Finding a shorter form for coplanar point in 3D space

*To*: mathgroup at smc.vnet.net*Subject*: [mg65135] Finding a shorter form for coplanar point in 3D space*From*: János <janos.lobb at yale.edu>*Date*: Wed, 15 Mar 2006 06:29:38 -0500 (EST)*Sender*: owner-wri-mathgroup at wolfram.com

Hi, Let say I have a Line in 3D with points A(xa,ya,0) and B(xb,yb,zb) and the distance between the two points in space is "l" -small "L". Then I move this line into another position where the end points will be C(xc,yc,0) and D(xd,yd,zd). The move is such that A,B and D are collinear, A and C remain in the xy-plane and A,B,C and D are coplanar. The distance between C and D are also "l". Let's say I know the coordinates of A, B, and C and I am looking for D(xd,yd,zd). From the condition of collinearity of A,B,C is coming that the determinant of the matrix created from the coordinates has to vanish so my first equation eqn1: In[1]:= m = {{xa, ya, 0}, {xb, yb, zb}, {xd, yd, zd}} In[2]:= eqn1 = Det[m] == 0 Out[2]= xd*ya*zb - xa*yd*zb - xb*ya*zd + xa*yb*zd == 0 The second equation is coming from the fact that the distance between C and D is "l": In[11]:= eqn2 = (xc - xd)^2 + (yc - yd)^2 + zd^2 == l^2 The third equation is given by similar triangles. Let say if the vertical projection of D down to the xy-plane is Dxy and the same for B is Bxy, then BBxy:BxyA == DDxy:DxyA. In coordinates: eqn3 = zb/Sqrt[(xa - xb)^2 + (ya - yb)^2] == zd/Sqrt[(xa - xd)^2 + (ya - yd)^2] Well, if I put all three into Solve, like this innocent looking sol=Solve[{eqn1,eqn2,eqn3},{xd,yd,zd}], Solve will take my G5 duo into the bushes forever. It looks to me that I should have up to max 16 solutions for {xd,yd,zd}. If I try to be cleaver and solve eqn3 for zd: In[22]:= solzd = Solve[eqn3, zd] Out[22]= {{zd -> (Sqrt[(xa - xd)^2 + (ya - yd)^2]*zb)/ Sqrt[(xa - xb)^2 + (ya - yb)^2]}} and punch it back to eqn1 and eqn2 then I get In[8]:= eqn1 Out[8]= xd*ya*zb - (xb*ya*Sqrt[(xa - xd)^2 + (ya - yd)^2]*zb)/ Sqrt[(xa - xb)^2 + (ya - yb)^2] + (xa*yb*Sqrt[(xa - xd)^2 + (ya - yd)^2]*zb)/ Sqrt[(xa - xb)^2 + (ya - yb)^2] - xa*yd*zb == 0 and In[13]:= eqn2 Out[13]= (xc - xd)^2 + (yc - yd)^2 + (((xa - xd)^2 + (ya - yd)^ 2)*zb^2)/ ((xa - xb)^2 + (ya - yb)^ 2) == l^2 Now these two still take my G5 Duo to the laundry if I try Solve [{eqn1,eqn2},{xd,yd}]. The next step to do is to solve eqn1 for yd: In[22]:= solyd = Flatten[Solve[eqn1, yd]] Out[22]= {yd -> ((-xb)*ya + xd*ya + xa*yb - xd*yb)/ (xa - xb), yd -> (xa^2*xb*ya + xa^2*xd*ya - 2*xa*xb*xd*ya + xb*ya^3 + xd*ya^3 - xa^3*yb + xa^2*xd*yb - xa*ya^2*yb - xd*ya^2*yb)/ (xa^3 - xa^2*xb + xa*ya^2 + xb*ya^2 - 2*xa*ya*yb)} and punch it back to eqn2. Eqn2 will become two equations, eqn21 and eqn22. In[28]:= eqn21 = eqn2 /. yd -> solyd[[1,2]] Out[28]= (xc - xd)^2 + (-(((-xb)*ya + xd*ya + xa*yb - xd*yb)/ (xa - xb)) + yc)^2 + (((xa - xd)^2 + (ya - ((-xb)*ya + xd*ya + xa*yb - xd*yb)/(xa - xb))^ 2)*zb^2)/ ((xa - xb)^2 + (ya - yb)^ 2) == l^2 and In[33]:= eqn22 = eqn2 /. yd -> solyd[[2,2]] Out[33]= (xc - xd)^2 + (-((xa^2*xb*ya + xa^2*xd* ya - 2*xa*xb*xd*ya + xb*ya^3 + xd*ya^3 - xa^3*yb + xa^2*xd* yb - xa*ya^2*yb - xd*ya^2*yb)/(xa^3 - xa^2*xb + xa*ya^2 + xb*ya^2 - 2*xa*ya* yb)) + yc)^2 + (((xa - xd)^2 + (ya - (xa^2*xb*ya + xa^2*xd*ya - 2*xa*xb*xd*ya + xb*ya^3 + xd*ya^3 - xa^3*yb + xa^2*xd* yb - xa*ya^2*yb - xd*ya^2*yb)/ (xa^3 - xa^2*xb + xa*ya^2 + xb*ya^2 - 2*xa*ya*yb))^2)* zb^2)/((xa - xb)^2 + (ya - yb)^2) == l^2 Solving these two is now a breeze for the G5 duo and both give two solutions. The third and fourth solutions for xd are quite a few pages long. I am wondering if there is another way - like using cylindrical coordinate system or using Reduce - to get shorter looking solutions. Does Solve do a FullSimplify on its findings ? Thanks ahead, János ---------------------------------------------- Trying to argue with a politician is like lifting up the head of a corpse. (S. Lem: His Master Voice)

**Follow-Ups**:**Re: Finding a shorter form for coplanar point in 3D space***From:*Daniel Lichtblau <danl@wolfram.com>

**Re: Finding a shorter form for coplanar point in 3D space***From:*"Carl K. Woll" <carlw@wolfram.com>