Finding an equation of the plane

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the_chemist

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I am having trouble with this homework problem of mine. It's quite basic but I can't seem to get it and I'm a little off. I would appreciate any help anyone can offer me.

Question:
Find an equation of the plane with x-intercept a, y-intercept b, z-intercept c.

The answer is
(x/a) + (y/b) + (z/c) = 1


I don't get this answer...=( This is what I'm doing and this is what I get:

Assuming that you have three points. P1 (a,0,0); P2 (0,b,0); P3 (0,0,c)
Finding two vectors with base (or..starting point) P1 and then finding their cross product to get a vector normal to both vectors I get <bc, ac, ab>

If I dot this vector with (x-a, y-b, z-c) and equal that to 0 I should come out with an equation with the x-,y-,z- intercepts I need...


(normal vector) dotten with (x-a, y-b, z-c) = 0

I think this is where I am going wrong. This is what I get for my dot product

bcx - abc + acy - abc + abz - abc = 0
bcx + acy +abz = 3abc

I simplify and get...
(x/3a) + (y/3b) + (z/3c) = 1

=( This problem is irritating me and has me stumped. Any help anyone can offer me would be greatly appreciated!
 
the_chemist said:
Question:
Find an equation of the plane with x-intercept a, y-intercept b, z-intercept c.
The answer is: (x/a) + (y/b) + (z/c) = 1
Click to expand...
There is really only one question for you to answer!
Does each of (a,0,0), (0,b,0) & (0,0,c) satisfy that equation?
If yes, then the equation must be right.
Three non-collinear points determine a unique plane.
 
Eh? I'm not quite sure I understand what you wrote. But I thought the equation for a plane wasn't unique? If they were collinear that means that all the points lie on the same line right? I don't really understand what it means to be collinear. I'm assuming that all the point would lie on the same plane?

But...the equation wouldn't work when a,b, and c equal 0. So in that instance...the equation isn't satisfied. All other points work. And...I think I'm just going in a circle. hahaha. Hmm...Is my work incorrect or am I looking at this problem differently than what I should be??
 
First, there are lines in \(\displaystyle R^3\).
The point \(\displaystyle \( a,0,0\)\) is on the x-axis; say \(\displaystyle a \not= 0\).
The point \(\displaystyle \( 0,b,0\)\) is on the y-axis; say \(\displaystyle b \not= 0\).
The point \(\displaystyle \( 0,0,c\)\) is on the z-axis; say \(\displaystyle c \not= 0\).
Therefore, those three points cannot be collinear!

Surely you know the axiom: Any three non-collinear points determine a plane.
Now we really do not need the word ‘unique”.
Planes are by definition unique!
 
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