Dear Gert:  For a line     y= m x +  b   Then by definition of Hugh coordinates  (r, theta)  in 2D, one can get easily the following connections:  m= -cotan theta  and  b=r cosec theta. I  prefer to use:   theta =[0,2Pi].

Dear Tarik, I mean the 3D plane. From my point of view the challange is the consideration of the pattern center :-). And a happy New year!

Dear Gert; you gave only one angle and one distance in your question. For the 3D representation as you know one employs vectorial distance of the plane with respect to the origin, then needs two angles {theta and phi in polar coordinates} and the radial distance . A person knows a vectorial geometry the answer to your question is not very tough.

The normal distance of plane is  d:  where  ıjk tri-pot orthonormal  set 

d = i  r sin Theta cos Phi  +  j  r sin Theta sin Phi  +  k  r cos Theta

x=r sin Theta cos Phi;   y=r sin Theta sin Phi  and z=r cos Theta

Regards.

Dear Tarik, my question is of course related to EBSD :-). There you have the origin of the signal (projection centre) which is described by the pattern centre on the screen and the projection plane distance DD (usually called detector distance). The Hough transformation is commonly independed from the pattern center and therefore often given for the image centre or the corner of the image. This makes the problem a bit challenging (from my point of view).  I.e. we have to coordinate systems, or at least a shift to consider. Attached is a slide containing a graph which displays the arrangement.

Dear Gert; I am not familiar with this projection system. I have had any time to deal with the developments have been taking place in Theory of X-ray and Electron Diffraction and their application since left Turkey for Germany in 1980. I thank you very much sending me that figure in RG. I would like to look at the problem to see what those people cooked up in introducing this transformation. Cheers

Gert , the vectorial calculation of  phi angle is perfectly  alright.  But what is necessary here to transform these  vectorial quantities into the  measurable quantities, which are all scaller. Scince I have no idea what are those measured quantities I can not put  that transformation into a mathematical format. You should give me some hint, I don't wish to go Google!!!

Hern Gert, I found a  very elagant solution of this problem using the vector and matrix calculus simultaneously, which doesn't need to use the vector product in the final numerical evaluation. All I need to know how you are representing R vector? in terms of its cartesian coordinates or its  spherical coordinates or what??? Please specify.

Finally it doesn't matter as you showed in one of your previous comments. It should be fast calculatable :-), so that trigonometric functions are usually not the best way. I would prefer carthesian coordinates since the aim is the calculation between these vectors. For this I can use then x1*x2+y1*y2+z1*z2=arccos(phi) if the vectors are nromalized. 

Dear Gert, what would you do if the carthesian coordinates are not available but the polar ones?  Here is the recipy:

H1. H2=  (T1XR)  (T2XR)=( T1.T2)  (R.R) -  (R. T1)  (R.T2),

Hi.2 = Ti2  R2  - (Ti. R )2.   All one  needs now is to get the Carthesian cordinates out of polar ones  with the recipy as I furnished below:  

Here, every thing is simple algebra! 

cos-1Zeta =H1. H2/{SQR(H12). SQR(H22)} ; where  Zeta is the angle between two planes 1 and 2 .

T1= rho1 ( i sin theta1 + J cos theta1],  polar coordinates  as defined by Gert Noize in 2D

T2= rho2 ( i sin theta2 - J cos theta2],, polar coordinates  as defined by Gert Noize in 2D

R  =R [ i sin thetaR cos PhiR +  j sin thetaR sin PhiR  + k  Cos thetaR ]; the  spherical coordinates use in 3D space 

I thank Dr. Noize in introducing me this very interesting topic in XRD which was completely strange to me since I haven't been involved in XRD last fourty years.