Let $X\subset\mathbb{P}^n$ be a **smooth** projective variety of dimension $\geq 2$ and assume that it is not contained in any hyperplane. Now, take some hyperplane $H\subset\mathbb{P}^n$ and consider the set $X\cap H$. Is it possible that there exists another hyperplane $H'$ containing the set $X\cap H$ **if we assume furthermore that $X\cap H$ does not contain ruled components?**


**EDIT1:** As @Lev Borisov pointed out in his answer, such examples exist if we don't put any restrictions on the geometry of the intersection set.


**EDIT2:** The example can be generalized in such a way that the set $X\cap H$ is almost arbitrary. So the assumption about ruled components does not make any difference.



Does every ample divisor "span" a hyperplane?
Take Veronese embedding of ${\mathbb P}^2$ into ${\mathbb P}^5$. Take the hyperplane given by 
$x\_0^2$, where $(x\_0:x\_1:x\_2)$ are coordinates on ${\mathbb P}^2$. Then $x\_0^2=0$ will also be 
(as a set) contained in $x\_0x\_1=0$.


