I am struggling to find binormal and tangential vector, which are both perpendicular to normal vector, which I have found.
So I have a right-hand system ($\mathbf{e_x,e_y,e_z}$), there is a sketch bellow.
As is seen, there is a surface. I need to find a normal vector, tangential vector and binormal vector on the surface ${\xi(x,z)}$. Bellow is another sketch:
Normal vector ${\bf{n}}$ can be found from the following formulas: $${\bf{n}=\frac{\nabla F}{|\nabla F|}},$$ where ${\nabla = \frac{\partial}{\partial x} + \frac{\partial}{\partial y} + \frac{\partial}{\partial z}}$ and ${F}$ is the equation of surface $${F=y-\nabla \xi(x,z)}$$.
So, components of the ${\bf{n}}$ is:
$${\mathbf{n}(-\frac{\frac{\partial \xi(x,z)}{\partial x}}{a}\mathbf{e_x},\frac{1}{a}\mathbf{e_y}, -\frac{\frac{\partial \xi(x,z)}{\partial z}}{a}\mathbf{e_z})},$$
where ${a=\sqrt{1+(\frac{\partial \xi}{\partial x})^2+(\frac{\partial \xi}{\partial z})^2}}$.
Now, I need to find ${\mathbf{ \gamma }}$, ${\mathbf {\tau} }$. Since all vectors are perpendicular to each other and one another can be found through the vector multiplication we could use the following set of equations:
$${{\mathbf{\tau} \cdot \mathbf{n} = n_x \tau_x + n_y \tau_y + n_z \tau_z =0}}$$ $${{\mathbf{\gamma} \cdot \mathbf{n} = n_x \gamma_x + n_y \gamma_y + n_z \gamma_z =0}}$$ $${{\mathbf{n} \times \mathbf{\gamma} =\mathbf{\tau}}}$$
The last vector equation can be expand into 3 equations if we equate corresponding projections:
$${n_y \gamma_z - n_z \gamma_y = \tau_x},$$
$${n_z \gamma_x - n_x \gamma_z = \tau_y},$$
$${n_x \gamma_y - n_y \gamma_x = \tau_z},$$
So we get a set of 5 equations and 6 variables: ${\tau_x , \tau_y, \tau_z, \gamma_x, \gamma_y, \gamma_z}$.
As far as I understand, vectors ${\mathbf{\tau}}$ and ${\mathbf{\gamma}}$ must be unit, so it gives two additional equations:
$${\tau_x^2 + \tau_y^2 + \tau_z^2} = 1,$$ $${\gamma_x^2 + \gamma_y^2 + \gamma_z^2} = 1.$$
That is all in common leads to the following equations set:
$${-\frac{\partial \xi}{\partial x} \tau_x + \tau_y - \frac{\partial \xi}{\partial z} \tau_z = 0 },$$
$${-\frac{\partial \xi}{\partial x} \gamma_x + \gamma_y - \frac{\partial \xi}{\partial z} \gamma_z = 0 },$$
$${\gamma_z + \frac{\partial \xi}{\partial z} \gamma_y = a \tau_x, }$$
$${ - \frac{\partial \xi}{\partial z} \gamma_x + \frac{\partial \xi}{\partial x} \gamma_z = a \tau_y, }$$
$${ -\frac{\partial \xi}{\partial x} \gamma_y - \gamma_x = a \tau_z, }$$
$${\tau_x^2 + \tau_y^2 + \tau_z^2} = 1,$$ $${\gamma_x^2 + \gamma_y^2 + \gamma_z^2} = 1.$$
It is not only 7 equations for 6 variables, but the system is incompatible..
What am I missing? Where is the wrong assumption?..
Given $$\vec n=(n_1,n_2,n_3)$$ we can take for example $$\vec \gamma=\left(\frac{n_2}{\sqrt{n_1^2+n_2^2}},-\frac{n_1}{\sqrt{n_1^2+n_2^2}},0\right)$$ such that $\vec n \cdot \vec \gamma =0$ and finally $$\vec \tau = \vec n \times \vec \gamma$$
More in general to obtain all the possible solutions, assuming wlog $n_3 \neq 0$, $\vec \tau$ and $\vec \gamma$ are unit vectors lying on the plane $n_1x+n_2y+n_3z=0$, we can then set $\vec \Gamma =(\Gamma_1,\Gamma_2,\Gamma_3)$ such that
$$n_1\Gamma_1+n_2\Gamma_2+n_3\Gamma_3=0 \implies \Gamma_3=-\frac{n_1\Gamma_1+n_2\Gamma_2}{n_3}$$
$$ \implies \vec \Gamma =\left(\Gamma_1,\Gamma_2,-\frac{n_1\Gamma_1+n_2\Gamma_2}{n_3}\right),\;\;\vec \gamma = \frac{\vec \Gamma}{\left|\vec \Gamma\right|}$$
and finally $$\vec \tau = \pm\; \vec n \times \vec \gamma$$