Finding Perpendicular Planes to a Given Plane

[amcavoy]

Hi,

I suppose my question has to do with planes in general, rather than just tangent planes. Say you have a plane given by the equation

$$z=\frac{\partial f}{\partial x}(x)+\frac{\partial f}{\partial y}(y).$$

How would you find the equation of the plane perpendicular to this one?

Thanks for your help.

 

[mathman]

General approach:

Equation of plane ax+by+cz+d=0, where at least one of (a,b,c) is not =0.
Assume it is a, then bx-ay+m=0 or cx-az+n=0, where m and n are arbitrary. The general idea is to set up a nonzero vector perpendicular to (a,b,c) and use it as coefficients for x,y,z. The constant term is arbitrary.

 

[amcavoy]

General approach:

Equation of plane ax+by+cz+d=0, where at least one of (a,b,c) is not =0.
Assume it is a, then bx-ay+m=0 or cx-az+n=0, where m and n are arbitrary. The general idea is to set up a nonzero vector perpendicular to (a,b,c) and use it as coefficients for x,y,z. The constant term is arbitrary.

Where did the $$bx-ay+m=0$$ and $$cx-az+n=0$$ come from?

 

[amcavoy]

Well, I guess I can understand that. If you have one vector on a plane "r-r0 = (x-x0)i + (y-y0)j + (z-z0)k" and the gradient vector "ai + bj +ck" tangent to f(x,y), would you compute the cross product between these two vectors to find that perpendicular vector, then use that as coefficients?

(ex.) <(x-x0), (y-y0), (z-z0)> x <a, b, c> = <p, q, t>

ax+by+cz=d

So... the tangent plane is px+qy+cz=d.

Is this correct?

 

[mathman]

I tried to give an answer for planes in general. I am a little rusty on your special case (tangent planes).
The plane equations in the previous post just simply reflect the fact that (b,-a,0) and (c,0,-a) are both non-zero and perpendicular to (a,b,c) as long as a is not 0.

 

[amcavoy]

Alright that makes sense then. By which method did you compute that (b, -a, 0) and (c, 0, -a) are perpendicular to (a, b, c)?

The reason I ask is that I have the equation $$z=\frac{\partial g}{\partial x}(x)+\frac{\partial g}{\partial y}(y)$$ and need to find the perpendicular plane (the problem actually has to do with projections).

Thanks again.

 

[mathman]

which method did you compute that (b, -a, 0) and (c, 0, -a) are perpendicular to (a, b, c)?

Inspection!

In your case a=dg/dx, b=dg/dy, c=-1. Since c not=0, you can use (1,0,dg/dx) or
(0,1,dg/dy) or any linear combination of the two vectors.

 

[amcavoy]

Alright I understand now. Thanks a lot for your help.