general position lemma for tangent lines of an algebraic curve 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. 
 A: 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.  
