# Finding the matrix of a reflection in a plane

So I was wondering how you would determine the matrix of the reflections in a plane in ℝ3. For example if you have a function like $x_1 + x_2 + x_3 = 0$ then how would you find the matrix of the reflection? I know that for lines what I did was I found a point in the line (which I call n) and then I used that in this formula: $e_1 -2\dfrac{e_1 · n}{||n||^2}n$ for e1 and e2 where e1 and e2 = $\begin{bmatrix}1\\0\end{bmatrix}$ and $\begin{bmatrix}0\\1\end{bmatrix}$ respectively. It worked for lines no matter what n vector I picked but for planes I'm using the same formula (except using e1, e2, and e3) but the answer seems to change depending on what n value I picked and none of the answers seem to be the correct one. So do I have to do something differently for finding reflections in planes as opposed to lines?

Also the answer to $x_1 + x_2 + x_3 = 0$ is $1/3\begin{bmatrix}1&-2&-2\\-2&1&-2\\-2&-2&1\end{bmatrix}$ but I can't seem to get that answer using the above formula.

• It should be the same: $$\operatorname{R_n}(v) = v-2\frac{v\cdot n}{\|n\|^2}n$$ where $n$ is any normal to the plane. Note that the plane must contain the origin (or else it's not even a linear transformation). In your case $(1,1,1)$ is one perfectly fine normal vector. Can you figure out how to construct the matrix from here? – user137731 Nov 16 '16 at 0:17
• Are you talking about the x1 + x2 + x3 = 0 question? If so then how would (1, 1, 1) work? – david mah Nov 16 '16 at 0:22
• $$x_1 + x_2 + x_3 = (1,1,1)\cdot (x_1-0,x_2-0,x_3-0)=0 \\ \implies (1,1,1)\ \bot\ \left[(x_1,x_2,x_3)-(0,0,0)\right] \\ \implies \begin{matrix}\bullet &(1,1,1) \text{ is normal^\dagger to the plane and} \\ \bullet &(0,0,0) \text{ is in the plane (as I said it has to be).}\end{matrix}$$ $\dagger$: Here's the Wikipedia page of normality in geometry in case you haven't heard that word before. – user137731 Nov 16 '16 at 0:30

## 1 Answer

If $n = (a, b, c)$ is a unit vector orthogonal to the plane $P$, then reflection in $P$ sends $(x, y, z)$ to $$(x, y, z) - 2\operatorname{proj}_{n}(x, y, z) = (x, y, z) - 2(ax + by + cz)(a, b, c).$$ The standard matrix is $$\left[\begin{array}{@{}ccc@{}} 1 - 2a^{2} & -2ab & -2ac \\ -2ab & 1 - 2b^{2} & -2bc \\ -2ac & -2bc & 1 - 2c^{2} \\ \end{array}\right].$$