Triangulation in three dimensions: Difference between revisions

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== Development ==
[[Image: triangulation_illust_02.gif|700px|thumb|Figure 1. The apex and its mirror reflection about the plane of ABC precipitate D and D'.]]
Three sticks of known lengths ''AD, BD, CD'' are anchored in the ground at known coordinates '''A, B, C'''. This development calculates the coordinates of the apex where the other ends of the three sticks will meet. These coordinates are given by the vector '''D'''. In the mirror case, '''D'''' is sub-apex where the three sticks would meet below the plane of '''A, B, C''' as well.
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[[image: triangulation_illust_02.gif]]
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By the [[law of cosines]],
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(CD)^2 = (BC)^2+(BD)^2-2(BC)(BD)\cos(\angle{CBD})
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[[image: facesABD_ACD_BCD_2.gif|700px|thumb|Figure 2. The normals are dropped on the sides from the apex and their intersections with AB, AC and BC are determined.]]
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The projection<ref>Borisenko, A. I. and Tarapov, I. E., (1968) "Vector and Tensor Analysis", General Publishing Company, p. 6. ISBN 0-486-63833-2</ref> of '''AD''' onto '''AB''' and '''AC''', and the projection of '''BD''' onto '''BC''' results in,
[[image: facesABD_ACD_BCD_2.gif]]
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:<math>\mathbf
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[[image: Intersect_normals_ABC_3.gif|700px|thumb|Figure 3. The red normals intersect at a common point.]]<br />
 
The three [[unit normal]]s to '''AB''', '''AC''' and '''BC''' in the [[Plane (geometry)|plane]] of ABC are:
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[[image: Intersect_normals_ABC_3.gif]]
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:<math>
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== Condition for intersection ==
 
If ''AD, BD, CD'' are arrangedassigned asaccording to the arrangement,
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:<math>AD \le BD \le CD</math><br />
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== Example ==
[[imageImage: Example_00_triang_02.gif|750px|thumb|Figure 4. The data is input into the equations to obtain the solution.]]
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[[Image:3_spheres_intersecting.gif|750px|thumb|Figure 5. Showing points D and D' as the result of 3 intersecting spheres from data given in Figure 4. The line adjoining D and D' imbedded in the interiors of all three spheres is the axis of symmetry of the three spheres.]]
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== See also ==
* [[Sphere]]