# The product γ is a rotation or a translation

• MHB
• mathmari
In summary: I thought to do \begin{align*}d_{\alpha}d_{\beta}\left (x-q\right )+d_{\alpha}(q-p)+p&=d_{\alpha}d_{\beta}x-d_{\alpha}d_{\beta}q+d_{\alpha}q-d_{\alpha}p+p\\ & =d_{\alpha}d_{\beta}x-d_{\alpha}d_{\beta}q+d_{\alpha}d_{\beta}d_{\beta}^{-1}q-d_{\alpha}d_{\beta}^{-
mathmari
Gold Member
MHB
Hey! :giggle:

For $p\in \mathbb{R}^2$ let $\delta_{p,\alpha}=\tau_p\circ \delta_{\alpha}\circ\tau_p^{-1}$.

Let $p,q\in\mathbb{R}^2$ and $\alpha,\beta\in \mathbb{R}$.

(a) Show that $\gamma=\delta_{p,\alpha}\circ\delta_{q,\beta}$ is a rotation of a translation (or both). Give the center of the rotation or the translation vector of $\gamma$ in respect to $p,q,\alpha, \beta$.

(b) Show analytically that the product $\gamma$ of two line reflections is a rotation or a translation. Give the geometric interpretation of the rotation angle/translation vector of $\beta$.
For (a) I have done the following :
\begin{align*}\left(\delta_{p,\alpha}\circ\delta_{q,\beta}\right )(x)&=\left (\tau_p\circ \delta_{\alpha}\circ\tau_p^{-1}\circ\tau_q\circ \delta_{\beta}\circ\tau_q^{-1}\right )(x)\\& =\tau_p\left ( \delta_{\alpha}\left (\tau_p^{-1}\left (\tau_q\left (\delta_{\beta}\left (\tau_q^{-1}(x)\right )\right )\right )\right)\right )\\ & =\tau_p\left ( \delta_{\alpha}\left (\tau_p^{-1}\left (\tau_q\left (\delta_{\beta}\left (x-q\right )\right )\right )\right)\right ) \\ & =\tau_p\left ( \delta_{\alpha}\left (\tau_p^{-1}\left (\tau_q\left (d_{\beta}\left (x-q\right )\right )\right )\right)\right )\\ & =\tau_p\left ( \delta_{\alpha}\left (\tau_p^{-1}\left (d_{\beta}\left (x-q\right )+q\right )\right)\right )\\ & =\tau_p\left ( \delta_{\alpha}\left (d_{\beta}\left (x-q\right )+q-p\right)\right )\\ & =\tau_p\left ( d_{\alpha}\left (d_{\beta}\left (x-q\right )+q-p\right)\right )\\ & = d_{\alpha}\left (d_{\beta}\left (x-q\right )+q-p\right)+p\\ & = d_{\alpha}d_{\beta}\left (x-q\right )+d_{\alpha}(q-p)+p\end{align*}
Is that correct so far? Do we have to substitute the rotation matrices $d_{\alpha}$ and $d_{\beta}$ ? Or is there an other way to continue?

:unsure:

Hey mathmari!

It looks correct so far. (Nod)

We want to prove that it is a rotation around some point or a translation.

So let's assume that it is a rotation with some angle $\phi$ around point $r$.
Then we should be able to equate what you have so far to this unknown rotation.
Can we find $\phi$ and $r$ in that case?

Klaas van Aarsen said:
So let's assume that it is a rotation with some angle $\phi$ around point $r$.
Then we should be able to equate what you have so far to this unknown rotation.
Can we find $\phi$ and $r$ in that case?

From the equation so far we have that we subtruct from $x$ the vector $q$, then we rotate by $\alpha$ and then again by $\beta$ then we add the vector $d_{\alpha}(q-p)+p$.
So... the angle is $\alpha+\beta$, right? But the point? Is the rotation around $q$ since we subtract it at the beginning? But shouldn't we add the same vector at the end as we subtract it at the beginning? :unsure:

mathmari said:
From the equation so far we have that we subtract from $x$ the vector $q$, then we rotate by $\alpha$ and then again by $\beta$ then we add the vector $d_{\alpha}(q-p)+p$.
So... the angle is $\alpha+\beta$, right? But the point? Is the rotation around $q$ since we subtract it at the beginning? But shouldn't we add the same vector at the end as we subtract it at the beginning?
Yep.
So we should set up an equation.

Klaas van Aarsen said:
Yep.
So we should set up an equation.

So does it have to hold that $d_{\alpha}(q-p)+p=q$ ? :unsure:

mathmari said:
So does it have to hold that $d_{\alpha}(q-p)+p=q$ ?
Nope. (Shake)

We should set up the full equation.
The argument to $d_{\alpha+\beta}$is not supposed to be $x-q$. Instead it should be $x-r$ for the as yet unknown center of rotation $r$.

Klaas van Aarsen said:
Nope. (Shake)

We should set up the full equation.
The argument to $d_{\alpha+\beta}$is not supposed to be $x-q$. Instead it should be $x-r$ for the as yet unknown center of rotation $r$.

Ah you mean to write $d_{\alpha}d_{\beta}\left (x-q\right )+d_{\alpha}(q-p)+p=d_{\phi}(x-r)+r$ ? :unsure:

mathmari said:
Ah you mean to write $d_{\alpha}d_{\beta}\left (x-q\right )+d_{\alpha}(q-p)+p=d_{\phi}(x-r)+r$ ?
Yep. (Nod)

Klaas van Aarsen said:
Yep. (Nod)

I thought to do
\begin{align*}d_{\alpha}d_{\beta}\left (x-q\right )+d_{\alpha}(q-p)+p&=d_{\alpha}d_{\beta}x-d_{\alpha}d_{\beta}q+d_{\alpha}q-d_{\alpha}p+p\\ & =d_{\alpha}d_{\beta}x-d_{\alpha}d_{\beta}q+d_{\alpha}d_{\beta}d_{\beta}^{-1}q-d_{\alpha}d_{\beta}d_{\beta}^{-1}p+p\\& =d_{\alpha}d_{\beta}(x+d_{\beta}^{-1}q -d_{\beta}^{-1}p)+
d_{\alpha}d_{\beta}q+p\\& =d_{\alpha}d_{\beta}(x-d_{\beta}^{-1}( p-q))+
d_{\alpha}d_{\beta}q+p\end{align*} but that is still not in the desired form. :unsure:

Suppose we solve the equation for $r$?

A rotation is in the form $d_{\phi}(x-r)+r=d_{\phi}x+(u_n-d_{\phi})r$ so in this case it must hold \begin{align*}&d_{\phi}=d_{\alpha}d_{\beta} \\ &(I_n-d_{\phi})r=-d_{\alpha}d_{\beta}q+d_{\alpha}q-d_{\alpha}p+p \Rightarrow (I_n-d_{\alpha}d_{\beta})r=-d_{\alpha}d_{\beta}q+d_{\alpha}q-d_{\alpha}p+p \\ & \Rightarrow r=(I_n-d_{\alpha}d_{\beta})^{-1}\left (-d_{\alpha}d_{\beta}q+d_{\alpha}q-d_{\alpha}p+p\right )\end{align*}
Itholdsthat $d_{\alpha}d_{\beta}=d_{\alpha+\beta}$.

Then we have a rotation around the point $(I_n-d_{\alpha}d_{\beta})^{-1}\left (-d_{\alpha}d_{\beta}q+d_{\alpha}q-d_{\alpha}p+p\right )$ with angle $\alpha+\beta$. We have a transltion if $d_{\alpha}d_{\beta}=I_n$, then \begin{equation*}d_{\alpha}d_{\beta}x-d_{\alpha}d_{\beta}q+d_{\alpha}q-d_{\alpha}p+p=x-q+d_{\alpha}q-d_{\alpha}p+p=x+\left ((d_{\alpha}-I_n)q-(d_{\alpha}-I_n)p\right )=x+(d_{\alpha}-I_n)(q-p)\end{equation*}
Then we have a translation about $(d_{\alpha}-I_n)(q-p)$. Is that correct? Do we have to consider the case that we have both translation and rotation? Or do allconditions must hold simultaneously in that case? :unsure:

mathmari said:
Is that correct? Do we have to consider the case that we have both translation and rotation? Or do allconditions must hold simultaneously in that case? :unsure:
Put otherwise, we've found that if $I_n-d_\alpha d_\beta$ is invertible, that we have a rotation around a point.
And if $d_\alpha d_\beta=I_n$, that we have a translation.
If there is another possibility, it must be when $I_n-d_\alpha d_\beta$ is not invertible. Note that it isn't in the case of a translation.
Are there such cases?

Last edited:

## 1. What is the difference between a rotation and a translation?

A rotation is a transformation that turns an object around a fixed point, while a translation is a transformation that moves an object from one location to another without changing its orientation.

## 2. How can I tell if the product γ is a rotation or a translation?

The product γ can be determined to be a rotation or a translation by looking at the type of mathematical operation used. If the product involves a rotation matrix, it is a rotation. If it involves a translation vector, it is a translation.

## 3. Can the product γ be both a rotation and a translation?

No, the product γ can only be either a rotation or a translation. These two transformations are distinct and cannot be combined into one.

## 4. What are some real-life examples of rotations and translations?

A real-life example of a rotation is the movement of a bicycle wheel, where the pedals rotate around the fixed point of the axle. A translation can be seen in the movement of a car on a straight road, where the car moves from one location to another without changing its orientation.

## 5. How are rotations and translations used in computer graphics?

Rotations and translations are fundamental operations used in computer graphics to manipulate and transform objects in 3D space. They are used to create animations, simulate movement, and position objects in a virtual environment.

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