Chapter 14 - Calculus of Vector-Valued Functions - 14.2 Calculus of Vector-Valued Functions - Exercises - Page 720: 34

$$ L(s)= \langle 0,1,9\rangle +s \langle1,-1,9\rangle .$$

Since we can write $ r (s)=\langle \ln s, s^{-1},9s \rangle, $ then we have the derivative vector $ r' (s)=\langle s^{-1},-s^{-2},9 \rangle $. Then, the parametrization of the tangent line at $ s=1$ is given by $$ L(s)=r (1)+sr'(1)=\langle 0,1,9\rangle +s \langle1,-1,9\rangle .$$