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Distance From a Point to a Plane

Date: 03/31/98 at 16:57:06
From: Keir Larock
Subject: Distance between a point (x0,y0,z0) and the plane 


In my calculus class the other day, we went over the proof that the 
distance from the point (x0,y0,z0) to the plane ax + by + cz = d is

     |ax0 + by0 + cz0|/((a^2 + b^2 + c^2)^(1/2))

I was wondering if you could do it out for me, as I have lost my 
notes. Thank you again for your time and help; it is greatly 

Keir Larock

Date: 04/01/98 at 10:24:43
From: Doctor Rob
Subject: Re: Distance between a point (x0,y0,z0) and the plane 

Hi Keir,

Thanks for writing to Dr. Math. 

Consider the plane a*x + b*y + c*z = d. The vector v = (a,b,c) is 
normal to the plane. Pick any point on the plane, P1 = (x1,y1,z1), and 
you are given the point P0 = (x0,y0,z0) in space. Then

     a*x1 + b*y1 + c*z1 = d.

The vector from P1 to P0 is:

     P0 - P1 = (x0 - x1, y0 - y1, z0 - z1)

This vector, projected onto v, will have length the distance from P0 
to the plane. That length is:

   d = |P0 - P1|*cos(theta)
     = |v.(P0 - P1)|/|v|
     = |a*(x0 - x1) + b*(y0 - y1) + c*(z0 - z1)|/sqrt(a^2 + b^2 + c^2)
     = |a*x0 + b*y0 + c*z0 - d|/sqrt(a^2 + b^2 + c^2)

-Doctor Rob, The Math Forum   

Date: 04/01/98 at 13:16:05
From: Keir Larock
Subject: Re: [Re: two problems (x^y = y^x and distance between a point 
(x0,y0,z0) and the plane ax+by+cz+d)]


Thank you very much for your help! Every time I write I am amazed at 
how quickly and thoroughly my questions are answered! I cannot thank 
you guys enough...

Thank you,
Keir Larock
Associated Topics:
High School Calculus
High School Geometry
High School Higher-Dimensional Geometry

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