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 > -- Daniel Huber Metrohm Ltd. Oberdorfstr. 68 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>