normal of plane

The perpendicular or normal line of a plane is a special case of the surface normal, but may be defined separately as follows:

A line $l$ is a normal of a plane $\pi$, if it intersects the plane and is perpendicular to all lines passing through the intersection point in the plane.  Then the plane $\pi$ is a normal plane of the line $l$.  The normal plane passing through the midpoint (http://planetmath.org/Midpoint3) of a line segment is the center normal plane of the segment.

There is the

Theorem.  If a line ($l$) a plane ($\pi$) and is perpendicular to two distinct lines ($m$ and $n$) passing through the cutting point ($L$) in the plane, then the line is a normal of the plane.

Proof.  Let $a$ be an arbitrary line passing through the point $L$ in the plane $\pi$.  We need to show that  $a⟂l$.  Set another line of the plane cutting the lines $m$, $n$ and $a$ at the points $M$, $N$ and $A$, respectively.  Separate from $l$ the equally segments $LP$ and $LQ$.  Then

$$PM=QM\mathit{\hspace{1em}}\text{and}\mathit{\hspace{1em}}PN=QN,$$

since any point of the center normal of a line segment ($PQ$) is equidistant from the end points of the segment.  Consequently,

$$\mathrm{\Delta }MNP\cong \mathrm{\Delta }MNQ\mathit{\hspace{1em}}(\text{SSS}).$$

Thus the segments $PA$ and $QA$, being corresponding parts of two congruent triangles, are equally long.  I.e., the point $A$ is equidistant from the end points of the segment $PQ$, and it must be on the perpendicular bisector (http://planetmath.org/CenterNormal) of $PQ$.  Therefore  $AL\perp PQ$, i.e.  $a\perp l$.

Proposition 1.  All of a plane are parallel.  If a line is parallel to a normal of a plane, then it is a normal of the plane, too.

Proposition 2.  All normal planes of a line are parallel.  If a plane is parallel to a normal plane of a line, then also it is a normal plane of the line.