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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
>
--
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>
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