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>
- easy plane normal vector question