I am struggling with this question. Given a vector field $F(x,y,z)$ and a surface $z = S(x, y)$.
Is there a tangent plane of $S$ that is parallel to $F$ (anywhere)?
For example $F(x, y, z) = (-x^3, -x - y, z)$ and $S(x, y) = x^2 + y^2$.
My attempt is first to get the tangent plane of $S$. The plane equation:
$$ z - z_0 = S_x(x_0, y_0)(x - x_0) + S_y(x_0, y_0)(y - y_0) $$
Gradient:
$$ S_x(x, y) = 2x $$ $$ S_y(x, y) = 2y $$
So that is
$$ z - z_0 = 2 x_0 (x - x_0) + 2 y_0 (y - y_0) $$ $$ q(x, y, z) = (-2x_0) x + (-2y_0) y + z = (z_0 - 2x_0^2 - 2y_0^2) $$
Again gradient:
$$ F_x = q_x \rightarrow -3x^2=-2x_0 $$ $$ F_y = q_y \rightarrow -1 = -2y_0 $$ $$ F_z = q_z \rightarrow 1 = 1 $$
So my answer would be that there is a tangent plane at $x = 2/3, y = 1/2$ ($z$ doesn't matter).
- Is this done right?
- Can I treat the $x_0$ and the $x$ like the same variable because I want to check all $x_0$? I am not sure because the $S_x(x_0, y_x)(x - x_0)$ would then be $0$ if $x_0 = x$.
- Do I have to insert $z = S(x, y)$ in $F(x, y, z) = F(x, y, S(x, y))$?
Per the comments under the question, we're looking for points $(u, v, w) \in \Bbb R^3$ and $(x, y, z) \in {\bf S}$ such that $$F_{(u, v, w)} \parallel T_{(x, y, z)} {\bf S} .$$ Here, we denote by $\bf S$ the surface $\{(x, y, S(x, y)) : (x, y) \in \Bbb R^2\}$.
This holds iff $F_{(u, v, w)}$ is orthogonal to a nonzero vector normal to $S$ at $(x, y, z)$. Since $\bf S$ is the level set $G^{-1}(0)$ of $$G(x, y, z) = z - S(x, y) = - x^2 - y^2 + z ,$$ one such normal vector at $(x, y, z) \in {\bf S}$ is $\nabla G (x, y, z) = (-2 x, -2 y, 1)$.
Thus we can write the orthogonality condition as $$F_{(u, v, w)} \cdot \nabla G(x, y, z) = 0 ,$$ or substituting, $$(-u^3, -u - v, w) \cdot (-2 x, -2 y, 1) = 0$$ This is a single equation in five variables, and so generically we expect solutions to depend on four parameters.
Remark It's perhaps more interesting to ask for points $(x, y, z)$ where $F_{(x, y, z)} \parallel T_{(x, y, z)} {\bf S}$, which amounts to specializing the above equation to $u = x, v = y, w = z = x^2 + y^2$, leaving a single equation in two variables.