Re: easy plane normal vector question

*To*: mathgroup at smc.vnet.net*Subject*: [mg89707] Re: [mg89690] easy plane normal vector question*From*: "Szabolcs HorvÃt" <szhorvat at gmail.com>*Date*: Wed, 18 Jun 2008 06:40:52 -0400 (EDT)*References*: <200806180824.EAA26154@smc.vnet.net>

On Wed, Jun 18, 2008 at 11:24, will parr <willpowers69 at hotmail.com> wrote: > > 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? Will, This is not really a Mathematica question, but a mathematics one ... I will use Mathematica notation in the following, but the equations are not meant to be entered to Mathematica ... The equation of a plane passing through v0 = {x0,y0,z0} with normal vector n = {p,q,r} is n.(v-v0) == 0 (this is the single equation that you need to figure out), where v = {x,y,z}. Expanding this equation we get p x - p x0 + q y - q y0 + r z - r z0 == 0 Matching coefficients with your equation z == c + a x + b y we get that p == a, q == b, r == -1.

**References**:**easy plane normal vector question***From:*will parr <willpowers69@hotmail.com>