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Intercept Equation


Date: 10/07/98 at 18:03:27
From: Sara
Subject: Algebra II - Intercept equation

I found a plane using the intercepts (4,0,0), (0,-5,0), and (0,0,3).
Now I want an equation for those points using Ax + By + Cz = D.

It can't be as easy as using (4,-5,3) because that doesn't work.


Date: 10/08/98 at 12:13:12
From: Doctor Peterson
Subject: Re: Algebra II - Intercept equation

Hi, Sara -

You're on the right track. Let's just plug each of your three points 
into the equation and see what we get:

    (4,0,0) ---> 4A + 0B + 0C = D
    (0,-5,0) --> 0A - 5B + 0C = D
    (0,0,3) ---> 0A + 0B + 3C = D

So we have these equations to solve to get the coefficients:

     4A = D
    -5B = D
     3C = D

That's three equations in four unknowns, so we can choose whatever we 
want for D. You could just use D = 1 and you'd get a fine equation, 
but I'll choose the LCM of 4, 5, and 3 to make it come out neatly:

    D = 60
    A = 60/4 = 15
    B = 60/-5 = -12
    C = 60/3 = 20

So an equation for the plane is

    15x - 12y + 20z = 60

How's that?

Your thought of using (4,-5,3) is close; another form of the equation 
uses the inverses of those for A, B, and C:

     x     y     z
    --- - --- + --- = 1
     4     5     3

Can you see where that came from? In fact, given the intercepts 
(a,0,0), (0,b,0), and (0,0,c), the plane is given by

     x     y     z
    --- + --- + --- = 1
     a     b     c

- Doctor Peterson, The Math Forum
  http://mathforum.org/dr.math/   
    


Date: 06/11/2004 at 02:54:29
From: Tom
Subject: equation of a plane

I read through your answer to Sara, but I'm a little confused by 
some parts of it.  Could you explain a few things again?  Looking 
at your equations:

x/A + y/B + z/C = 1 
 4A = D
-5B = D
 3C = D

For the first equation where does the 1 come from?  And for the set 
of three equations, you said that because you have three equations for 
four unknowns you can assign any value for D.  Why is this?


Date: 06/11/2004 at 10:59:46
From: Doctor Peterson
Subject: Re: equation of a plane

Hi, Tom.

There are a few different ideas mixed together on this page, because 
of the way the question was asked.  Let's try untangling them.

First, the "patient" asked for the equation of the plane through 
the three given points, in the general form Ax + By + Cz = D, where 
A, B, C, and D have no specific meaning (though we can relate them 
to the normal vector to the plane).  To find the coefficients, I 
suggested simply plugging in each point (x,y,z), giving three 
equations to solve.  

Normally you can't solve three equations for four unknowns, (A, B, 
C, and D); but with this equation it doesn't matter.  That's 
because you can multiply the whole equation by any constant and it 
still represents the same plane.  For example, if we multiply by 
1/D, we get

  A/D x + B/D y + C/D z = 1

and since A/D, B/D, and C/D are all still arbitrary constants, we 
could just rename them as A, B, and C.  So we could say that the 
general form of the equation is

  Ax + By + Cz = 1

since that would cover all planes.  The only reason we don't express 
it that way is that for many planes (at least those used in school 
problems!) we can make all the coefficients integers if we don't 
require a 1 on the right side.  So the form with D often looks nicer.

To put it another way, we can choose any value for D (I chose 1 just 
now) and adjust the equation so that D will have that value.  My 
suggestion was to choose D so that all the coefficients would be 
integers; if you don't see why I chose what I did, you can just take 
D = 1 and see what you get.

After going through that, I introduced another formulation of the 
equation of a plane that is specifically adapted for the problem Sara 
was solving.  When you are given the intercepts a, b, and c, rather 
than three arbitrary points, you can use the equation

  x/a + y/b + z/c = 1

You're asking where this comes from.  One approach to it is to do 
exactly what we had just done with her specific problem: take the 
intercepts as (a,0,0), (0,b,0), and (0,0,c), and plug those into the 
general equation.  Do the same algebra, and you find that

  A = 1/a, B = 1/b, C = 1/c

If you have any further questions, feel free to write back.

- Doctor Peterson, The Math Forum
  
Associated Topics:
High School Basic Algebra
High School Equations, Graphs, Translations
Middle School Algebra
Middle School Graphing Equations

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