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Given a pair of functions $f(x,y,z)$ and $g(x,y,z)$.

Now make the three equations:

$$ \nabla f(x,y,z) \times \nabla g(x,y,z) = 0 $$

This gives equations for three 2D surfaces. At first inspection this seems like the intersection of all three should be a point. But playing around it seems they all intersect on a 1D curve. Why?

For example try:

$$f(x,y,z) = x^2+(y-5)^2+z^2$$ $$g(x,y,z) = (x+3)^2+y^2+(z-1)^2$$

this gives three planes which cross each other on a single line between the points $(0,5,0)$ and $(-3,0,1)$.

Is this always the case? Or in what circumstances does this happen?

Edit: Another question is, can you direct me to a book (Fluid Mechanics?) that has this equation? Also apparently it implies there is a function $h(f,g)=0$ how to find this function $h$?

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    I'd try linearizing the equations near a solution - assume $\nabla f \times \nabla g = 0$ at the origin and compute the matrix with $i,j$ coefficient $\nabla_i (\nabla f \times \nabla g)_j|_0$. If this has a one-dimensional kernel, then your equations are "dependent" in some sense and we expect to see curves. If this is indeed true I'm guessing it will involve the commutativity of second derivatives. – 2017-02-13
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    This answer may be helpful: http://math.stackexchange.com/a/1309778/259085 – 2017-02-13
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    Actually, think about things geometrically - the condition is that the vector fields are parallel, which (assuming everything is non-zero) is saying that their direction fields $\mathbb R^3 \to \mathbb {RP}^2$ are equal. Since the space $\mathbb{RP}^2$ of directions is two-dimensional, you morally only have two equations worth of constraints. – 2017-02-13
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    You can turn this in to a rigorous argument using the implicit function theorem: assuming the condition at a single point and some transversality condition on the derivatives at that point, the set condition will hold on some curve through that point. – 2017-02-13
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    Are you saying that it is because vectors (ignoring length) are essentially 2 dimensional? – 2017-02-13
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    @zooby: yeah, that's the gist of it. Your answer is a nice elementary way to see it - so long as one of the gradients is everywhere non-zero, there are only two independent equations. – 2017-02-13

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I just realised the third equation can be written in terms of the first two!

$$A = \partial_x f \partial_y g - \partial_y f \partial_x g$$

$$B = \partial_y f \partial_z g - \partial_z f \partial_y g$$

$$\frac{\partial_z g}{\partial_y g}A +\frac{\partial_x g}{\partial_y g}B = \partial_y f \partial_z g - \partial_z f \partial_y g$$

So its actually only two equations.

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Taylor expanding, $$ f(a+x) = f(a) + x \cdot \nabla f(a) + \frac{1}{2} x \cdot Hf(a) \cdot x + o(x^2), $$ so the Taylor expansion of $\nabla f$ is $$ \nabla f(a+x) = \nabla f(a) + Hf(a) \cdot x + o(x), $$ remembering that $Hf(a)$ is symmetric. Thus, $$ (\nabla f \times \nabla g)(a+x) = (\nabla f \times \nabla g)(a) + (Hf(a)x) \times \nabla g(a) + \nabla f(a) \times (Hg(a)x) + o(x), $$ or setting $n=\nabla f(a)$, $m=\nabla g(a)$, $$ (\nabla f \times \nabla g)(a+x) = n \times m + (Hf(a)x) \times m + n \times (Hg(a)x) + o(x) $$ provided that $\nabla f(a)$ and $\nabla g(a)$ are not both zero at the same time (if they were, we would have to go to $(Hf(a)x) \times (Hg(a)x)$, and the whole thing is more complicated, although not especially difficult).

Now, $Hf$, being symmetric, can be decomposed into three linear maps, $ Hf(a) = \alpha n n^T + an^T + n^T a + A $, where $\alpha $ is a number, $a$ is vector perpendicular to $n$, and $A$ is symmetric with $ An = 0$. Likewise, $Hg(a) = \beta m + bm^T + mb^T + Bk$, $b^Tm=0$, $Bm = 0$. Then $$ (\nabla f \times \nabla g)(a+x) \\ = n \times m + (\alpha nn^Tx + an^Tx + na^Tx+Ax) \times m + n \times (\beta mm^Tx + am^Tx + ma^Tx+Bx ) + o(x) \\ = n \times m (1+(\alpha n + a + \beta m + b )^T x) + (n^Tx)a \times m + (m^Tx)n \times b + (Ax) \times m + n \times (Bx) + o(x) $$

If we now take $n \times m=0$, the first term vanishes, and there is a constant $\lambda$ so that $m=\lambda n$. Hence $$ (\nabla f \times \nabla g)(a+x) = n\times \Big( \lambda(b-a)n^T x + (B-\lambda A)x\Big) + o(x). $$

This was rather messy, but the moral of the story is that the rank of the linear map acting on $x$ on the right-hand side is at most $2$, because the result is always perpendicular to $n$. Hence there is always at least one direction for $x$ so that $(\nabla f \times \nabla g)(a+x)$ remains zero for $x$ sufficently small, so the set $\nabla f \times \nabla g = 0$ is locally a line, and may be higher-dimensional if $f$ and $g$ are chosen appropriately (obviously $f=g$ gives all of space, for example).