How Is a Point Reflected Across a Plane in Vector Form?

[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?

 

[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 independent. (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.

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

 

[Charles49]

Thanks quasar987!

 

[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

 

[Charles49]

Thanks qbert!