= Solution
Event $A$ is the chosen origin. At $B$, the leading-end <world line> reaches $x=3$, so
$$
3=\frac{12}{13}s_B+\frac{20}{13},\qquad
\boxed{(ct_B,x_B)=\left(\frac{19}{12},3\right)\ \mathrm m.}
$$
The <Lorentz transformation> gives
$$
\boxed{(ct'_B,x'_B)=\left(-\frac{37}{12},4\right)\ \mathrm m.}
$$
For example $s'_B=(13/5)(19/12-36/13)=-37/12$. The squared spatial separation minus squared time separation is
$$
x_B^2-s_B^2=9-\frac{361}{144}=\frac{935}{144}>0,
$$
equally $16-1369/144$ in the other frame. Thus \b[$A$ and $B$ have spacelike separation].
In the barn frame, for $0<s<19/12$ the trailing end has entered while the leading end has not yet exited, so the whole javelin is inside at each simultaneous time slice. In the javelin frame the barn is shorter than the javelin and there is no simultaneous slice on which it contains the whole javelin. Also $B$ precedes $A$ there. This is consistent because <relativity of simultaneity> changes which endpoint events are simultaneous; the ordering of events with <spacelike separation> is not invariant. The physical intersections of the four <world lines> are the same in both diagrams.
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