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Consider an $(x, y, z)$ system where positive $x$ points to the right, positive $y$ points upwards, and positive $z$ points outside of the screen.

I create a new system $(x', y', z')$ by applying two rotations:

  1. Counter-clockwise rotation of $(x, y, z)$ around the $z$ axis, an angle $\theta$. This generates an intermediate system $(x_i, y_i, z_i)$, where $z_i=z$.
  2. Clockwise rotation of the $(x_i, y_i, z_i)$ system around the $x_i$ axis, an angle $\phi$. This generates the final $(x', y', z')$ system.

I know both $\theta, \phi$ angles. I also know the transformation equations between both systems $(x, y, z)$ and $(x', y', z')$:

$$ x' = x\,cos\theta + y\,sin\theta \\ y' = -x\,sin\theta\,cos\phi + y\,cos\theta\,cos\phi - z\,sin\phi \\ z' = -x\,sin\theta\,sin\phi + y\,cos\theta\,sin\phi + z\,cos\phi $$

What I need is the equation for the plane defined by $(x', y')$, in the form:

$$ax+by+cz+d=0$$

How can I calculate this equation?

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    $\begingroup$ Do you simply mean the plane $z' = 0$? In this case, substituting this in the equation for $z'$ above gives you the equation in the desired form, no? $\endgroup$ Jan 15, 2016 at 15:57
  • $\begingroup$ I believe you are absolutely right. I'll check with Emilio's answer to see if I get the same result. $\endgroup$
    – Gabriel
    Jan 15, 2016 at 17:50

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enter image description here

The first rotation around the $z$ axis transform the versor $\hat x=(1,0,0)^T$ of the $x$ axis to $\hat u=(u_1,u_2,u_3)^T=(\cos \theta, \sin \theta,0)^T$.

If I well understand you want the equation of the plane $x,y$ rotated of an angle $\phi$ around the axis oriented by such versor.

You can find such equation rotating in the same way the versor $\hat z=(0,0,1)^T$, orthogonal to the plane $x,y$, so to obtain the versor $\hat v=R( \hat z)$:

Given: $\hat u=(u_1,u_2,u_3)^T$ the rotation matrix is:

$$ R_{\hat u,\phi}=\left ( \begin{array}{cccc} \cos 2\psi + 2u_1^2 \sin^2 \psi &-u_3\sin 2\psi+2u_1u_2 \sin^2 \psi & u_2\sin 2\psi+2u_1u_3 \sin^2 \psi \\ u_3 \sin 2\psi +2u_1u_2 \sin^2 \psi& \cos 2\psi+ 2u_2^2 \sin^2 \psi& -u_1 \sin 2 \psi + 2u_2u_3 \sin^2 \psi\\ -u_2 \sin 2 \psi + 2u_1u_3 \sin^2 \psi & u_1 \sin 2 \psi + 2u_2u_3 \sin^2 \psi& \cos 2\psi+ 2u_3^2 \sin^2 \psi \end {array} \right) $$ where $2\psi=-\phi$ (I prefer this form of the matrix that is derived by representing rotation with quaternions, but it is the same as you can see here, and there is a minus sign because the angle $\phi$ is clockwise)

So, find $\hat v=R_{\hat u,\phi} \hat z$ and the equation of the rotated plane is: $$ \hat v \cdot \vec x=0 $$ with $\vec x=(x,y,z)^T$. Note that the plane passes thorough the origin that is the common fixed point of the two rotations.

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  • $\begingroup$ Great answer Emilio, thank you! Quick question: I'm getting a different sign compared to using the eq $z'=0$ as Travis suggested. Are you sure the element 33 of the matrix shouldn't be $-cos2\psi+...$? $\endgroup$
    – Gabriel
    Jan 15, 2016 at 19:20
  • $\begingroup$ It seems correct. Lock at the same term in the wiki reference. $\endgroup$ Jan 15, 2016 at 22:34
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    $\begingroup$ Your $\phi$ is clockwise, but my matrix is for counterclockwise angles. So if you change $\phi \to (-\phi) \Rightarrow \sin(-\phi)=- \sin (\phi)$ you have the result that you want. ( I'll fix this in my answer). $\endgroup$ Jan 16, 2016 at 10:55
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    $\begingroup$ I've added also a figure that shows the angles counterclockwise. $\endgroup$ Jan 16, 2016 at 11:12
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    $\begingroup$ I use ''tikz'' package in Latex. $\endgroup$ Jan 16, 2016 at 18:01

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