Plane–plane intersection

The line of intersection between two planes $\backslash Pi\_1\; :\; \backslash boldsymbol\; \{n\}\_1\; \backslash cdot\; \backslash boldsymbol\; r\; =\; h\_1$ and $\backslash Pi\_2\; :\; \backslash boldsymbol\; \{n\}\_2\; \backslash cdot\; \backslash boldsymbol\; r\; =\; h\_2$ where $\backslash boldsymbol\; \{n\}\_i$ are normalized is given by

:$\backslash boldsymbol\; \{r\}\; =\; (c\_1\; \backslash boldsymbol\; \{n\}\_1\; +\; c\_2\; \backslash boldsymbol\; \{n\}\_2)\; +\; \backslash lambda\; (\backslash boldsymbol\; \{n\}\_1\; \backslash times\; \backslash boldsymbol\; \{n\}\_2)$

where

:$c\_1\; =\; \backslash frac\{\; h\_1\; -\; h\_2(\backslash boldsymbol\; \{n\}\_1\; \backslash cdot\; \backslash boldsymbol\; \{n\}\_2)\; \}\{\; 1\; -\; (\backslash boldsymbol\; \{n\}\_1\; \backslash cdot\; \backslash boldsymbol\; \{n\}\_2)^2\; \}$

:$c\_2\; =\; \backslash frac\{\; h\_2\; -\; h\_1(\backslash boldsymbol\; \{n\}\_1\; \backslash cdot\; \backslash boldsymbol\; \{n\}\_2)\; \}\{\; 1\; -\; (\backslash boldsymbol\; \{n\}\_1\; \backslash cdot\; \backslash boldsymbol\; \{n\}\_2)^2\; \}.$

This is found by noticing that the line must be perpendicular to both plane normals, and so parallel to their cross product $\backslash boldsymbol\; \{n\}\_1\; \backslash times\; \backslash boldsymbol\; \{n\}\_2$ (this cross product is zero if and only if the planes are parallel, and are therefore non-intersecting or entirely coincident).

The remainder of the expression is arrived at by finding an arbitrary point on the line. To do so, consider that any point in space may be written as $\backslash boldsymbol\; r\; =\; c\_1\backslash boldsymbol\; \{n\}\_1\; +\; c\_2\backslash boldsymbol\; \{n\}\_2\; +\; \backslash lambda(\backslash boldsymbol\; \{n\}\_1\; \backslash times\; \backslash boldsymbol\; \{n\}\_2)$, since $\backslash \{\; \backslash boldsymbol\; \{n\}\_1,\; \backslash boldsymbol\; \{n\}\_2,\; (\backslash boldsymbol\; \{n\}\_1\; \backslash times\; \backslash boldsymbol\; \{n\}\_2)\; \backslash \}$ is a basis. We wish to find a point which is on both planes (i.e. on their intersection), so insert this equation into each of the equations of the planes to get two simultaneous equations which can be solved for $c\_1$ and $c\_2$.

If we further assume that $\backslash boldsymbol\; \{n\}\_1$ and $\backslash boldsymbol\; \{n\}\_2$ are orthonormal then the closest point on the line of intersection to the origin is $\backslash boldsymbol\; r\_0\; =\; h\_1\backslash boldsymbol\; \{n\}\_1\; +\; h\_2\backslash boldsymbol\; \{n\}\_2$. If that is not the case, then a more complex procedure must be used.[http://mathworld.wolfram.com/Plane-PlaneIntersection.html Plane-Plane Intersection - from Wolfram MathWorld]. Mathworld.wolfram.com. Retrieved 2013-08-20.

{{main|Dihedral angle}}

Given two intersecting planes described by $\backslash Pi\_1\; :\; a\_1\; x\; +\; b\_1\; y\; +\; c\_1\; z\; +\; d\_1\; =\; 0$ and $\backslash Pi\_2\; :\; a\_2\; x\; +\; b\_2\; y\; +\; c\_2\; z\; +\; d\_2\; =\; 0$, the dihedral angle between them is defined to be the angle $\backslash alpha$ between their normal directions:

:$\backslash cos\backslash alpha\; =\; \backslash frac\{\backslash hat\; n\_1\backslash cdot\; \backslash hat\; n\_2\}$

\hat n_1

\hat n_2

= \frac{a_1 a_2 + b_1 b_2 + c_1 c_2}{\sqrt{a_1^2+b_1^2+c_1^2}\sqrt{a_2^2+b_2^2+c_2^2}}.

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Category:Euclidean geometry

Category:Computational physics

Category:Geometric algorithms

Category:Geometric intersection

Category:Planes (geometry)