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Homenormal of plane

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# 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 of a line segment is the center normal plane of the segment.

There is the

Theorem. If a line ($l$) cuts 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 $a\perp l$. 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 long line segments $LP$ and $LQ$. Then

$PM\;=\;QM\quad\mbox{and}\quad PN\;=\;QN,$ |

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

$\Delta MNP\cong\Delta MNQ\quad(\mbox{SSS}).$ |

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 of $PQ$. Therefore $AL\bot PQ$, i.e. $a\bot l$.

Proposition 1. All normals 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.

## Mathematics Subject Classification

51M04*no label found*

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