Solution
= Solution
First,
$$
\nabla_T(T\cdot S)=(\nabla_TT)\cdot S+T\cdot\nabla_TS
=T\cdot\nabla_ST=\frac12\nabla_S(T\cdot T)=0.
$$
Hence orthogonality at one point persists. Substituting the constant-curvature tensor into geodesic deviation gives
$$
R^\alpha{}_{\beta\mu\nu}T^\beta T^\mu S^\nu
=K\{T^\alpha(T\cdot S)-S^\alpha(T\cdot T)\}=KS^\alpha,
$$
because $T\cdot S=0$ and $T\cdot T=-1$. Therefore
$$
(\nabla_T\nabla_TS)^\alpha-KS^\alpha=0
$$
along the whole geodesic.