MHB How can we determine the reflection across a given line in Hesse normal form?

[mathmari]

Hey!

Let $g$ be a line with equation $g:ax+by+c=0$ in Hesse normal form. I want to show that the reflection across $g$ is described by \begin{equation*}\binom{x}{y}\mapsto \binom{x}{y}-2(ax+by+c)\binom{a}{b}\end{equation*}

At the reflection across $g$ it holds the following for the image $P'$ of each point $P$:

• $P'$ lies on the perpendicular to $g$ through $P$.
• $g$ bisects $PP'$.

So to show the desired result, do we have to find the perpendicular line to $g$ ? (Wondering)

 

[I like Serena]

Hey mathmari! (Smile)

Let's draw a picture:
\begin{tikzpicture}[>=stealth]
\coordinate (C) at (4,3);
\coordinate (P) at (4.5,6.5);
\coordinate (Q) at (2.5,5);
\coordinate (P') at (0.5,3.5);
\draw[->, green, ultra thick] (Q) -- node[above left] {$(\vec{CP} \cdot \mathbf n)\mathbf n$} (P);
\draw (Q) -- (P');
\draw[->, blue] (0,0) -- node[below right] {$-c\mathbf n$} (C) node

{$C$};
\draw[->, blue, ultra thick] (0,0) -- node[below right] {$\mathbf n = \binom ab$} (4/5,3/5);
\draw[blue, ultra thick] (1,7) -- (7, -1) node

{$g:\mathbf n \cdot \mathbf x + c = ax+by+c=0$};
\draw[->, red, ultra thick] (0,0) -- (P);
\draw[->, red, ultra thick] (0,0) -- (P');
\draw[->, green, ultra thick] (C) -- (P);
\fill (P) circle (0.05) node

{$P$};
\fill (Q) circle (0.05) node [below] {$Q$};
\fill (P') circle (0.05) node

{$P'$};
\fill circle (0.1) node [below] {$O$};
\end{tikzpicture}
We are given $\vec{OP}$ and we want to find $\vec{OP'}$ yes?

Can we construct the vector $\vec{OP'}$ using $\vec{OP}$, $\mathbf n$, and $c$? (Wondering)​

 

[mathmari]

Let's draw a picture:
\begin{tikzpicture}[>=stealth]
\coordinate (C) at (4,3);
\coordinate (P) at (4.5,6.5);
\coordinate (Q) at (2.5,5);
\coordinate (P') at (0.5,3.5);
\draw[->, green, ultra thick] (Q) -- node[above left] {$(\vec{CP} \cdot \mathbf n)\mathbf n$} (P);
\draw (Q) -- (P');
\draw[->, blue] (0,0) -- node[below right] {$-c\mathbf n$} (C) node

{$C$};
\draw[->, blue, ultra thick] (0,0) -- node[below right] {$\mathbf n = \binom ab$} (4/5,3/5);
\draw[blue, ultra thick] (1,7) -- (7, -1) node

{$g:\mathbf n \cdot \mathbf x + c = ax+by+c=0$};
\draw[->, red, ultra thick] (0,0) -- (P);
\draw[->, red, ultra thick] (0,0) -- (P');
\draw[->, green, ultra thick] (C) -- (P);
\fill (P) circle (0.05) node

{$P$};
\fill (Q) circle (0.05) node [below] {$Q$};
\fill (P') circle (0.05) node

{$P'$};
\fill circle (0.1) node [below] {$O$};
\end{tikzpicture}
We are given $\vec OP$ and we want to find $\vec OP'$ yes?

Can we construct the vector $\vec OP'$ using $\vec OP$, $\mathbf n$, and $c$? (Wondering)​

It holds that $$\vec OP'+(-c\mathbf n)=\vec OP$$ doesn't it? (Wondering)​

 

[I like Serena]

It holds that $$\vec OP'+(-c\mathbf n)=\vec OP$$ doesn't it? (Wondering)

Ah no, it doesn't.
It may look like it, but that's just due to a bit of unfortunate choices for the coordinates in the drawing. (Blush)
OCPP' is generally not a parallellogram.

Let me rectify that:
\begin{tikzpicture}[>=stealth]
\coordinate (O) at (-2,-1.5);
\coordinate (C) at (4,3);
\coordinate (P) at (4.5,6.5);
\coordinate (Q) at (2.5,5);
\coordinate (P') at (0.5,3.5);
\draw[->, green, ultra thick] (Q) -- node[above left] {$(\vec{CP} \cdot \mathbf n)\mathbf n$} (P);
\draw (Q) -- (P');
\draw[->, blue] (O) -- node[below right] {$-c\mathbf n$} (C) node

{$C$};
\draw[->, blue, ultra thick] (O) -- node[below right] {$\mathbf n = \binom ab$} +(4/5,3/5);
\draw[blue, ultra thick] (1,7) -- (7, -1) node

{$g:\mathbf n \cdot \mathbf x + c = ax+by+c=0$};
\draw[->, red, ultra thick] (O) -- (P);
\draw[->, red, ultra thick] (O) -- (P');
\draw[->, green, ultra thick] (C) -- (P);
\fill (P) circle (0.05) node

{$P$};
\fill (Q) circle (0.05) node [below] {$Q$};
\fill (P') circle (0.05) node

{$P'$};
\fill (O) circle (0.1) node [below] {$O$};
\end{tikzpicture}​

 

[mathmari]

I don't really know how we could get $\vec{OP'}$. Could you give me a hint? (Wondering)

 

[I like Serena]

I don't really know how we could get $\vec{OP'}$. Could you give me a hint?

Isn't $\vec{OP'}=\vec{OP}-2(\vec{CP}\cdot\mathbf n)\mathbf n$ and $\vec{CP}=\vec{OP}-(-c\mathbf n)$?
Can we solve that? (Wondering)