# Using axioms of incidence to show whether two lines meet in space

I have this question:

$L_1$ and $L_2$ are lines, and $P$ is a plane. $L_1$ does not lie in $P$, but meets the plane in a point $p$. $L_2$ lies in $P$, but does not contain $p$. Does $L_1$ meet $L_2$? Which of the axioms of incidence are used in solving this question?

I think $L_1$ does not meet $L_2$ because $L_1$ and $L_2$ are not coplanar ?!

The axioms are:

1. (line axiom) Through any two distinct point there is exactly one line,
2. (plane axiom) Through any $3$-non collinear points there is exactly one plane
3. (dimension axiom) Any line contains at least two distinct points, any plane contains at least two distinct lines, there are at least two distinct planes.
4. (line-plane intersection) If two distinct points of a line lie in some plane $P$ then the whole line lies on P
5. (the parallel axiom) Let $L$ be a line and $p$ be a point then there is exactly one line that passes through $p$ and is parallel to $L$.
6. (plane plane intersection) If two distinct planes meet then their intersection is a line.
• "... because $L_1$ and $L_2$ are not coplanar" Well, you know that $L_1$ and $L_2$ aren't both on plane $P$, but it's possible that they share a some other plane. (Whether they do is equivalent to the question at hand.) Anyway ... Ask yourself these questions: If the lines meet at a point $q$ (other than $p$), then where is $q$? Can $q$, as a point on $L_1$, be on plane $P$? Can $q$, as a point on $L_2$, be outside of plane $P$?
– Blue
Oct 30, 2016 at 7:08
• @Blue what about line-plane intersection : if two distinct points of a line lie in some plane P then the whole line lies on P i.e. there is no points on L2 outside the plane P Oct 30, 2016 at 7:30
• @Tasneem: Correct, so this mystery point $q$, which presumably is on $L_2$, cannot be outside of $P$. What does that same axiom (together with what you know about $L_1$) say about the possibility of $q$ being on $P$?
– Blue
Oct 30, 2016 at 8:06
• @Blue i.e. q must be point on P and q cannot equal p so, L1 does not meet L2 using line plane intersection ..... may also l can said that " let q1 and q2 be any two points on L2, q1, q2 and p can make a plane so q1,q2 and p are 3 non -collinear points i.e. L1 does not meet L2....... Thank you very very much 😉 Oct 30, 2016 at 8:22

The intersection of a line $L_1$ with a plane $P$ in $\mathbb{R}^3$ is either empty $L_1 \cap P = \emptyset$, contains one point $L_1 \cap P = \{ p \}$ or infinite many points $L_1 \cap P = L_1$.
(This reasoning is directly from Linear Algebra, it translates into the solution of a system of linear equations, one equation expressing the plane, two equations expressing the line)

If $L_1$ lies not in $P$ and has at least one point in common you must have the case with one intersection point $p$ and you know there can not be another point from $L_1$ on $L_2$ as that one has to lie in $P$ as well and we have only one in $P$.

Update: Now let us try the axioms:

1. (line-plane intersection) If two distinct points of a line lie in some plane P then the whole line lies on P

We apply $(A \Rightarrow B) \iff (\neg B \Rightarrow A)$: $$(\exists p, q \in L: p \ne q \wedge p, q \in P ) \Rightarrow L \subset P \iff \\ \neg (L \subset P) \Rightarrow \neg (\exists p, q \in L: p \ne q \wedge p, q \in P ) \iff \\ \neg (L \subset P) \Rightarrow (\forall: p, q \in L: p = q \vee p \notin P \vee q \notin P)$$ The last means: If $L$ does not lie in $P$ then for all two distinct points of $L$ at least one of them is not in $P$.

As $L_1$ does not lie in $P$ and $p \in P$ we know from the above $q \notin P$.

This too allows at most one point of $L_1$ to be in $P$: If we had two distinct ones both from $P$, at least one of them had not be in $P$, contradiction.

Again: We know there can not be another point $q \in L_1$ on $L_2$ as that one has with $L_2$ to lie in $P$ as well and we have only one in $P$.

• You should write that the intersection of line and plane "either is empty, contains exactly one point, or contains the entire line" (not just "infinite[ly] many points"), so as to explicitly exclude the possibility that the intersecting line contains points outside of the plane.
– Blue
Oct 30, 2016 at 7:16
• ok you mean that L1 does not meet L2, how ai can use the axiom of incidence ?? Oct 30, 2016 at 7:35
• @mvw oh this question is from the book " elementary geometry John Roe "I will write all axioms Oct 30, 2016 at 7:54
• @mvw Axiom1(line axiom) through any two distinct point there is exactly one line, Axiom2(plane axiom) "through any 3-non collinear points there is exactly one plane " Axiom3(dimension axiom) " any line contains at least two distinct point, any plane contains at least two distinct lines, there are at least two distinct planes. Axiom4(line-plane intersection)"if two distinct points if a line lie in some plane P then the whole line lies on P" axiom5(the parallel axiom)"let L be a line and p be a point then there is exactly one line thar pass through p and parallel to L" Oct 30, 2016 at 8:04
• @mvw Axiom6 (plane plane intersection)" if two distinct planes meet then their intersection is a line " this is the last axiom we take then we have this question. Thanks for help me Oct 30, 2016 at 8:07