Gram-Schmidt process when metric is not an Euclidean Can we use a Gram-Schmidt process for finding an orthogonal vector if a metric on a space is not an Euclidean one?
These are the vectors and the metric.
If we have the next two vectors:
\begin{align}
\notag
A_{1}=c_{1}y_{1}^{2}y_{2}^{2}u(u+2t)(u+y_{2}^{2})\frac{\partial}{\partial x_{2}},
\end{align}
\begin{align}
\notag
A=-c_{11}\frac{\partial}{\partial x_{1}}-c_{12}\frac{\partial}{\partial x_{2}}-\frac{u(u+2t)}{u+y_{2}^{2}}(c_{8}(u+y_{1}^{2})+c_{1}y_{1}y_{2})\frac{\partial}{\partial y_{1}},
\end{align}
and the metric
\begin{align}
\notag
G=
\begin{pmatrix}
u+vy_{1}^{2} & vy_{1}y_{2} & 0& 0 \\
vy_{1}y_{2} & u+vy_{2}^{2} & 0& 0 \\
0 & 0 & \frac{u+vy_{2}^{2}}{u(u+2tv)}& -\frac{vy_{1}y_{2}}{u(u+2tv)}\\
0& 0& -\frac{vy_{1}y_{2}}{u(u+2tv)} & \frac{u+vy_{1}^{2}}{u(u+2tv)} ,
\end{pmatrix}
\end{align}
If we use Gram-Schmidt in this case, we get a vector which is orthogonal to $A_{1}$ but not to $A$.\
On the other hand, we can find vectors that span a normal bundle of a submanifold generated by $A_{1}$ and $A$, using an almost complex structure.\
We need to find a normal component of a given vector with respect to $A_{1}$ and $A$, that is why we need a vector that is orthogonal to $A_{1}$ and $A$. So, my other question would be is one of the vectors in the normal bundle enough for finding mentioned component?
Thank you!

In the exact same way, but instead of using the dot product, you use whatever inner product you're given.
Example: Consider $\mathbb{R}^2$ with the inner product 
$$\langle (x_1, x_2), (y_1, y_2)\rangle = 2x_1y_1 + x_1y_2 + x_2y_1 + 3x_2y_2.$$ 
Given the basis $\{(1, 0), (0, 1)\}$ of $\mathbb{R}^2$, what do we get when we apply the Gram-Schmidt process? As 
$$\|(1, 0)\|^2 = \langle (1, 0), (1, 0)\rangle = 2(1)(1) + (1)(0) + (0)(1) + 3(0)(0) = 2,$$ 
$u_1 = \frac{1}{\sqrt{2}}(1, 0)$. Now
\begin{align*}
u_2' &= (0, 1) - \left\langle (0, 1), \frac{1}{\sqrt{2}}(1, 0)\right\rangle\frac{1}{\sqrt{2}}(1, 0)\\ 
&= (0, 1) - \left(2(0)\left(\frac{1}{\sqrt{2}}\right)+(0)(0)+(1)\left(\frac{1}{\sqrt{2}}\right)+3(1)(0)\right)\frac{1}{\sqrt{2}}(1,0)\\ 
&= (0,1)-\frac{1}{\sqrt{2}}\frac{1}{\sqrt{2}}(1,0)\\
&= (0,1) - \frac{1}{2}(1, 0)\\
&= \left(-\frac{1}{2}, 1\right)
\end{align*}
and 
\begin{align*}
\|u_2'\|^2 &= \left\|\left(-\frac{1}{2},1\right)\right\|^2\\ 
&= \left\langle\left(-\frac{1}{2},1\right), \left(-\frac{1}{2},1\right)\right\rangle\\ 
&= 2\left(-\frac{1}{2}\right)\left(-\frac{1}{2}\right) + \left(-\frac{1}{2}\right)(1) + (1)\left(-\frac{1}{2}\right)+3(1)(1)\\ 
&= \frac{5}{2}
\end{align*}
so $u_2 = \frac{\sqrt{2}}{\sqrt{5}}(-\frac{1}{2}, 1)$. So an orthonormal basis for $\mathbb{R}^2$, with the inner product above is 
$$\{u_1, u_2\} = \left\{\frac{1}{\sqrt{2}}(1, 0), \frac{\sqrt{2}}{\sqrt{5}}\left(-\frac{1}{2}, 1\right)\right\}.$$
