0

I want to rotate a plane represented by the equation z = 6 , by n degrees along y axis and find the new equation of the plane. how can this be done?

Thanks

NiladriBose
  • 1,849
  • 2
  • 16
  • 31

1 Answers1

1

Base point (0,0,6) after rotation will lie in XZ plane with coordinates

(x0, y0, z0) = (-6*sin(Fi), 0, 6*cos(Fi))

normal vector

n = (A,B,C) = (-sin(Fi), 0, cos(Fi))

so new plane equation is (for explanation see beginning of the article)

A*(x-x0)+B*(y-y0)+C*(z-z0)=0
or
-sin(Fi)*x + cos(Fi)*z - 6 = 0
MBo
  • 77,366
  • 5
  • 53
  • 86
  • By base point do you mean the center of the plane ? Is there any reference you can point me to to understand your answer? How will the base point be in xz plane? – NiladriBose Aug 17 '13 at 11:18
  • Base point is any point in plane. I chose intersection of plane with Z-axis as initial base point, then rotated it. About rotation - read about affine transformations. About plane defining - link added to the answer. – MBo Aug 17 '13 at 12:29
  • Hi MBo , A bit confused about this (xo,y0,z0) should be (6sin(fi),0,6cps(fi) ) given rotation matrix around y of ([cos(fi),0,sin(fi))],[0,1,0],[-sin(fi), 0 , cos(fi)]). Is that correct? the extra -ve for x should not be there? – NiladriBose Aug 19 '13 at 10:38
  • vector (0,0,6) multiplied by matrix you described - gives (-6*sin(Fi), 0, 6*cos(Fi) vector – MBo Aug 19 '13 at 11:01
  • Not if (0,0,6) is a colum vector? – NiladriBose Aug 19 '13 at 12:02
  • V is row vector when V*M multiplication order is used – MBo Aug 19 '13 at 12:08
  • ok got you ! last one how did you get the normal vector from (x0,y0,z0)? – NiladriBose Aug 19 '13 at 14:08
  • normal vector is independent from base point. – MBo Aug 19 '13 at 16:44
  • Hi Sorry I meant from (Xo , Y0, z0 ) in your answer how did you derive normal to be ( -sin(Fi) , 0 , cos(fi) ) – NiladriBose Aug 20 '13 at 07:50
  • @MBo What would happen if the plane is also translated? Should we add the translations to the base point? Say is we had Translation `(-5,1,-3)` should we do something like `base_point = (x0-5,y0+1,z0-3)`. If this is true the equation of the plane should be `new_norm*(R-new_base_point)=6`. `R=(x,y,z)` – Alexander Cska May 24 '19 at 16:06
  • @Alexander Cska If translation goes after rotation - yes, you are right – MBo May 24 '19 at 16:23
  • @MBo and what about the opposite order? – Alexander Cska May 24 '19 at 16:56
  • @Alexander Cska Base point will have another coordinates – MBo May 24 '19 at 17:01
  • @MBo I Have an equation of a plane x=9. Then I try to run 3D translation and rotation on the point (9,0,0). However I am not able to get the correct position, unless I do only translation. Say I can translate the origin, but any other point involving translation and rotation fails. – Alexander Cska May 24 '19 at 17:12
  • @Alexander Cska What is translation vector? How is rotation defined? – MBo May 24 '19 at 17:34
  • @MBo I have a plane say x=6. The rotation is clockwise -30 degrees. I accomplish this via two steps with numpy `n2=R.dot(np.array(n0))` `pfin= np.array(p0) + np.array(Tr) + n2*dx`. `n0` is the original normal, `p0` the base pint and `Tr` is the translation by which i change the base point – Alexander Cska May 24 '19 at 18:54
  • @Alexander Cska Too much for comments. If you will create a question, describe rotation better - I did not catch. – MBo May 24 '19 at 18:58