Show the vector field $F(\mathbf{x}) = \begin{pmatrix} x \\ y \\ z \\ \end{pmatrix}$ is tangent to the curve $z^2=x^2+y^2$ everywhere.
Geometrically, this is quite straightforward: the curve is made of two cones joined at their tips at the origin with aperture $\frac{\pi}{2}$, whilst $F(\mathbf{x})$ is a vector field which is going radially outwards, so the conclusion follows. Is there a way to show this algebraically, maybe using grad? What if the vector field and curve in question were not as easy to visualise?
Instead of writing $z^2=x^2+y^2$ I can let $S=x^2+y^2-z^2$, and then the equation becomes $S=0$. Think of $S=0$ as a surface in 3D; we can find a normal vector to that surface by computing $\nabla S$, \begin{equation} \nabla S=\begin{pmatrix} 2x\\2y\\-2z \end{pmatrix}. \end{equation}
Now, $\nabla S$ is normal to the surface $S=0$, which is equivalent to saying $\nabla S$ is perpendicular to a vector $v$ if and only if $v$ is tangent to $S=0$. You can easily check that $\nabla S\cdot F=0$, which is enough to prove $F$ is a tangent vector everywhere.