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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