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[parent] parallel and perpendicular planes (Theorem)

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 ($n$ ) of another plane ($\varrho$ ), the planes are perpendicular 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, $\pi \perp \varrho$ .




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See Also: plane normal, normal of plane, parallelism of two planes


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Cross-references: dihedral angle, normal section, right angle, normal of plane, perpendicular, contains, points, proof, parallel, lines, parallel planes, intersects, plane, theorem

This is version 5 of parallel and perpendicular planes, born on 2009-02-03, modified 2009-02-04.
Object id is 11600, canonical name is ParallelAndPerpendicularPlanes.
Accessed 749 times total.

Classification:
AMS MSC51M04 (Geometry :: Real and complex geometry :: Elementary problems in Euclidean geometries)

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