OpenStudy (anonymous):

It is given that at the point \[ [x_0,y_0,z_0] \] the normal vector to the plane is \[ n= [p,r,s] \]. You can deduce from this that the equation of the plane is \[ p(x-x_0)+r(y-y_0)+s(z-z_0) =0\] How?

5 years ago
OpenStudy (turingtest):

I would walk you through this personally, but my connection sucks right now, so... http://tutorial.math.lamar.edu/Classes/CalcIII/EqnsOfPlanes.aspx

5 years ago
OpenStudy (anonymous):

think of it as a dot product...

5 years ago
OpenStudy (anonymous):

(x -xo), (y-yo) etc are the components of any vector in the plane....

5 years ago
OpenStudy (anonymous):

if <p,r,s > dot < any vector in the plane>=0, then <p,r,s> is normal to the plane and defines the surface..

5 years ago
OpenStudy (anonymous):

or the orientation of the surface, rather

5 years ago
OpenStudy (anonymous):

Yes, POMN says similarly. I think I've got the intuition, thanks both.

5 years ago
OpenStudy (anonymous):

what's POMN?

5 years ago
OpenStudy (anonymous):

oh, Paul's online math notes... never mind:)

5 years ago
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