In analytic geometry, the intersection of two planes in three-dimensional space is a line.
The line of intersection between two planes
\Pi1:\boldsymbol{n}1 ⋅ \boldsymbolr=h1
\Pi2:\boldsymbol{n}2 ⋅ \boldsymbolr=h2
\boldsymbol{n}i
\boldsymbol{r}=(c1\boldsymbol{n}1+c2\boldsymbol{n}2)+λ(\boldsymbol{n}1 x \boldsymbol{n}2)
where
c1=
h1-h2(\boldsymbol{n | |
1 |
⋅ \boldsymbol{n}2)}{1-(\boldsymbol{n}1 ⋅ \boldsymbol
2 | |
{n} | |
2) |
}
c2=
h2-h1(\boldsymbol{n | |
1 |
⋅ \boldsymbol{n}2)}{1-(\boldsymbol{n}1 ⋅ \boldsymbol
2 | |
{n} | |
2) |
}.
This is found by noticing that the line must be perpendicular to both plane normals, and so parallel to their cross product
\boldsymbol{n}1 x \boldsymbol{n}2
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
\boldsymbolr=c1\boldsymbol{n}1+c2\boldsymbol{n}2+λ(\boldsymbol{n}1 x \boldsymbol{n}2)
\{\boldsymbol{n}1,\boldsymbol{n}2,(\boldsymbol{n}1 x \boldsymbol{n}2)\}
c1
c2
If we further assume that
\boldsymbol{n}1
\boldsymbol{n}2
\boldsymbolr0=h1\boldsymbol{n}1+h2\boldsymbol{n}2
See main article: Dihedral angle. Given two intersecting planes described by
\Pi1:a1x+b1y+c1z+d1=0
\Pi2:a2x+b2y+c2z+d2=0
\alpha
\cos\alpha=
\hatn1 ⋅ \hatn2 | |
|\hatn1||\hatn2| |
=
a1a2+b1b2+c1c2 | ||||||
|
2}}. | |
\sqrt{a | |
2 |