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?

Question

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?

turingtest · Answer

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

anonymous · Answer

think of it as a dot product...

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

or the orientation of the surface, rather

anonymous · Answer

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

anonymous · Answer

what's POMN?

anonymous · Answer

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