# Equation of a plane

• Sep 9th 2006, 06:45 PM
Ranger SVO
Equation of a plane
I just finished my home work and I have a question

First, I had to find the equation of a plane containing the points (A,0,0), (0,B,0) and (0,0,C)

The Equation I came up with is 2BCx + ACy + ABz = 2ABC

I'm sure I am supposed to see something here, but I don't. Is the equation right?
• Sep 9th 2006, 07:18 PM
ThePerfectHacker
Here is a method that you should have used.
Let the points be, (assuming non-colinear)
$P(A,0,0)$
$Q(0,B,0)$
$R(0,0,C)$
Consider the vectors,
$\bold{QP}=A\bold{i}-B\bold{j}$
$\bold{RP}=-B\bold{j}+C\bold{k}$
Now the cross product,
$\bold{QP}\times\bold{RP}$ is a vector normal to the plane determined by these vectors.
Thus, you know a normal vector and a point the plane contains, everything else is elementary.
• Sep 9th 2006, 07:25 PM
Soroban
Hello,Ranger SVO!

Quote:

I had to find the equation of a plane containing the points (A,0,0), (0,B,0), and (0,0,C).

The equation I came up with is: .2BCx + ACy + ABz = 2ABC .
. . . Why the 2's ?

There is a formula for this very problem (which you can derive yourself) . . .

The plane with intercepts (A,0,0), (0,B,0), and (0,0,C) has the equation:
Code:

    x  y  z     - + - + -  =  1     A  B  C
• Sep 10th 2006, 12:19 AM
CaptainBlack
Quote:

Originally Posted by Ranger SVO
I just finished my home work and I have a question

First, I had to find the equation of a plane containing the points (A,0,0), (0,B,0) and (0,0,C)

The Equation I came up with is 2BCx + ACy + ABz = 2ABC

I'm sure I am supposed to see something here, but I don't. Is the equation right?

You should hear alarm bells ringing when you look at the equation you
have here.

In the problem A, B and C have the same sort of standing and importance,
you would expect the resulting equation to have some symmetry in the way
A, B and C are treated. So you should expect to see the same numerical
coefficients in the second and third terms as you see for the first term on
the left hand side of this equation.

You can see this symmetry in the equation in Sorobans solution, which is
very close to what you have, other than you have lost coefficients of 2
in the second and third terms. Check you algebra, you will probably find
a minor error which when corrected will give you the correct solution.

RonL
• Sep 10th 2006, 12:28 AM
Glaysher
Quote:

Originally Posted by Ranger SVO
I just finished my home work and I have a question

First, I had to find the equation of a plane containing the points (A,0,0), (0,B,0) and (0,0,C)

The Equation I came up with is 2BCx + ACy + ABz = 2ABC

I'm sure I am supposed to see something here, but I don't. Is the equation right?

Easiest version to get from this is the vector equation of a plane

r = (point on plane) + t(direction vector in plane) + s(non parallel vector in plane)

r = (A,0,0) + t(-A,B,0) + s(-A,0,C)

Together with Soroban's and Perfect Hacker's versions (which are much easier to transfer between) you have the main three ways of expressing the equation.
• Sep 11th 2006, 02:23 PM
Ranger SVO
Thanks to everyone.

This Vector stuff may not be to bad
• Sep 11th 2006, 05:29 PM
ThePerfectHacker
Quote:

Originally Posted by Ranger SVO
This Vector stuff may not be to bad

I find it simple, and useful for 3 dimensions.
What you learnining in college? Calculus III how they call it here in America.
• Sep 11th 2006, 07:19 PM
topsquark
Quote:

Originally Posted by ThePerfectHacker
I find it simple, and useful for 3 dimensions.
What you learnining in college? Calculus III how they call it here in America.

I like vectors, just as long as I don't have to visualize them in more than 2D. (Generally I avoid visualizing them at all, if possible.) My visualization skills are pretty stinky for a Physicist. :(

-Dan