Why do these intersections of surfaces give a line not a point?

Question

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 $h$?

Answer 1

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.

Answer 2

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).