By von Oniscik A.L., Sulanke R.

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**Example text**

1 they would be the same line, contradicting the fact that they are distinct. Therefore L can be taken to be either of the two lines not containing A. 6 (Two intersecting lines determine a plane). Given lines L and M such that L ¤ M and L \ M ¤ ;, there exists one and only one plane E such that L Â E and M Â E. Proof. There are two things to be proved: (1) there is such a plane E (existence), and (2) there is not more than one such plane (uniqueness). We first prove that there is such a plane E.

C) Two planes P and P 0 are parallel (notation: P k P 0 ) iff P \ P 0 D ;. (D) A set E of two or more distinct lines on a plane P is a pencil iff either (1) the members of E are concurrent at some point O, or (2) every member of E is parallel to every other member of E. In case (1), the point O is the focal point of E. 1), and by (C), two planes are intersecting iff they are nonparallel. But beware that two nonparallel lines do not necessarily intersect—there may not be a plane that contains them both.

2 there is a plane E such that fA; B; Cg Â E. 3, L Â E and M Â E. 30 1 Preliminaries and Incidence Geometry (I) To prove that there is not more than one such plane, suppose on the contrary that there were a second plane E 0 containing both L and M. Then all of the points A, B, and C defined above would belong to E 0 . 2, E D E 0 . 7. Let E be a plane. There exists a point P such that P … E. Proof. 5 there exist points A, B, C, and D which are noncoplanar. If fA; B; C; Dg were a subset of E, then A, B, C, and D would be coplanar.