BBS: Inland Empire Archive Date: 05-01-92 (23:32) Number: 179 From: RON MCDERMOTT Refer#: NONE To: RICH GELDREICH Recvd: NO Subj: 3-D Graphics Conf: (2) Quik_Bas

Rich, RG>What we need: RG>A point, outside of the plane, that is perpendicular to the RG>plane when a line is drawn from the point through the middle RG>of the plane. The point also has to be certain distance from RG>the plane (64 units or so). We need the X, Y, & Z coordinates RG>of this point. Not sure if this would help, but one test for perpendicularity is a dot product of the two vectors defined by the two lines to be tested. This would involve defining a line drawn from one of the corners, passing through the center point, and calculating the change in x, y, and z values. The resulting differences are then the coefficients of a vector in i,j,k format. The second vector would be the one passing through the center point and the point external (which is unknown). The delta x of the first vector times the delta x of the unknown vector, added to the product of the delta y's, added to the product of the delta z's must equal zero. For example, consider a corner of 1,1,1; a center point of 3,4,5; and an unknown point out of plane of x,y,z..... (3-1)*(x-3) + (4-1)*(y-4) + (5-1)*(z-5) = 0 . You'd need other information to solve, but this might be one piece to add to the puzzle!? --- * Origin: * Info-Center BBS * (914)567-1814 >>HST<< * (1:272/26)

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