Let $\boldsymbol{\gamma}: [0,A] \to S$ be a unit-speed geodesic, where $S$ is the sphere of radius $R$ centered at the origin. Define $u=\gamma(0) \times \gamma'(0)$. I want to show that $u$ is a non-zero vector.
I know that since $\gamma$ is unit-speed, we must have $||\gamma'(t)||=1$ for all $t \in [0,A]$ and thus we have in particular that $\gamma'(0) \neq0$. However, I'm not sure how I can conclude that $\gamma(0)\neq0$.
The sphere is $\{x:x \cdot x = R^2 \}$. $\gamma$ is a curve in this set, so $$\gamma \cdot \gamma = R^2.$$ Differentiating this gives $$ \gamma \cdot \gamma' = 0. $$ The first equation tells you $\gamma(0) \neq 0$, and then $$ \lvert \gamma \times \gamma' \rvert^2 = (\gamma \cdot \gamma)(\gamma' \cdot \gamma') - (\gamma \cdot \gamma')^2 = (\gamma \cdot \gamma)(\gamma' \cdot \gamma') \neq 0. $$