Why 7 points on a twisted cubic is enough to fix a quadratic? 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. 
 A: 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 $Q_3\cap L\subset C$.
A: Let a homogeneous polynomial of degree 2 on $\mathbb{P}^3$ be written as $$ a_{00}z_0^2 + a_{01}z_0z_1 + \cdots + a_{33}z_3^2$$ and $p=[p_0:p_1 : p_2: p_3], q=[q_0:q_1:q_2:q_3] \in \{p_1, \dots, p_7\} \subset C$. Since to give a quadric hypersurface some $a_{ij}$ must be non-zero, and since $F,\lambda F=0$ define the same quadric hypersurface in $\mathbb{P}^3$, we have a projective space $\mathcal{P}$ of dimension $\binom{4}{2} + 4-1=9$ of quadric hypersurfaces in $\mathbb{P}^3$ as stated above.
The condition that a given quadric contain $p$ is given by $H_p := \sum_{i,j}a_{ij}p_ip_j=0$ which is linear in $a_{ij}$ (as the $p_i,p_j$ are constants), so gives a linear hyperplane in $\mathcal{P} \simeq \mathbb{P}^9$. Since the $\{p_i\}_{i=1}^7$ lie in general position, we may choose $\{a_{ij}\}$ in each $H_p$ so that $H_p, H_q$ intersect transversely for all $p,q\in \{p_i\}_{i=1}^7$. Therefore, the subspace of $\mathcal{P}$ given by homogeneous quadratic polynomials vanishing at all $\{p_i\}_{i=1}^7$ has projective dimension $9-7=2$, as stated above. But this subspace is already spanned by $Q_0,Q_1,Q_2$ defining $C$ as stated above, since their coefficients lie in general position in $\mathcal{P}$.
[there is not much novelty in my answer beyond previous responses, but this is my understanding of the details]
