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MathGroup Archive 2006

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Re: Finding a shorter form for coplanar point in 3D space

  • To: mathgroup at smc.vnet.net
  • Subject: [mg65168] Re: [mg65135] Finding a shorter form for coplanar point in 3D space
  • From: "Carl K. Woll" <carlw at wolfram.com>
  • Date: Wed, 15 Mar 2006 23:59:59 -0500 (EST)
  • References: <200603151129.GAA23276@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

János wrote:
> 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).
> 

Why wouldn't you define point D as

{xa+(xb-xa)t,ya+(yb-ya)t,zb t}

Then collinearity of D with A and B is guaranteed, and you only have one 
unknown to solve for instead of 3.

>  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
> 

This equation is unnecessary when using the parametric form of D I gave 
above.

> 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
> 

With one unknown, this is the only equation you need. Defining the points:

pa = {xa, ya, 0};
pb = {xb, yb, zb};
pc = {xc, yc, 0};
pd = {xa + (xb - xa) t, ya + (yb - ya) t, zb t};

Solve the resulting distance equation, where I suppress the output as 
its a little messy:

In[129]:=
Solve[(pb - pa) . (pb - pa) == (pd - pc) . (pd - pc), t];

We get two solutions, as expected for a quadratic equation in t. We can 
obtain the coordinates of pd by substituting this value of t into it. 
Perhaps I'm misunderstanding your problem, as you state that you 
expected 16 solutions.

Carl Woll
Wolfram Research

> 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,
> 
> Jç?£os
> 
> 
> ----------------------------------------------
> Trying to argue with a politician is like lifting up the head of a  
> corpse.
> (S. Lem: His Master Voice)


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