
I didn't get an answer, I knew the problem would reduce to finding the incenter, which I didn't try to find. I want you to figure out why the origin is on that intersection, keep thinking about that, draw a picture if necessary.
I think origin is obvious because the bisector planes pass through the axes. :-p
I don't know how I would find the incenter. I would try to solve it in two dimensions first. There may be no easy way to answer it.

There is a formula I have used before for finding the incentre in two dimensions. I tried to extend it to three dimensions and it seems to work. The formula can be found here:
http://mathworld.wolfram.com/Incenter.html

Thank you verty! :)
 

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