Triangulation in three dimensions: Difference between revisions

Revision as of 12:37, 16 May 2011

Revision as of 16:30, 16 May 2011

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</math><br />

<br />

[[image: FacesABD_ACD_BCD.gif]]

::<math>

(BD)^2 = (AB)^2+(AD)^2-2(AB)(AD)\cos(\angle{BAD})

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\mathbf{M_{BC}}+m_{BC}\mathbf{N_{BC}}</math>

<br /><br />

[[image: Intersect_normals_ABC.gif]]

Solving for ''m_AB'', ''m_AC'' and ''m_BC''

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