Union of one-dimensional family of straight lines has zero measure

Question

Let $\gamma:[0,1]\to \mathbb S^2$ be a smooth curve. Consider the following set $$ \bigcup_{t\in [0,1]}\mathrm{span}(\gamma(t))\subset \mathbb R^3 $$ My question is: does this set have Hausdorff dimension at most $2$, and as a result it has zero 3-dimensional Lebesgue measure?. Note that if the smoothness condition is removed then the answer is trivially false.

Answer 1

It seems that the answer is pretty easy......

The set $$ \bigcup_{t\in [0,1]}\mathrm{span}(\gamma(t)) $$ is just the graph of the smooth mapping $\Gamma(x,t):=x\gamma(t)$ over $\mathbb R\times [0,1]$. Therefore its Hausdorff dimension is exactly 2.