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Re: easy plane normal vector question

  • To: mathgroup at smc.vnet.net
  • Subject: [mg89738] Re: easy plane normal vector question
  • From: dh <dh at metrohm.ch>
  • Date: Thu, 19 Jun 2008 05:44:33 -0400 (EDT)
  • References: <g3agp3$pjo$1@smc.vnet.net>


Hi Will,

write your equation in the form:

ax+by==-c

or

{a,b}.{x,y}==-c

you see that the scalar product of {a,b} with any vector on (from origin 

to a point on the surface) the surface is -c. Therefore, as a vector in 

(between two points on) the surface is the difference between two 

vectors on the surface, the scalar product of a vector in the surface 

with {a,b} is zero. {a,b} is perpendicular to the surface.

hope this helps, Daniel



will parr wrote:

> Dear Math Forum,

> 

> sorry but i'm not very good at vector maths, so this is probably a very easy question:

> 

> if i have the equation for a plane in the form:

> 

> plane = c + ax + by

> 

> (for example plane=4.07919+ 0.0764884 x - 0.0632675 y)

> 

> as you get if you use the Fit function to fit some 3D data (data) in the following way:

> 

> Fit[data,{1, x, y},{x, y}],

> 

> How do i calculate the normal vector to this plane?

> 

> best wishes,

> 

> Will

> 





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Daniel Huber

Metrohm Ltd.

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CH-9100 Herisau

Tel. +41 71 353 8585, Fax +41 71 353 8907

E-Mail:<mailto:dh at metrohm.com>

Internet:<http://www.metrohm.com>




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