# The product γ is a rotation or a translation

#### mathmari

##### Well-known member
MHB Site Helper
Hey!!

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?

#### Klaas van Aarsen

##### MHB Seeker
Staff member
Hey mathmari !!

It looks correct so far.

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?

#### mathmari

##### Well-known member
MHB Site Helper
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?

#### Klaas van Aarsen

##### MHB Seeker
Staff member
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.

#### mathmari

##### Well-known member
MHB Site Helper
Yep.
So we should set up an equation.
So does it have to hold that $d_{\alpha}(q-p)+p=q$ ?

#### Klaas van Aarsen

##### MHB Seeker
Staff member
So does it have to hold that $d_{\alpha}(q-p)+p=q$ ?
Nope.

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 $r$.

#### mathmari

##### Well-known member
MHB Site Helper
Nope.

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 $d_{\alpha}d_{\beta}\left (x-q\right )+d_{\alpha}(q-p)+p=d_{\phi}(x-r)+r$ ?

#### Klaas van Aarsen

##### MHB Seeker
Staff member
Ah you mean to write $d_{\alpha}d_{\beta}\left (x-q\right )+d_{\alpha}(q-p)+p=d_{\phi}(x-r)+r$ ?
Yep.

#### mathmari

##### Well-known member
MHB Site Helper
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.

#### Klaas van Aarsen

##### MHB Seeker
Staff member
Suppose we solve the equation for $r$?

#### mathmari

##### Well-known member
MHB Site Helper
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?

#### Klaas van Aarsen

##### MHB Seeker
Staff member
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?
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?

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