How do we obtain the twisted cubic? According to the book of Harris, Algebraic geometry, A first course, page $9$, the twisted cubic is defined to be the image $C$ of the map $ v : \mathbb{P}^1 \to \mathbb{P}^3 $ given by $$ v : [X_0 , X_1 ] \to [ X_{0}^{3} , X_{0}^{2} X_{1} , X_{0} X_{1}^{2} , X_{1}^{3} ].$$ 
The author of this book says then, that $C$ lies on the $3$ quadric surfaces $ Q_0 $ , $ Q_1 $ and $ Q_2 $ given as the zero locus of the polynomials $ F_0 (Z) = Z_0 Z_2 - Z_{1}^2 $ , $ F_1 (Z) = Z_0 Z_3 - Z_{1} Z_2 $ and $ F_2 (Z) = Z_0 Z_3 - Z_{2}^2 $. 
Could you explain to me why do we have this? How do we obtain those polynomials in that form?
Thanks in advance for your help.
 A: One can note that the twisted cubic $C$ is a determinantal variety; that is $C$ is the zero locus defined as:
\begin{equation}
C=\left\{[z_0:z_1:z_2:z_3]\in\mathbb{P}^3_{\mathbb{C}}\mid rank\begin{pmatrix}
z_0 & z_1 & z_2\\
z_1 & z_2 & z_3
\end{pmatrix}=1\right\}=Q_0\cap Q_1\cap Q_2
\end{equation}
as you denoted.
Then if:
\begin{equation}
z_0\neq0,\,\exists t\in\mathbb{C}\mid\begin{cases}
z_1=tz_0\\
z_2=tz_1\\
z_3=tz_2
\end{cases}\iff\begin{cases}
z_1=tz_0\\
z_2=t^2z_0\\
z_3=t^3z_0
\end{cases}
\end{equation}
that is:
\begin{equation}
C\cap U_0=\{[1:t:t^2:t^3]\in\mathbb{P}^3_{\mathbb{C}}\mid t\in\mathbb{C}\}
\end{equation}
where:
\begin{equation}
U_0=\left\{[z_0:z_1:z_2:z_3]\in\mathbb{P}^3_{\mathbb{C}}\mid z_0\in\mathbb{C}^{\times}\right\};
\end{equation}
in other words, one can state that:
\begin{equation}
C\cap U_0=\left\{\left[1:\frac{t}{s}:\frac{t^2}{s^2}:\frac{t^3}{s^3}\right]\in\mathbb{P}^3_{\mathbb{C}}\mid t\in\mathbb{C},s\in\mathbb{C}^{\times}\right\}=\left\{\left[s^3:s^2t:st^2:t^3\right]\in\mathbb{P}^3_{\mathbb{C}}\mid t\in\mathbb{C},s\in\mathbb{C}^{\times}\right\}
\end{equation}
and therefore:
\begin{equation}
C=\left\{\left[s^3:s^2t:st^2:t^3\right]\in\mathbb{P}^3_{\mathbb{C}}\mid[s:t]\in\mathbb{P}^1_{\mathbb{C}}\right\}=v_{1,3}\left(\mathbb{P}^1_{\mathbb{C}}\right).
\end{equation}
An analogous reasoning holds for the general Veronese map:
\begin{equation}
v_{n,d}:[x_0:x_1:\dots:x_n]\in\mathbb{P}^n_{\mathbb{C}}\to\left[\mathbf{x}^d\right]\in\mathbb{P}^N_{\mathbb{C}}
\end{equation}
where $\displaystyle N=\binom{n+d}{d}-1$ and:
\begin{equation}
\mathbf{x}^d=\left(x_0^{e_0}\cdot\dots\cdot x_n^{e_n}\right)\mid\sum_{k=0}^ne_k=d,\,e_k\geq0.
\end{equation}
