Re: Finding a shorter form for coplanar point in 3D space
- To: mathgroup at smc.vnet.net
- Subject: [mg65159] Re: Finding a shorter form for coplanar point in 3D space
- From: "Valeri Astanoff" <astanoff at yahoo.fr>
- Date: Wed, 15 Mar 2006 23:59:33 -0500 (EST)
- References: <dv9021$nil$1@smc.vnet.net>
- Sender: owner-wri-mathgroup at wolfram.com
This is the way I would do it :
In[1]:=
OA={xa,ya,0};
OB={xb,yb,zb};
OC={xc,yc,0};
AB=OB-OA;
CA=OA-OC;
AD=k AB;
CD=CA+AD;
OD=OA+AD;
In[9]:=sol=First at Solve[CD.CD == l^2,k];
In[10]:=OD/.sol//Simplify
Out[10]=
{xa - ((xa - xb)*(xa^2 - xa*xb - xa*xc + xb*xc + ya^2 - ya*yb - ya*yc +
yb*yc -
(1/2)*Sqrt[4*(xa^2 + xb*xc - xa*(xb + xc) + (ya - yb)*(ya -
yc))^2 -
4*(-l^2 + xa^2 - 2*xa*xc + xc^2 + ya^2 - 2*ya*yc + yc^2)*(xa^2
- 2*xa*xb + xb^2 + ya^2 - 2*ya*yb + yb^2 +
zb^2)]))/(xa^2 - 2*xa*xb + xb^2 + ya^2 - 2*ya*yb + yb^2 +
zb^2),
ya - ((ya - yb)*(xa^2 - xa*xb - xa*xc + xb*xc + ya^2 - ya*yb - ya*yc
+ yb*yc -
(1/2)*Sqrt[4*(xa^2 + xb*xc - xa*(xb + xc) + (ya - yb)*(ya -
yc))^2 -
4*(-l^2 + xa^2 - 2*xa*xc + xc^2 + ya^2 - 2*ya*yc + yc^2)*(xa^2
- 2*xa*xb + xb^2 + ya^2 - 2*ya*yb + yb^2 +
zb^2)]))/(xa^2 - 2*xa*xb + xb^2 + ya^2 - 2*ya*yb + yb^2 +
zb^2),
(zb*(xa^2 - xa*xb - xa*xc + xb*xc + ya^2 - ya*yb - ya*yc + yb*yc -
(1/2)*Sqrt[4*(xa^2 + xb*xc - xa*(xb + xc) + (ya - yb)*(ya - yc))^2
-
4*(-l^2 + xa^2 - 2*xa*xc + xc^2 + ya^2 - 2*ya*yc + yc^2)*(xa^2
- 2*xa*xb + xb^2 + ya^2 - 2*ya*yb + yb^2 +
zb^2)]))/(xa^2 - 2*xa*xb + xb^2 + ya^2 - 2*ya*yb + yb^2 +
zb^2)}
hth
v.a.