# parallel and perpendicular planes

Theorem 1. If a plane ($\pi $) intersects two parallel planes^{} ($\varrho $, $\sigma $), the intersection lines are parallel^{}.

Proof. The intersection lines cannot have common points, because $\varrho $ and $\sigma $ have no such ones. Since the lines are in a same plane $\pi $, they are parallel.

Theorem 2. If a plane ($\pi $) contains the normal (http://planetmath.org/PlaneNormal) ($n$) of another plane ($\varrho $), the planes are perpendicular^{} (http://planetmath.org/DihedralAngle) to each other.

Proof. Draw in the plane $\varrho $ the line $l$ cutting the intersection line perpendicularly and cutting also $n$. Then $l$ must be perpendicular to $n$ and thus to the whole plane $\pi $ (see the Theorem in the entry normal of plane). Consequently, the right angle^{} formed by the lines $n$ and $l$ is the normal section of the dihedral angle formed by the planes $\pi $ and $\varrho $. Therefore, $\pi \perp \varrho $.

Title | parallel and perpendicular planes |
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Canonical name | ParallelAndPerpendicularPlanes |

Date of creation | 2013-04-19 15:18:51 |

Last modified on | 2013-04-19 15:18:51 |

Owner | pahio (2872) |

Last modified by | pahio (2872) |

Numerical id | 10 |

Author | pahio (2872) |

Entry type | Theorem |

Classification | msc 51M04 |

Related topic | PlaneNormal |

Related topic | NormalOfPlane |

Related topic | ParallelismOfTwoPlanes |