Calculate projection to tangent plane using first fundamental form 
Let $X : (0,\infty)^2 \to \mathbb{R}^3$ be a surface and $q = (1,1)$. If $E(q)=2$, $F(q)=1$ and $G(q)=5$ are the coefficients of the first fundamental form at the point $q$ find the projection of $X_{uv}(q)$ to the tangent plane $T_qX = \langle X_u(q), X_v(q) \rangle$.

Attempt: I really couldn't make any significant progress on this one. Perhaps it needs something that I am not yet familiar with. I know the formula for the projection of a vector into a plane but I can't use it here since it would involve calculations with $N(q)$ (the perpendicular vector) which we cannot deduce from the first fundamental form.
EDIT: Actually, in the past exam where this question comes from it is given that $E(u,v)=2u^3v$, $F(u,v)=u^2v^2$ and $G(u,v)=5uv^3$ though I'm not sure if this helps.
 A: Here's an attempt now that know the first fundamental form not only at a point but as a function of $u,v$.
The idea is to perform Gram-Schmidt to the basis $\{X_u ,X_v\}$ of vector fields.
$$v_1' := X_u$$
$$v_1:= \frac{v_1'}{\|v_1'\|} = \frac{1}{\sqrt{E}}X_u$$
$$v_2':= X_v - (X_v\cdot v_1) v_1 = X_v - \frac{F}{E}X_u$$
$$v_2 := \frac{v_2'}{\|v_2'\|} = \dots = -\frac{F/E}{\sqrt{G-F^2/E}}X_u+\frac{1}{\sqrt{G-F^2/E}}X_v$$
Then we have that $\{v_1,v_2\}$ is an orthonormal frame for the tangent bundle. Then, the projection of $X_{uv}$ can be expressed by (projection onto orthonormal basis)
$$
\pi_{TM}(X_{uv}) = (X_{uv}\cdot v_1) v_1 + (X_{uv}\cdot v_2) v_2
$$
The dot products can be computed by plugging in the expressions of $v_1$ and $v_2$ in terms of $X_u,X_v$ and by doing this nice trick:
$$
\partial_{v} (X_u\cdot X_u) = 2 (\partial_v X_u)\cdot X_u \implies X_{uv}\cdot X_u = \frac{1}{2} \partial_v (E)
$$
$$
\partial_{u} (X_v\cdot X_v) = 2(\partial_u X_v)\cdot X_v +  \implies \underbrace{X_{vu}}_{X_{uv}}\cdot X_v = \frac{1}{2} \partial_u (G)
$$
were we assume the parametrization is $C^2$ to interchange order of derivatives.
With this in mind the products can be computed:
$$
X_{uv}\cdot v_1 = \frac{1}{2\sqrt{E}}\partial_v(E)
$$
$$
X_{uv}\cdot v_2 = \frac{F/E}{2\sqrt{G-F^2/E}}\partial_v(E) + \frac{1}{2\sqrt{G-F^2/E}}\partial_u(G)
$$
Evaluating everything at $q\equiv(1,1)$ gives $E=2,F=1,G=5,\partial_v(E)=2,\partial_u(G)=5$ from the formulas for $E,F,G$ in terms of $u,v$.
$$
\pi_{TM}(X_{uv})|_q = \frac{1}{\sqrt{2}} v_1(q) + \sqrt{2} v_2(q) 
$$
or, if you prefer, in terms of $X_u(q),X_v(q)$:
$$
\pi_{TM}(X_{uv}(q)) = \frac{1}{6}X_u(q) + \frac{2}{3}X_v(q)
$$
Check the numbers twice before believing me, but this is the method I came up with. There are probably more elegant ways of doing this, though, in terms of Christoffel symbols and their relations to the derivatives of $E,F,G$.
