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)

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