Pane Rotation - Computer Graphics

Date: 11/14/96 at 16:14:15
From: Bart
Subject: Rotating a pane in 3d space

Dear Dr. Math:

Hello!  I am looking to see if anyone can help me solve a math problem 
that I have self-imposed :-).  It has to do with 3d vectors.

All right, let's say you have a pane in 3d space defined by four 
points.  You can rotate it using variables rx, ry, and rz.  The 
rotation formula is as follows: (The formula has to be used for all 
the points with the same rx, ry, rz values)

s1 = SIN(ry)
s2 = SIN(rz)
s3 = SIN(rx)
c1 = COS(ry)
c2 = COS(rz)
c3 = COS(rx)
rotated x = x(s1s2s3 + c1 * c3) + y(c2s3) + z(c1s2s3 - c3s1)
rotated y = x(c3s1s2 - c1s3) + y(c2c3) + z(c1c3s2 + s1s3)
rotated z = x(c1s2s3 - c1s1) + y(-s2) + z(c1c2)

Using the right values, you can rotate the pane so that all of the 
rotated z values are equal and the pane is flat in front of you.  Now, 
what I would just love to have is  a formula that computes the correct
rx, ry, and rz values to make the Z's equal!

Any  help will be greatly appreciated! Thanks in advance.

Date: 11/19/96 at 18:10:00
From: Doctor Tom
Subject: Re: Rotating a pane in 3d space 
them

Hi Bart,

It's a little hard to guess exactly what your rx, ry, and rz stand for 
geometrically, and solving the equations by brute force would be a 
pretty ugly proposition.

However, I can tell you how this sort of thing is done in computer
graphics.

First off, you don't want to deal with the points on the corners of
the pane.  Instead, you want to deal with a vector that's 
perpendicular to the pane and rotate that vector until it lines up 
with the z-axis.

If p0, p1, and p2 are three of the four corners, then p0-p1 and p1-p2 
are vectors parallel to the edges.  The cross product of those 
differences will be perpendicular to the pane.

To make the problem concrete, I'll also assume that after the 
rotation, you want to have the edges of the pane parallel to the x and 
y axes.

Let n = (p0-p1)x(p1-p2) be the perpendicular vector.  "Normalize" n
to make it have length 1 by dividing its components by its length.
Let N be the normalized n.

Similarly, normalize p0-p1 (giving P) and normalize p1-p2 (giving Q).

So the goal is to rotate the triple of vectors P, Q, and N to 
(1,0,0), (0,1,0), and (0,0,1).

I'll write P = (px,py,pz), Q = (qx,qy,qz) and N = (nx,ny,nz).  We
need to find a 3 by 3 matrix M so that:

| px py pz |     | 1  0  0 |
| qx qy qz | M = | 0  1  0 |.
| nx ny nz |     | 0  0  1 |

M is just the inverse of the matrix on the far left, and will be a
pure rotation matrix.

-Doctor Tom,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/