BBS: Inland Empire Archive
Date: 05-10-92 (10:01) Number: 158
From: PAUL LEONARD Refer#: NONE
To: RICH GELDREICH Recvd: NO
Subj: projection Onto Plane. Conf: (2) Quik_Bas
On or about <May 09 16:05>, Paul Leonard (1:105/48.111) scribbled:
PL> Given three points to describe the plane P1(x1,y1,z1), P2(x2,y2,z2)
PL> and P3(x3,y3,z3), the center point of the rectangle they form has
PL> coordinates
PL> ( (x1+2*x2+x3)/4, (y1+2*y2+y3)/4, (z1+2*z2+z3)/4 )
Bzzzzt - wrong answer. Sorry, Rich. The center point
coordinates are determined by finding the midpoint of the
diagonal formed by P1P3...
( (x1+x3)/2, (y1+y3)/2, (z1+z3)/2 )
...Please substitute these coordinates for Pm(xm,ym,zm).
PL> z = zm + (-z1ym+x1y2+xmy1-x2y1)*t
Also, there's a typo in the parametric equation for z. It should be...
z = zm + (-x1ym+x1y2+xmy1-x2y1)*t
...The first term in the coefficient of t isn't "-z1ym."
If you solve the equations for the example i gave
[P1(0,2,0), P2(2,2,0) and P3(2,0,0)] (which is how i found
the "bugs"), you get Pm(1,1,0) and
64^2 = (x-1)^2 + (y-1)^2 + (z-0)^2
and
x = 1
y = 1
z = 6*t
By substitution, 64^2=(6*t)^2 or 6*t=64, so z=64. So the
point you're looking for in this example is...
P(1,1,64)
...(-64 is also valid for z). Again, ta-dah (which is English for QED). :)
ptl
--- msged 2.07
* Origin: PTL Pointwork (1:105/48.111)
