From Joe Harris, Algebraic Geometry, Page 10.
Show that if seven points $p_{1},\cdots,p_{7}$ on a twisted cubic curve, then the common zero locus of the quadratic polynomials vanishing at the $p_{i}$ is the twisted cubic.
I am looking for a hint to solve this problem (and the more general case of rational normal curves). I know $p_{i}$ must be in general position, such that any four of them span $\mathbb{P}^{3}$. But this alone does not give an answer immediately. Here is an argument based on Harris' argument on page 7.
Consider $p_{1},p_{2},p_{3}$. It must span a hypersurface of dimension 2 in $\mathbb{P}^{3}$. And similarly is $p_{4},p_{5},p_{6}$. If $p_{7}$ lies on all quadratic polynomial passing through them, then $p_{7}$ has to lie on the intersection of two hypersurfaces – therefore must be lying on at least one hyperplane spanned by the above 2 sets. So we can write $p_{7}$ to be the linear combination of $p_{1},p_{2},p_{3}$ without loss of generality. But this contradicts the fact that they lying on general position (such that $p_{1},p_{2},p_{3},p_{7}$ span $\mathbb{P}^{7}$). So $p_{7}$ must not lie on all quadratic polynomials passing through them.
Since we know every quadratic in $\mathbb{P}^{3}$ is determined by $4+6-1=9$ coefficients, two quadratics pass through the same 6 points should give us 3 dimension "wiggle room" left. But then I am lost as what to proceed.
Best Answer
Let $C$ be the twisted cubic. The space $V$ of quadric surfaces containing $p_1,...,p_7$ has projective dimension $9-7 = 2$. Now, any quadric $Q$ containing $p_1,...,p_7$ intersects $C$ in $7 > 2deg(C) = 6$ points. Therefore $C\subset Q$.
We have three independet quadrics $Q_1,Q_2,Q_3\in V$ such that $C\subset Q_i$ for any $i = 1,2,3$. Now, $Q_1\cap Q_2 = C\cup L$ where $L$ is a line. Therefore $Q_1\cap Q_2\cap Q_3 = C$ because $Q_3\cap L\subset C$.