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
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