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

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Discussion Overview

The discussion revolves around determining the reflection of a point across a line defined in Hesse normal form, specifically the line given by the equation \(g: ax + by + c = 0\). Participants explore the geometric properties of reflection, the relationship between points and their images, and the mathematical formulation of the reflection process.

Discussion Character

  • Exploratory
  • Mathematical reasoning

Main Points Raised

  • One participant suggests that the reflection of a point \(P\) across the line \(g\) can be expressed as \(P' = P - 2(ax + by + c)\begin{pmatrix} a \\ b \end{pmatrix}\).
  • Another participant questions whether finding the perpendicular line to \(g\) is necessary to show the reflection properties.
  • Several participants discuss the construction of the vector \(\vec{OP'}\) using \(\vec{OP}\), the normal vector \(\mathbf{n}\), and the constant \(c\).
  • One participant corrects an earlier assumption about the relationship between \(\vec{OP}\) and \(\vec{OP'}\), indicating that the previous geometric interpretation was flawed due to coordinate choices.
  • A participant expresses uncertainty about how to derive \(\vec{OP'}\) and asks for hints, suggesting a potential formulation involving the dot product with the normal vector.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the exact formulation for \(\vec{OP'}\) or the necessary steps to derive it. There are multiple competing views and ongoing questions regarding the geometric interpretation and mathematical relationships involved.

Contextual Notes

There are limitations in the assumptions made regarding the geometric configuration, particularly in the drawing representations. The relationship between the vectors and the line's normal may depend on specific coordinate choices that have not been fully resolved.

mathmari
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Hey! :o

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

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