MHB Rotation around an arbitrary point

[mathmari]

Hey!

I want to rotate the point $(10, 11)$ with rotation angle $\frac{\pi}{3}$ around the point $P=(-1,-1)$.

I have done the following:

• We translate about $-\vec{OP}$: \begin{equation*}P'=(-1\mid -1)-(-1\mid -1)=(0\mid 0)\end{equation*}
• We rotate around the origin: \begin{equation*}R_{\frac{\pi}{3}}\cdot \binom{10}{11}=\begin{pmatrix}\cos \frac{\pi}{3} & -\sin \frac{\pi}{3} \\ \sin \frac{\pi}{3} & \cos\frac{\pi}{3}\end{pmatrix}\binom{10}{11}=\begin{pmatrix}\frac{1}{2} & -\frac{\sqrt{3}}{2} \\ \frac{\sqrt{3}}{2} & \frac{1}{2}\end{pmatrix}\binom{10}{11}=\begin{pmatrix}\frac{1}{2}\cdot 10 -\frac{\sqrt{3}}{2}\cdot 11 \\ \frac{\sqrt{3}}{2}\cdot 10 + \frac{1}{2}\cdot 11\end{pmatrix}=\begin{pmatrix}5 -\frac{11\sqrt{3}}{2} \\ 5\sqrt{3} + \frac{11}{2}\end{pmatrix}\end{equation*}
• We translate about $\vec{OP}$ :
  \begin{equation*}\begin{pmatrix}5 -\frac{11\sqrt{3}}{2} \\ 5\sqrt{3} + \frac{11}{2}\end{pmatrix}+\begin{pmatrix}-1 \\ -1\end{pmatrix}=\begin{pmatrix}4 -\frac{11\sqrt{3}}{2} \\ 5\sqrt{3} + \frac{9}{2}\end{pmatrix}\end{equation*}

Therefore, the image of $(10\mid 11)$ is $\left (4 -\frac{11\sqrt{3}}{2}, 5\sqrt{3} + \frac{9}{2}\right )$. Is everything correct? Have I done the translation right? (Wondering)

 

[Ackbach]

Hey!

I want to rotate the point $(10, 11)$ with rotation angle $\frac{\pi}{3}$ around the point $P=(-1,-1)$.

I have done the following:

• We translate about $-\vec{OP}$: \begin{equation*}P'=(-1\mid -1)-(-1\mid -1)=(0\mid 0)\end{equation*}

This translation isn't quite right. You need to move the point $(10,11)$ thus:

$\displaystyle (10,11) \to (10,11) - (-1,-1) \to (11, 12)$. So you apply your rotation matrix to the point $(11,12)$. Does that make sense?

The other steps look right to me.

• We rotate around the origin: \begin{equation*}R_{\frac{\pi}{3}}\cdot \binom{10}{11}=\begin{pmatrix}\cos \frac{\pi}{3} & -\sin \frac{\pi}{3} \\ \sin \frac{\pi}{3} & \cos\frac{\pi}{3}\end{pmatrix}\binom{10}{11}=\begin{pmatrix}\frac{1}{2} & -\frac{\sqrt{3}}{2} \\ \frac{\sqrt{3}}{2} & \frac{1}{2}\end{pmatrix}\binom{10}{11}=\begin{pmatrix}\frac{1}{2}\cdot 10 -\frac{\sqrt{3}}{2}\cdot 11 \\ \frac{\sqrt{3}}{2}\cdot 10 + \frac{1}{2}\cdot 11\end{pmatrix}=\begin{pmatrix}5 -\frac{11\sqrt{3}}{2} \\ 5\sqrt{3} + \frac{11}{2}\end{pmatrix}\end{equation*}
• We translate about $\vec{OP}$ :
  \begin{equation*}\begin{pmatrix}5 -\frac{11\sqrt{3}}{2} \\ 5\sqrt{3} + \frac{11}{2}\end{pmatrix}+\begin{pmatrix}-1 \\ -1\end{pmatrix}=\begin{pmatrix}4 -\frac{11\sqrt{3}}{2} \\ 5\sqrt{3} + \frac{9}{2}\end{pmatrix}\end{equation*}

Therefore, the image of $(10\mid 11)$ is $\left (4 -\frac{11\sqrt{3}}{2}, 5\sqrt{3} + \frac{9}{2}\right )$. Is everything correct? Have I done the translation right? (Wondering)

 

[mathmari]

This translation isn't quite right. You need to move the point $(10,11)$ thus:

$\displaystyle (10,11) \to (10,11) - (-1,-1) \to (11, 12)$. So you apply your rotation matrix to the point $(11,12)$. Does that make sense?

Let $M$ be the point that we want to rotate. Do we have to translate the vector $\vec{PM}$ to the origin, so that it becomes a vector of the form $\vec{OM'}$, with $|\vec{PM}|=|\vec{OM'}|$ ? (Wondering)

 

[Ackbach]

Let $M$ be the point that we want to rotate. Do we have to translate the vector $\vec{PM}$ to the origin, so that it becomes a vector of the form $\vec{OM'}$, with $|\vec{PM}|=|\vec{OM'}|$ ? (Wondering)

Exactly!

 

[mathmari]

Exactly!

I got stuck right now. Why is the differnece $M-P$ a vector from the origin? (Wondering)

 

[Ackbach]

I got stuck right now. Why is the differnece $M-P$ a vector from the origin? (Wondering)

Because if you move a vector around, keeping its length and direction the same, you have not changed the vector. The vector $M-P$ is precisely the vector you need to rotate. If you draw the points $M$ and $P$ on plot, and draw the vector from $P$ to $M$, then take that exact vector and translate it so that its tail is at the origin, you can then rotate it with the standard rotation matrices (which assume the tail of the vector to be rotated is already at the origin). Does that help?

 

[mathmari]

Because if you move a vector around, keeping its length and direction the same, you have not changed the vector. The vector $M-P$ is precisely the vector you need to rotate. If you draw the points $M$ and $P$ on plot, and draw the vector from $P$ to $M$, then take that exact vector and translate it so that its tail is at the origin, you can then rotate it with the standard rotation matrices (which assume the tail of the vector to be rotated is already at the origin). Does that help?

So we have the following:

View attachment 6985

A vector i defined by its length and direction. We have the vector PM and the vector OA. We have created the vector OA so that it has the same length and direction as PM. Therefore, OA=PM. is it correct so far? (Wondering)

After the rotation we get the vector OB. Then we have to translate it. How do we do that? (Wondering)

 

[Ackbach]

So we have the following:



A vector i defined by its length and direction. We have the vector PM and the vector OA. We have created the vector OA so that it has the same length and direction as PM. Therefore, OA=PM. is it correct so far? (Wondering)

After the rotation we get the vector OB. Then we have to translate it. How do we do that? (Wondering)

By doing this:

• We translate about $\vec{OP}$ :
  \begin{equation*}\begin{pmatrix}5 -\frac{11\sqrt{3}}{2} \\ 5\sqrt{3} + \frac{11}{2}\end{pmatrix}+\begin{pmatrix}-1 \\ -1\end{pmatrix}=\begin{pmatrix}4 -\frac{11\sqrt{3}}{2} \\ 5\sqrt{3} + \frac{9}{2}\end{pmatrix}\end{equation*}

 

[mathmari]

By doing this:

Ah ok! Thank you very much! (Smile)