Let $\sigma\_d:\mathbb{P}^2\to\mathbb{P}^n$ be the d-th Veronese map and let $X=\sigma\_d(\mathbb{P}^2)$. Let $W\subset\mathbb{P}^n$ be a 2-plane such that $W\cap X=\emptyset$. For a line $L\subset \mathbb{P}^2$ let $X\_L=\sigma\_d(L)$.
Since $\dim(L)=1$ the space $U\_L$ spanned by $X\_L$ has dimension $d$ which is basically the Veronese of $\mathbb{P}^1$ in different coordinates.


If $d\ge 4$, is it true that for generic $L$ the projection of $W$ onto $U\_L$ does not intersect $X\_L$ and still has dimension 2?


Dimensionwise this should be true since in $U\_L$, $X\_L$ has dimension 1 and $W$ has dimension at most 3.



Projecting onto the span of a generic Veronese variety
This is not true as stated in the comments. Consider $W=x\_3^{d-2}\mathbb{C}[x\_1,x\_2]\_2$ and $l$ a generic linear form. Then $W(x\_1,x\_2,l)=l^{d-2}\mathbb{C}[x\_1,x\_2]\_2$ and then setting $x\_3=0$ gives $\mathbb{C}[x\_1,x\_2]\supseteq W'=l'^{d-2}\mathbb{C}[x\_1,x\_2]\_2$. But $W'$ contains $l'^d$. If $l$ does not contain $x\_3$ then the dimension of $W'$ is not three anymore.


