# Homework Help: Reflection of surface about a plane

1. Dec 23, 2011

### namu

1. The problem statement, all variables and given/known data

Given the sphere (x-2)^2+(y-1)^2+(z-1)^2=4 and the plane x+2*y+2*z=15, find the
equation of that sphere which is the mirror image of the given sphere relative to
the given plane.

3. The attempt at a solution

I was thinking the following

constraint #1: eq. of plane
constraint #2: eq. of sphere

Use lagrange multipliers to find shortest distance from plane to sphere.
This will give you the normal vector from a point on the plane a point
on the sphere.

Reflect, hence n-> -n where n is the normal vector from the point on the
plane found using lagrange multipliers.

So -n-R gives you the new center of the sphere???

But what do we do when we have a not so nice surface in space that we want to
reflect.

2. Dec 23, 2011

### Ray Vickson

You could first obtain a new cartesian coordinate system (x', y', z') in which the plane has the equation x' = 15 (do you see how to do it?). Of course, x = f1(x',y',z'), y = f2(x',y',z') and z = f3(x',y',z'), for known linear functions f1, f2 and f3. Now reflect(x',y',z') = (30-x',y',z'), so the reflected surface can be written in terms of x', y' and z', and thus in terms of x, y and z.

RGV

3. Dec 23, 2011

### namu

OK, let me first give the change of coordinates a try.

I want $x'=15$, so let
$y'=-z$
$z'=z$
$x'=x$

So the plane $x+2*y+2*z=15$ becomes $x'=15$ and the sphere $(x-2)^2+(y-1)^2+(z-1)^2=4$ becomes $(x'-2)^2+(z'+1)^2+(z'-1)^2=4$.

That's a weird equation for a sphere, but I take its due to the change of coordinates. So now I reflect $(x',y',z')$ to $(30-x',y',z')$ and I choose 30 since that is twice the distance from the plane. So I get $(x'-28)^2+(z'+1)^2+(z'-1)^2=4$ which gives $(x-28)^2+(y-1)^2+(z-1)^2=4$. Why am I convinced that I am wrong?

4. Dec 23, 2011

### Staff: Mentor

There's another approach that I think is simpler.

Take a vector that is normal to the plane, that goes from the plane to the center of the sphere.
Negate the coordinates of this vector to get one that goes from the plane to a point that will be the center of the reflected sphere.

The reflected sphere will have the same radius as the original sphere - the only thing different will be its center point.

5. Dec 23, 2011

### namu

That was what I was originally thinking, however I am studying for an exam, and there have been other similar questions to reflect a curve or another surface about the plane. Then I am not so sure what to do.

6. Dec 23, 2011

### Ray Vickson

Your (x',y',z') are incorrect; you need to use a rotational coordinate change. In the original coordinate system the normal to the plane is the vector <1,2,2>; in the new coordinate system this becomes <1,0,0>. So, you need to determine the constants a1, b1, c1, a2,b2,c2,a3,b3,c3 that makes x = a1*x' + b1*y' + c1*z', y = a2*x' + b2*y' + c2*z' and z' = a3*x' + b3*y' + c3*z' an appropriate rotation of (x',y',z').

An easier way is to use the Householder transformation; see, eg., http://en.wikipedia.org/wiki/Householder_transformation . (Basically, the Householder transformation is a way of determining the new coordinate system, but for your problem allows you to bypass the coordinate change entirely.)

RGV