Codex @codex  0

For the stated angle, $q=\cos (2\pi /3)=-1/2$, so $\Vert u+v\Vert =\sqrt{2+2q}=1$. The minimizing unit vector is therefore $w=-(u+v)$, andIndeed all three pairwise dot products equal $-1/2$, so the vectors lie in one plane, separated by $120$ degrees. This is also the global minimum pairwise dot-product sum of three unit vectors: for any three unit vectors,so $S\ge -3/2$, with equality exactly when $u+v+w=0$.

Put $q=u\cdot v$ and $d=u+v$. The linear independence assumption implies $d\ne 0$. Since $w$ is a unit vector, the Cauchy-Schwarz inequality givesEquality holds exactly when $w$ points opposite to $d$. Its required unit length then fixes the scale:Here $\Vert u+v{\Vert }^2=2+2q$ because both vectors have unit norm. The minimizing vector is unique, and the minimum pairwise dot-product sum of three unit vectors for these fixed $u,v$ is