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Symmetries of cube

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
9,591
If we take as $w$ the cross product of $u$ and $v$ we get $w=u\times v=(1,1,-1)\times (-1,0,1)=(1,0,1)$.
So using these vectors we can now determine the matrix, which is \begin{equation*}\begin{pmatrix}-\frac{1}{2}-\frac{1}{2\sqrt{3}} & 2 & \frac{1}{2}-\frac{1}{2\sqrt{3}} \\ \frac{1}{2}-\frac{1}{2\sqrt{3}} & 0 & -\frac{1}{2}-\frac{1}{2\sqrt{3}} \\ -\frac{1}{\sqrt{3}} & -1 & -\frac{1}{\sqrt{3}}\end{pmatrix}\end{equation*} right? :unsure:
That does not look good. :eek:

On closer examination I see that $u$ and $v$ are not orthogonal.
Their dot product is -2 instead of 0. :oops:

I checked this with geogebra 3d.
If we enter:
Code:
Cube((-1,-1,-1),(1,-1,-1),(1,1,-1))
u=(1,1,-1)
v=(-1,0,1)
and if we rotate the view a bit, then we can see that $u$ and $v$ are not orthogonal. :geek:
 

mathmari

Well-known member
MHB Site Helper
Apr 14, 2013
4,713
I corrected it...

Do we have the following for each case?


1. Rotation around the axis $\mathbb{R}\begin{pmatrix}0 \\ 0 \\ 1\end{pmatrix}$ with angle $\frac{\pi}{2}$ :

The matrix is \begin{equation*}\begin{pmatrix}\cos\frac{\pi}{2}&-\sin\frac{\pi}{2} & 0 \\ \sin\frac{\pi}{2}&\cos\frac{\pi}{2} & 0 \\ 0 & 0 & 1\end{pmatrix}=\begin{pmatrix}0 &-1 & 0 \\ 1& 0 & 0 \\ 0 & 0 & 1\end{pmatrix}\end{equation*}


2. Rotation around the axis $\mathbb{R}\begin{pmatrix}0 \\ 1 \\ 0\end{pmatrix}$ with angle $\frac{\pi}{2}$

The matrix is \begin{equation*}\begin{pmatrix}\cos\frac{\pi}{2} & 0&\sin\frac{\pi}{2}\\ 0 & 1 & 0 \\ -\sin\frac{\pi}{2} & 0 & \cos\frac{\pi}{2} \end{pmatrix}=\begin{pmatrix}0 & 0& 1\\ 0 & 1 & 0 \\ -1 & 0 & 0 \end{pmatrix}\end{equation*}


3. Rotation around the axis $\mathbb{R}\begin{pmatrix}1 \\ 1 \\ -1\end{pmatrix}$ with angle $\frac{2\pi}{3}$

A vector $v$ perpendicular to $u= (1,1, -1)$ is for example $(-1,0,-1)$ with $\|v\|=\sqrt{2}$. An image is $v'=\left (1, -1, 0\right )$.
We consider the vector $w=u\times v=(1,1,-1)\times (-1,0,-1)=(-1,2,1)$ and the image $\displaystyle{w'=\|w\|\frac{u\times v'}{\|u\times v'\|}=\sqrt{6}\frac{(1,1,-1)\times (1, -1, 0)}{\|(1,1,-1)\times (1, -1, 0)\|}=\sqrt{6}\frac{(-1,-1,-2)}{\sqrt{6}}=(-1,-1,-2)}$.

Then we get the matrix \begin{equation*}\begin{pmatrix}0 & 0 & -1 \\ 1 & 0 & 0 \\ 0 & -1 & 0\end{pmatrix}\end{equation*}


4. Rotation around the axis $\mathbb{R}\begin{pmatrix}1 \\ -1 \\ -1\end{pmatrix}$ with angle $\frac{2\pi}{3}$

A vector $v$ perpendicular to $u=(1,-1,-1)$ is for example $(-1,0,-1)$ with $\|v\|=\sqrt{2}$. An image is $v'=\left (1, 1, 0\right )$.
We consider the vector $w=u\times v=(1,-1,-1)\times (-1,0,-1)=(1,2,-1)$ and the image $\displaystyle{w'=\|w\|\frac{u\times v'}{\|u\times v'\|}=\sqrt{6}\frac{(1,-1,-1)\times (1, 1, 0)}{\|(1,-1,-1)\times (1, 1, 0)\|}=\sqrt{6}\frac{(1,-1,2)}{\sqrt{6}}=(1,-1,2)}$.

Then we get the matrix \begin{equation*}\begin{pmatrix}0 & 0 & -1 \\ -1 & 0 & 0 \\ 0 & 1 & 0\end{pmatrix}\end{equation*}


5. Reflection to the plance that is spanned by the vectors $\begin{pmatrix}1 \\ -1 \\ 0\end{pmatrix}$ und $\begin{pmatrix}0 \\ 0 \\ 1\end{pmatrix}$

We have that \begin{equation*}D=\begin{pmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1\end{pmatrix}\end{equation*}
and \begin{equation*}P=\begin{pmatrix}\frac{1}{\sqrt{2}} & 0 & -\frac{1}{\sqrt{2}} \\ -\frac{1}{\sqrt{2}} & 0 & -\frac{1}{\sqrt{2}} \\ 0 & 1 & 0\end{pmatrix}\end{equation*}
and so we get\begin{equation*}A=PDP^{-1}=\begin{pmatrix}\frac{1}{\sqrt{2}} & 0 & -\frac{1}{\sqrt{2}} \\ -\frac{1}{\sqrt{2}} & 0 & -\frac{1}{\sqrt{2}} \\ 0 & 1 & 0\end{pmatrix}\begin{pmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1\end{pmatrix}\begin{pmatrix}\frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} & 0\\ 0 & 0 & 1 \\ -\frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} & 0\end{pmatrix}=\begin{pmatrix}0 & -1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & 1\end{pmatrix}\end{equation*}


:unsure:
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
9,591
1. Rotation around the axis $\mathbb{R}\begin{pmatrix}0 \\ 0 \\ 1\end{pmatrix}$ with angle $\frac{\pi}{2}$ :

The matrix is \begin{equation*}\begin{pmatrix}\cos\frac{\pi}{2}&-\sin\frac{\pi}{2} & 0 \\ \sin\frac{\pi}{2}&\cos\frac{\pi}{2} & 0 \\ 0 & 0 & 1\end{pmatrix}=\begin{pmatrix}0 &-1 & 0 \\ 1& 0 & 0 \\ 0 & 0 & 1\end{pmatrix}\end{equation*}


2. Rotation around the axis $\mathbb{R}\begin{pmatrix}0 \\ 1 \\ 0\end{pmatrix}$ with angle $\frac{\pi}{2}$

The matrix is \begin{equation*}\begin{pmatrix}\cos\frac{\pi}{2} & 0&\sin\frac{\pi}{2}\\ 0 & 1 & 0 \\ -\sin\frac{\pi}{2} & 0 & \cos\frac{\pi}{2} \end{pmatrix}=\begin{pmatrix}0 & 0& 1\\ 0 & 1 & 0 \\ -1 & 0 & 0 \end{pmatrix}\end{equation*}
Correct. (Nod)

3. Rotation around the axis $\mathbb{R}\begin{pmatrix}1 \\ 1 \\ -1\end{pmatrix}$ with angle $\frac{2\pi}{3}$

A vector $v$ perpendicular to $u= (1,1, -1)$ is for example $(-1,0,-1)$ with $\|v\|=\sqrt{2}$. An image is $v'=\left (1, -1, 0\right )$.
We consider the vector $w=u\times v=(1,1,-1)\times (-1,0,-1)=(-1,2,1)$ and the image $\displaystyle{w'=\|w\|\frac{u\times v'}{\|u\times v'\|}=\sqrt{6}\frac{(1,1,-1)\times (1, -1, 0)}{\|(1,1,-1)\times (1, -1, 0)\|}=\sqrt{6}\frac{(-1,-1,-2)}{\sqrt{6}}=(-1,-1,-2)}$.

Then we get the matrix \begin{equation*}\begin{pmatrix}0 & 0 & -1 \\ 1 & 0 & 0 \\ 0 & -1 & 0\end{pmatrix}\end{equation*}
How did you get the matrix? 🤔

Now we get to the orientation.
The map from $v$ to $v'$ must be according to the right hand rule with respect to the axis vector $u$.
It means that $v\times v'$ must be in the same direction as $u$.
That is equivalent to $(u\times v) \cdot v'\ge 0$.
Since we already had $w=u\times v$, we can check $w\cdot v' = (-1,2,1)\cdot (1, -1, 0) = -3\not\ge 0$.
So the orientation is wrong: we need the transpose of the matrix. 🤔

4. Rotation around the axis $\mathbb{R}\begin{pmatrix}1 \\ -1 \\ -1\end{pmatrix}$ with angle $\frac{2\pi}{3}$

A vector $v$ perpendicular to $u=(1,-1,-1)$ is for example $(-1,0,-1)$ with $\|v\|=\sqrt{2}$. An image is $v'=\left (1, 1, 0\right )$.
We consider the vector $w=u\times v=(1,-1,-1)\times (-1,0,-1)=(1,2,-1)$ and the image $\displaystyle{w'=\|w\|\frac{u\times v'}{\|u\times v'\|}=\sqrt{6}\frac{(1,-1,-1)\times (1, 1, 0)}{\|(1,-1,-1)\times (1, 1, 0)\|}=\sqrt{6}\frac{(1,-1,2)}{\sqrt{6}}=(1,-1,2)}$.

Then we get the matrix \begin{equation*}\begin{pmatrix}0 & 0 & -1 \\ -1 & 0 & 0 \\ 0 & 1 & 0\end{pmatrix}\end{equation*}
Checking the orientation of the rotation... $w\cdot v' = (1,2,-1)\cdot (1, 1, 0) = 3 \ge 0$, so this one does have the correct rotation. 😅

5. Reflection to the plance that is spanned by the vectors $\begin{pmatrix}1 \\ -1 \\ 0\end{pmatrix}$ und $\begin{pmatrix}0 \\ 0 \\ 1\end{pmatrix}$

We have that \begin{equation*}D=\begin{pmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1\end{pmatrix}\end{equation*}
and \begin{equation*}P=\begin{pmatrix}\frac{1}{\sqrt{2}} & 0 & -\frac{1}{\sqrt{2}} \\ -\frac{1}{\sqrt{2}} & 0 & -\frac{1}{\sqrt{2}} \\ 0 & 1 & 0\end{pmatrix}\end{equation*}
and so we get\begin{equation*}A=PDP^{-1}=\begin{pmatrix}\frac{1}{\sqrt{2}} & 0 & -\frac{1}{\sqrt{2}} \\ -\frac{1}{\sqrt{2}} & 0 & -\frac{1}{\sqrt{2}} \\ 0 & 1 & 0\end{pmatrix}\begin{pmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1\end{pmatrix}\begin{pmatrix}\frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} & 0\\ 0 & 0 & 1 \\ -\frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} & 0\end{pmatrix}=\begin{pmatrix}0 & -1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & 1\end{pmatrix}\end{equation*}
The explanation seems to be missing... but the result is correct! (Nod)
 

mathmari

Well-known member
MHB Site Helper
Apr 14, 2013
4,713
How did you get the matrix? 🤔

Now we get to the orientation.
The map from $v$ to $v'$ must be according to the right hand rule with respect to the axis vector $u$.
It means that $v\times v'$ must be in the same direction as $u$.
That is equivalent to $(u\times v) \cdot v'\ge 0$.
Since we already had $w=u\times v$, we can check $w\cdot v' = (-1,2,1)\cdot (1, -1, 0) = -3\not\ge 0$.
So the orientation is wrong: we need the transpose of the matrix. 🤔
Why is the orientation wrong? I did the same procedure at the next one, which was correct?
Is the vector of $v$ that I chose wrong? :unsure:
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
9,591
Why is the orientation wrong? I did the same procedure at the next one, which was correct?
Is the vector of $v$ that I chose wrong?
The vector $v$ is fine.
The problem is that we have 2 possible choices for $v'$ and we need to pick the right one.
That is, the one such that $v\times v'$ is in the same direction as $u$. 🤔
 

mathmari

Well-known member
MHB Site Helper
Apr 14, 2013
4,713
The vector $v$ is fine.
The problem is that we have 2 possible choices for $v'$ and we need to pick the right one.
That is, the one such that $v\times v'$ is in the same direction as $u$. 🤔
Taking $v'=\left (0, 1, 1\right )$ we have the following:

We have the vector $w=u\times v=(1,1,-1)\times (-1,0,-1)=(-1,2,1)$ and we get $\displaystyle{w'=\|w\|\frac{u\times v'}{\|u\times v'\|}=\sqrt{6}\frac{(1,1,-1)\times (0, 1, 1)}{\|(1,1,-1)\times (0, 1, 1)\|}=\sqrt{6}\frac{(2,-1,1)}{\sqrt{6}}=(2,-1,1)}$.

The condition is now satisfied, isn't it? We have $w\cdot v'=3$.

So we get the matrix \begin{equation*}\begin{pmatrix}0 & 1 & 0 \\ 0 & 0 & -1 \\ -1 & 0 & 0\end{pmatrix}\end{equation*}

:unsure:
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
9,591
The condition is now satisfied, isn't it? We have $w\cdot v'=3$.

So we get the matrix \begin{equation*}\begin{pmatrix}0 & 1 & 0 \\ 0 & 0 & -1 \\ -1 & 0 & 0\end{pmatrix}\end{equation*}
Yep. All correct. (Nod)