 Solution

The event separation is timelike precisely whenA simultaneous pair in an inertial frame would instead have $(D-B{)}^2=-|\Delta {\mathbf{x}}^{\prime }{|}^2\le 0$. Since a Lorentz transformation preserves this interval, no such simultaneous frame exists. A timelike pair can be brought to the same spatial position: the required velocity is $\mathbf{u}=\Delta \mathbf{x}/\Delta t$, whose magnitude is less than $c$.

The separate condition $D\cdot B=0$ says that the two event-position four-vectors, measured from the chosen spacetime origin, are orthogonal in the Minkowski inner product. It is invariant under Lorentz transformations about that origin but not under arbitrary translations. It is not a general criterion for their causal separation.

For a concrete example, put $B=(cT,0,0,0)$ and $D=(0,R,0,0)$. Their inner product vanishes, but their separation has squared interval $c^2{T}^2-R^2$, which can be positive, zero or negative. If $B$ is timelike, a rest frame for $B$ makes $D^0=0$; this means that $D$ and the chosen origin are simultaneous in that frame, not that $D$ and $B$ are simultaneous.