A theorem about three lines, any two of them being coplanar
I am pretty convinced the following result holds in dimension 3:
If three non-coplanar lines are such that any two of them are coplanar, then either all three lines are parallel, or they all intersect at the same point.
But I failed to come with a demonstration. Note that this is not homework as I need this in a problem of physics whose solution I highlighted on the stackexchange for physics, in this answer (the part until QED).
$\endgroup$ 12 Answers
$\begingroup$If $g_1$, $g_2$, $g_3$ are not pairwise parallel two of them have to intersect. Assume that $g_1\wedge g_2=P$. As $P\in g_1\subset g_1\vee g_3$, and similarly $P\in g_2\vee g_3$ it follows that $(g_1\vee g_3)\cap (g_2\vee g_3)\ne\emptyset$. In fact $(g_1\vee g_3)\cap (g_2\vee g_3)=g_3$, which then implies $P\in g_3$.
$\endgroup$ 4 $\begingroup$If you accept that this picture shows all ways in which three planes can intersect, then I have some elementar reasoning for you.
Lets call the lines $a,b$ and $c$. Any two of theme define a plane (they are coplanar). Call the planes $E_{ab},E_{bc}$ and $E_{ca}$. So any two of these planes intersect in a common line, e.g. $E_{ab}$ and $E_{bc}$ intersect in $b$. This excludes two of the five pictures above (the first and the third). In the second picture all lines are coplanar (actually even idnetical), hence we have to exclude it too.
In the two remaining ones, either all (intersection-)lines are parallel (fourth pic) or intersect in a common point (last pic).
$\endgroup$ 3
