# Reflection across a Plane

1. Aug 10, 2010

### Charles49

Suppose you have the plane given by

$$\bold n \cdot (\bold r-\bold r_0)=0$$

where

$$\bold n$$ is the normal vector to the plane which passes through the point $$\bold r_0$$.

What is the reflection $$x'$$ of a point $$x$$ across this plane?

2. Aug 11, 2010

### quasar987

Consider $\mathbf{r}'_1, \mathbf{r}'_2$ two points in the plane such that the vectors $\mathbf{r}_1:=\mathbf{r}_1-\mathbf{r}_0, \mathbf{r}_2:=\mathbf{r}_2-\mathbf{r}_0$ are linearly independant. (Think of it this way: the vectors r_1,r_2 based at r_0 form a basis for the plane.) Then any point $\mathbf{x}$ of R^3 corresponds to a triple (a,b,c), where

$$\mathbf{x}=\mathbf{r}_0+a\mathbf{r}_1+b\mathbf{r}_2+c\mathbf{n}$$

The reflection of x with respect to the plane is the point x' corresponding to the triple (a,b,-c). I.e.

$$\mathbf{x}'=\mathbf{r}_0+a\mathbf{r}_1+b\mathbf{r}_2-c\mathbf{n}$$

3. Aug 11, 2010

### Charles49

Thanks quasar987!

4. Aug 11, 2010

### qbert

you can take this a step further, and get rid of the coordinates.

(x-r0) . n = a (r'1 - r0) . n + b (r'2 - r0) . n + c n . n = c

and
x' = r0 + a r1 + b r2 - c n
=r0 + a r1 + b r2 + c n - 2 c n
= x - 2 c n
= x - 2 ((x-r0) . n) n

/assuming we normalized n . n = 1

5. Aug 15, 2010

### Charles49

Thanks qbert!