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Given two noncoplanar lines $p$ and $q$, and a point $A$, does there always exist a line that passes through $p$, $q$ and $A$?

This should be solved using Hilbert's axioms.

Intuitively, that line doesn't always exist, but I don't know how to formally prove it.

I tried the following: Let's assume that there always exist a line that passes through $p$, $q$ and $A$. We denote that line as $a$. There exists excatly one plane that contains a line and a point that doesn't belong to it. So, let $\alpha$ = $\alpha$($p$, $A$) and $\beta$ = $\beta$($q$, $A$) be planes that are defined by the terms in brackets. Clearly, $\alpha$ $\cap \beta$ = {$a$} (if two planes interesect, their intersection is a line). But this doesnt' seem to yield anything useful.

Ranko
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1 Answers1

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You just need one example of a situation where the required line does not exist. Once you have the plane $\alpha(p,A)$, this happens if $q$ belongs to a plane which is parallel to $\alpha$.

For example, let $p$ be the $x$ axis and $A=(0,1,0)$: the plane containing them is $z=0$. Any line passing through $p$ and $A$ is confined on this plane (axiom I.6: "the line lies on the plane"). Now imagine $q$ being on the plane $z=1$ and you have your counterexample.

lesath82
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