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Calculation of the intersection of two 3D lines in space.

Posted on 2012-04-12
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Last Modified: 2012-04-24
If I have two lines in 3 dimensional space defined by:

Line 1:  (x1,y1,z1) and (x2,y2,z2)
Line 2: (x3,y3,z3) and (x4,y4,z4)

Does anyone have the source code to solve this?
I am working in C#.Net
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Question by:cupper
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6 Comments
 
LVL 27

Accepted Solution

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d-glitch earned 500 total points
ID: 37839817
The method of solution is described here:  http://mathforum.org/library/drmath/view/63719.html
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Author Comment

by:cupper
ID: 37839967
The proposed solution is for a point-vector pair. I'm looking for the equations for two lines, each defined by two points in space.

Line 1:  (x1,y1,z1) and (x2,y2,z2)
Line 2: (x3,y3,z3) and (x4,y4,z4)
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LVL 27

Expert Comment

by:d-glitch
ID: 37840046
For each of your line segments, you can convert to point-vector form by using either point, and the difference between the two points as the direction vector.

       (x ,y ,z ) =  (x1,y1,z1)  =  t*( x2-x1), (y2-y1), (z2-z1))   for the first line.    t=0  gives  (x1,y1,z1)     and    t=1  gives  (x2,y2,z2)
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Author Comment

by:cupper
ID: 37860770
Sorry for getting back to this so late.
In the example provided, the final answer is unclear to me (i.e. the (X,Y,Z) location of the intersection point).
Is the final intersection point (X,Y,Z) = (5,6,2)?

Also as an aside, I'm not getting the cross product he is getting in his example (see below)

from article
"Using Bensegueni's method, we find

  a(V1 x V2) = a(-10,-11,-13)"


I get  a(V x V2) =  a (-10, 11, -13)   (i.e. middle term positive).
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LVL 18

Expert Comment

by:JoseParrot
ID: 37874532
In 2D, if two lines aren't parallel, it exists, for sure, an interception point.
In 3D there is another requirement: besides to be not parallel, the lines must be coplanar, say, they must be in a same plane.

The John Taylor's article Intersecting Lines in 3D: describes these conditions and gives some examples.

Jose
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LVL 18

Expert Comment

by:JoseParrot
ID: 37878141
Assuming the lines are defined by their end points (x1, y1, z1) and (x2, y2, z2) lets use the general parametric equation:

      x = x1 + (x2 - x1)*t
      y = y1 + (y2 - y1)*t
      z = z1 + (z2 - z1)*t

if lines are
Line A --> (xa1,ya1, za1,  xa2,ya2, za2) and
Line B --> (xb1,yb1, zb1,  xb2,yb2, zb2),

first define the parametric equations

Line A
x = xa1 + (xa2 - xa1) * ta
y = ya1 + (ya2 - ya1) * ta
z = za1 + (za2 - za1) * ta

Line B
x = xb1 + (xb2 - xb1) * tb
y = yb1 + (yb2 - yb1) * tb
z = zb1 + (zb2 - zb1) * tb

To check if they are parallel,  get the vectors for each line.

Line A
(the components of the vector are the coefficients of ta):
M1 = (xa2 - xa1) * i + (ya2 - ya1) * j + (za2 - za1) * k

Line B
(do the same as for Line A, by using the coefficients of tb):
M2 = (xb2 - xb1) * i + ...

If M1 and M2 are equal or proportional, the lines are parallel, so there is no interception point.
If not, go ahead.

Now set the x and y values.
Of course, x and y must be the same in both lines equations, so,

x = xa1 + (xa2 - xa1) * ta = xb1 + (xb2 - xb1) * tb
and
y = ya1 + (ya2 - ya1) * ta = yb1 + (yb2 - yb1) * tb

By solving tha above equations, we have values for ta and tb.
Now, just apply these values in the z equations to chek is they are true.
If so, you found the interception point, if not, they don't intercept each another.

Jose
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