= Solution
For a massive particle on a timelike <world line>, the <four-velocity> is $U^\mu=dX^\mu/d\tau=\gamma(c,\mathbf v)$, with $U^2=c^2$. Its <four-momentum> is $P^\mu=mU^\mu=(E/c,\mathbf p)$ and its <four-force> is $F^\mu=dP^\mu/d\tau$. Here $\tau$ is <proper time>. A massless trajectory does not admit this massive-particle proper-time parametrization.
\b[The asserted classification of arbitrary orthogonal vectors is false.] With the chosen <Minkowski metric>, $D=(0,1,0,0)$ and $G=(0,0,1,0)$ are nonzero, orthogonal and both <spacelike vectors>. Also $D=(1,1,0,0)$ and $G=2D$ are orthogonal <null vectors>. The valid statement is that a nonzero vector orthogonal to a <timelike vector> is spacelike. To prove it, choose the rest frame of the <timelike vector>, $D=(d,0,0,0)$; then $D\cdot G=dG^0=0$ forces $G^0=0$, so $G^2=-|\mathbf G|^2<0$.
The <relativistic energy-momentum relation> gives
$$
P\cdot P=\frac{E^2}{c^2}-|\mathbf p|^2=m^2c^2.
$$
Therefore $\boxed{P^2=0\ \Longrightarrow\ m=0}$. A <photon> is an example: $E=c|\mathbf p|$. A collection of <photons> need not itself have zero <invariant mass>.
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