## Two parallel planes

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Svavar
Posts: 3
Joined: Tue Dec 04, 2007 7:10 pm UTC

### Two parallel planes

I have been trying to proof this rule for quite a bit now, our math teacher gave us this rule to solve as homework but I have gone through several books, and searched the internet for any help but I am stuck on what to do. if anyone would like to contribute then please do.

the rule is for the distance between two parallel planes(Ax+By+Cz=D1 and Ax+By+Cz=D2) in 3D space and can be written as the following:

|D1-D2|
------------ =d1-2
|Ai+Bj+Ck|

where (A,B,C) is the normal vector for the parallel planes, i, j andk are unit vectors and d1-2 is the distance between the planes

so.. I know that |Ai+Bj+Ck| is the length of the normal vector and
|D1-D2| has something to do with the distance between them in a certain direction.

the usual method I use to do this is

Ax+By+Cz+D
------------ =d1-2
|Ai+Bj+Ck|

where x,y,z are the coordinates of a point I choose

so please can anyone help me prove this rule?
Last edited by Svavar on Tue Dec 04, 2007 8:54 pm UTC, edited 1 time in total.

mike-l
Posts: 2758
Joined: Tue Sep 04, 2007 2:16 am UTC

### Re: Two parallel planes

The shortest distance from a point on one plane to the other will be along the normal vector. So I would pick a point P on one plane, and make a line P + t(N/|N|) where N is the normal vector, and solve for t so that this lies on the second plane. Since the vector multiplied by t is unit length, the value of t will be the distance.

Mike
addams wrote:This forum has some very well educated people typing away in loops with Sourmilk. He is a lucky Sourmilk.

Svavar
Posts: 3
Joined: Tue Dec 04, 2007 7:10 pm UTC

### Re: Two parallel planes

thanks for the input. but I have thought of this, but it does not help me prove the given rule =/
but I figured D=Ax+By+Cz so the rule can be rewritten as:

|Ax1+By1+Cz1-Ax2-By2-Cz2|
-------------------------------------- =d1-2
|Ai+Bj+Ck|

mike-l
Posts: 2758
Joined: Tue Sep 04, 2007 2:16 am UTC

### Re: Two parallel planes

The method I outlined proves the rule in only a couple of lines.

Call P (x1, y1, z1). What is P + t(N/|N|)?
Plug this into the equation for the second plane. Use the equation of the first plane to eliminate P. Solve for t.

Your method looks promising as well. Write that guy in the numerator as |A(x1 - x2) + B(y1 - y2) + C(z1 - z2)|.

Now the vector (x1 - x2, y1 - y2, z1 - z2) should be a constant multiple of N. In fact, it should be d1-2N/|N|. Go from there.

Mike
addams wrote:This forum has some very well educated people typing away in loops with Sourmilk. He is a lucky Sourmilk.

Svavar
Posts: 3
Joined: Tue Dec 04, 2007 7:10 pm UTC

### Re: Two parallel planes

thanks man that really helped me a lot, now all I have left are some damn polar coordinates exercises

rafa
Posts: 1
Joined: Thu Dec 06, 2007 9:07 pm UTC

### Re: Two parallel planes

nice

TemperedMartensite
Posts: 34
Joined: Tue Nov 20, 2007 4:32 am UTC

### Re: Two parallel planes

Svavar wrote:thanks man that really helped me a lot, now all I have left are some damn polar coordinates exercises

For (Amount of fun had in polar coordinates) x 2 fun, do everything in spherical coordinates

Tryptophan
Posts: 4
Joined: Mon Dec 03, 2007 1:26 pm UTC
Location: Iceland
Contact:

### Re: Two parallel planes

Hmm... Ertu í háskólanum eða?
Ég var í einhverju svipuðu í stæ533 í MH á þessari önn, en við áttum ekki að sanna svona dót. :- p

Gaman að rekast á annan íslending hérna.

-
edit: (english translation, personal question to Svavar)

Are you in the university or something?
We were studying something similar in a certain course in my school (which is not university), but we didn't go much into proofs like this.

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