On page 232 of Do Carmo's Differential Geometry of curves and surfaces, we take the inner product of the first four relations below with $X_u$ and $X_v$: $$X_{uu} = \Gamma^1_{11}X_u + \Gamma^2_{11}X_v+L_1N$$ $$X_{uv} = \Gamma^1_{12}X_u + \Gamma^2_{12}X_v+L_2N$$ $$X_{vu} = \Gamma^1_{21}X_u + \Gamma^2_{21}X_v + \overline{L_2}N$$ $$X_{vv} = \Gamma^1_{22}X_u + \Gamma^2_{22}X_v + L_3N$$
to get \begin{cases} \langle X_u, X_{uu} \rangle = \Gamma^1_{11}\cdot \langle X_u, X_u \rangle + \Gamma^2_{11}\cdot \langle X_v, X_u \rangle + 0 = \Gamma^1_{11}E + \Gamma^2_{11}F \\ \langle X_v, X_{uu}\rangle = \Gamma^1_{11}\cdot \langle X_v, X_u \rangle + \Gamma^2_{11}\cdot \langle X_v, X_v \rangle + 0 = \Gamma^1_{11}F + \Gamma^2_{11}G \end{cases}
\begin{cases} \langle X_u, X_{uv} \rangle = \Gamma^1_{12}\cdot\langle X_u, X_u \rangle + \Gamma^2_{12}\cdot \langle X_u, X_v\rangle + 0 = \Gamma^1_{11}E + \Gamma^2_{12}F \\ \langle X_v, X_{uv}\rangle = \Gamma^1_{12}\cdot \langle X_v, X_u \rangle + \Gamma^2_{12}\cdot \langle X_v, X_v\rangle +0= \Gamma^1_{12}F+ \Gamma^2_{12}G \end{cases}
\begin{cases} \langle X_u, X_{vv}\rangle = \Gamma^1_{22}\cdot \langle X_u, X_u\rangle + \Gamma^2_{22}\cdot \langle X_u, X_v \rangle + 0 = \Gamma^1_{22}E + \Gamma^2_{22}F \\ \langle X_v, X_{vv} \rangle = \Gamma^1_{22}\cdot \langle X_v, X_u \rangle + \Gamma^2_{22}\cdot \langle X_v, X_v\rangle + 0 = \Gamma^1_{22}F + \Gamma^2_{22}G \end{cases}
which Do Carmo says can be written as $$\frac{1}{2} E_u$$, $$F_u-\frac{1}{2}E_v$$, $$\frac{1}{2}E_v$$, $$\frac{1}{2} G_u$$, $$F_v - \frac{1}{2}G_u$$, and $$\frac{1}{2}G_v$$ in that order.
Now I don't understand why this is the case. For example, for the very first equation, shouldn't the inner product between $X_u$ and $X_{uu}$ be $$\frac{\partial X}{ \partial u} \cdot \frac{\partial X}{\partial uu} = \frac{\partial E}{\partial u} = E_u ? $$ Where does the half come from?
Thanks.