= Solution
{c}
A nonzero <vector field> has <hypersurface orthogonality> when it is locally normal to a family of <hypersurfaces>. In terms of its associated <differential one-form> $K^\flat=g_{\mu\nu}K^\nu dx^\mu$, the <Frobenius theorem> gives the equivalent conditions
$$
K^\flat=f\,dh,\qquad K^\flat\wedge dK^\flat=0,
\qquad K_{[\mu}\nabla_\nu K_{\rho]}=0,
$$
locally with nonzero $f$ and $dh$; square brackets denote <tensor index antisymmetrization>.
All coefficients of the <Reissner-Nordstrom metric> are independent of $t$, and $K=\partial_t$ has constant coordinate components. Thus the <Lie derivative> obeys $\mathcal L_Kg=\partial_tg=0$: $K$ is a <Killing vector field>. Moreover,
$$
K^\flat=-F(r)\,dt,\qquad dK^\flat=-F'(r)\,dr\wedge dt,
\qquad \boxed{K^\flat\wedge dK^\flat=0.}
$$
Thus $K$ has <hypersurface orthogonality>, with the $t=\text{constant}$ <hypersurfaces> providing the orthogonal family wherever this chart is regular. It is timelike where $F>0$, since $g(K,K)=-F$. Its <hypersurface orthogonality> also holds in a regular extension across a horizon; singular static <coordinates> do not invalidate the <Frobenius theorem> test.
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