Homework Help: 3-D Geometry problem

1. Sep 15, 2013

Saitama

1. The problem statement, all variables and given/known data
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.

2. Relevant equations

3. 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!

2. Sep 15, 2013

verty

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: Sep 15, 2013
3. Sep 15, 2013

Saitama

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.

$$\frac{x}{\sqrt{b^2+c^2}}=\frac{y}{\sqrt{a^2+c^2}}=\frac{z}{\sqrt{b^2+a^2}}$$
Is this what you get?

4. Sep 15, 2013

verty

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.

5. Sep 15, 2013

Saitama

I think origin is obvious because the bisector planes pass through the axes. :tongue2:
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! :)