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

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The discussion revolves around finding the line of intersection of bisector planes for a triangle formed by the intersection of the plane ax+by+cz=1 with the coordinate axes. Participants clarify that the task does not require finding the bisector planes themselves but rather two points on the intersection line, with the origin identified as one such point. The coordinates of the triangle's vertices A, B, and C are provided, and the incenter is suggested as a potential second point. A formula for the incenter in two dimensions is mentioned, with an attempt to extend it to three dimensions. Overall, the focus remains on understanding the geometry of the problem and finding the necessary points for the intersection line.
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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.


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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.
 
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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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