The effect of rotated coordinates on level surfaces

Question

Consider the function $$ f(x, y, z) = \begin{cases} \dfrac{|(x, y, z) \times v|}{|(x, y, z)|} & (x, y, z) \neq (0, 0, 0), \\ 0 & (x, y, z) = (0, 0, 0). \end{cases} $$ Why, without loss of generality, can we choose $v$ to be $\hat{k} = (0,0,1)$ or any other non-zero vector? How can it be promised that all level surfaces of this function will be the same?

Answer 1

$\newcommand{\Vec}[1]{\mathbf{#1}}$Let's write $\Vec{x} = (x, y, z)$. If $A$ is a $3 \times 3$ orthogonal matrix of determinant $1$ (geometrically, a rotation of Euclidean three-space), then for every vector $\Vec{v}$ we have $$ A(\Vec{x} \times \Vec{v}) = A\Vec{x} \times A\Vec{v} $$ thanks to the geometric characterization of the cross product.

To emphasize how the function $f$ depends on $\Vec{v}$, let's write $$ f_{\Vec{v}}(\Vec{x}) = \begin{cases} \frac{|\Vec{x} \times \Vec{v}|}{|\Vec{x}|} & \text{if $\Vec{x} \neq \Vec{0}$,} \\ 0 & \text{otherwise.} \end{cases} $$

If $\Vec{v}$ is a unit vector and $A$ is a rotation with $A\Vec{v} = \Vec{e}_{3} = (0, 0, 1)$, then for all non-zero $\Vec{x}$ we have $$ f_{\Vec{v}}(A^{-1}\Vec{x}) = \frac{|A^{-1}\Vec{x} \times \Vec{v}|}{|A^{-1}\Vec{x}|} = \frac{|A\bigl[A^{-1}\Vec{x} \times \Vec{v}\bigr]|}{|A[A^{-1}\Vec{x}]|} = \frac{|\Vec{x} \times A\Vec{v}|}{|\Vec{x}|} = f_{A\Vec{v}}(\Vec{x}) = f_{\Vec{e}_{3}}(\Vec{x}). $$ In words, up to rotation we may as well pick $\Vec{v} = \Vec{e}_{3}$: The level sets of $f$ will merely be rotated as dictated by the preceding equation.