Let $X$ be a smooth irreducible algebraic curve in $\mathbb{P} V$. 
The general position lemma states that the points given by general hyperplane section of $X$ are "in general position". I'm considering a similar situation for tangent lines. For each point of the general hyperplane section, the tangent line at that point corresponds to a point in $\mathbb{P} ( \wedge^2 V)$. Is it true that these tensors in $\wedge^2V$ span the whole ambient space $\wedge^2 V$? (Of course, one should assume that the degree of $X$ is bigger than $\dim (\wedge^2 V)$). I suspect if this question is related to (the surjectivity of) the Gauss map. 



general position lemma for tangent lines of an algebraic curve
No, that is not true. Let $V$ be the $4$-dimensional vector space with ordered basis $(\mathbf{e}\_0,\mathbf{e}\_1,\mathbf{e}\_2,\mathbf{e}\_3)$, and let $[X\_0,X\_1,X\_2,X\_3]$ be the corresponding homogeneous coordinates on $\mathbb{P}V$. Let $[T\_0,T\_1]$ be homogeneous coordinates on $\mathbb{P}^1$. For every integer $r\geq 3$, consider the morphism, $$ u\_r:\mathbb{P}^1 \to \mathbb{P}V, \ \ [T\_0,T\_1] \mapsto [T\_0^r\mathbf{e}\_0 + T\_0^{r-1}T\_1\mathbf{e}\_1 + T\_0T\_1^{r-1}\mathbf{e}\_2 + T\_1^r\mathbf{e}\_3].$$ The corresponding Gauss map is $$\widetilde{u}\_r: \mathbb{P}^1 \to \mathbb{P}(\bigwedge^2V),$$ $$[T\_0,T\_1] \mapsto [rT\_0^{2r-2}\mathbf{e}\_0\wedge \mathbf{e}\_1 + r(r-1)T\_0^rT\_1^{r-2}\mathbf{e}\_0\wedge\mathbf{e}\_2 + r^2T\_0^{r-1}T\_1^{r-1}\mathbf{e}\_0\wedge \mathbf{e}\_3$$ $$ + r(r-2)T\_0^{r-1}T\_1^{r-1}\mathbf{e}\_1\wedge \mathbf{e}\_2 + r(r-1)T\_0^{r-2}T\_1^r\mathbf{e}\_1\wedge\mathbf{e}\_3 + rT\_1^{2r-2}\mathbf{e}\_2\wedge \mathbf{e}\_3]. $$ In particular, the coefficients of $\mathbf{e}\_0\wedge\mathbf{e}\_3$ and $\mathbf{e}\_1\wedge\mathbf{e}\_2$ are proportionate, i.e., the following "linear form" vanishes on the image curve, $$(r-2)X\_0\wedge X\_3 - rX\_1\wedge X\_2.$$ Notice that $r$ can be arbitrarily positive. 


