# Reflection in the plane

Let $\bf{A}$ be a linear transformation in a 3D vector space that represents a reflection in the plane $$x_1 \sin \theta =x_2 \cos \theta$$ Find the matrix that represents this linear transformation

As the question didn't specify the basis to be standard basis, I am thinking of a basis on that plane and apply two points through that transformation. But I'm unfamiliar to the equation that represents the plane, usually it's the form of $\bf{n}.(\bf{x-x_0})=0$, and I don't know where to start with.

• Usually, when the basis is unspecified, the standard basis is implied. – amd Dec 29 '17 at 1:01

The equation of the plane is:

$$x_1 \sin \theta -x_2 \cos \theta=0$$

To write the matrix for the reflection it is convenient at first assume as basis the normal vector to the plane and two linearly independent vectors in the plane:

$v_1=(\sin \theta,-\cos \theta,0)$

$v_2=(\cos \theta,\sin \theta,0)$

$v_3=(0,0,1)$

with respect to this basis to the canonical the reflection matrix is

$$A=\left( \begin{array}{cc} -\sin \theta & \cos \theta & 0 \\ \cos \theta & \sin \theta & 0 \\ 0& 0 & 1 \end{array} \right)$$

to find the matrix with respect to the canonical basis you need a change of basis.

For the change of basis let's consider the matrix $M$ which columns are the basis vectors $v_i$ which components are expressed with respect to the canonical basis:

$$M=\left( \begin{array}{cc} \sin \theta & \cos \theta & 0 \\ -\cos \theta & \sin \theta & 0 \\ 0& 0 & 1 \end{array} \right)$$

thus any vector $v$, given with respect to the $v_i$ basis, can be expressed with respect to the canonical basis as follow:

$$w=Mv\implies v=M^{-1}w$$

Finally, since the reflection from the $v_i$ basis to the canonical is given by:

$$u=Av$$

in the canonical basis the reflection is given by:

$$u=AM^{-1}w$$

• @amd ops...you're absolutely right, thanks! – user Dec 29 '17 at 1:03
• Well, w/r to the chosen basis, the reflection matrix is actually $\operatorname{diag}(-1,1,1)$. Your matrix $A$ has as its “input” basis the special basis, but it outputs vectors with coordinates relative to the standard basis, which is not what’s usually meant by a matrix w/r to a (single) basis. – amd Dec 29 '17 at 1:36
• @amd yes you're right, I've explained after that the matrix A is from basis $v_i$ to the canonical. – user Dec 29 '17 at 1:40

Your plane can be written in the form

$$\begin{pmatrix}\sin\theta\\-\cos\theta\\0\end{pmatrix}\cdot\begin{pmatrix}x_{1}\\x_{2}\\x_{3}\\\end{pmatrix}=0$$

which is the usual form with $\boldsymbol{x}_{0}=0$ and

$$\boldsymbol{n}=\begin{pmatrix}\sin\theta\\-\cos\theta\\0\end{pmatrix}$$