Find the equation of the tangent line to the intersection curve of the surfaces $x=2+\cos(\pi yz)$ and $y=1+\sin(\pi xz)$ at the point $(3,1,2)$ Find the equation of the tangent line to the intersection curve of the two following surfaces at the point $(3,1,2)$:
$$x=2+\cos(\pi yz)$$
$$y=1+\sin(\pi xz)$$

The only way that I know to find the intersection curve is by plugging one of the surfaces into the other one:
$$x=2+\cos(\pi (1+\sin(\pi xz))z)$$
However I really doubt if this is how I should solve the problem.
 A: We find normal vectors to the tangent planes of both surfaces at $(3, 1, 2)$.
$S1: f(x, y, z) = x - \cos(\pi yz) - 2 = 0$
$S2: g(x, y, z) = y - \sin(\pi xz) - 1 = 0$
$f'(x, y, z) = (1, \pi z \sin (\pi y z), \pi y \sin (\pi y z))$
$g'(x, y, z) = (- \pi z \cos (\pi x z), 1, -\pi x \cos (\pi x z))$
$f'(3, 1, 2) = (1, 0, 0)$
$g'(x, y, z) = (- 2 \pi, 1, - 3 \pi)$
The tangent line to the intersection curve of both surfaces at point $(3, 1, 2)$ should be on tangent planes of both surfaces. In other words, it is the intersection line of both planes.
To find direction vector of the intersection line, we take cross product of both normal vectors and that is $(0, 3 \pi, 1)$.
So the equation of tangent line is $ \ (3, 1, 2) + (0, 3 \pi, 1) t $.
A: This is what I would do:

*

*The function that describes the geometric field is $$f(x,y,z)=x-cos(\pi yz)$$

*The gradient of the field is the vector of highest rate of change, with components $$f_x=\frac{\partial f}{\partial x};f_y=\frac{\partial f }{\partial y}; f_z=\frac{\partial f}{\partial z}; \vec n=(f_x, f_y, f_z)$$

*The equation of the plane tangent to any surface generated by a constant value of the field is $$(\vec r -\vec r_0)\cdot \vec n=0$$

*By intersecting the two plane equations of both surfaces I will obtain a line.

And now the work:

*

*$f(x,y,z)=x-cos(\pi yz)$

*$f_x=1; f_y=\pi z sin(\pi yz); f_z=\pi ysin(\pi yz); \vec{n_{f_0}}=(1,0,0)$

*$(\vec r-\vec r_0)\cdot \vec{n_{f_0}}=0\Rightarrow (x-3)\cdot 1+(y-1)\cdot 0+(z-2)\cdot 0=0$

*$g(x,y,z)=y-sin(\pi zx)$

*$g_x=-\pi zcos(\pi zx); g_y=1; g_z=-\pi xcos(\pi zx); \vec{n_{g_0}}=(-2\pi, 1, -3\pi)$

*$(\vec r -\vec r_0)\cdot \vec{n_{g_0}}=0\Rightarrow (x-3)\cdot (-2\pi)+(y-1)\cdot 1 +(z-2)\cdot (-3\pi)=0$

*The tangent line is the solution of linear system $$\left \{\begin{array}{rr} x-3=0 \\(x-3)\cdot(-2\pi)+y-1+(z-2)\cdot(-3\pi)=0\end{array}\right .$$

*The line tangent to the surfaces of equations $x=2+cos(\pi yz)$ and $y=1+sin(\pi zx)$ at point (3,1,2) is $x=3,y=3\pi z+1-6\pi (\therefore )$
