Let $f: \mathbb{R}^2 \supset M \rightarrow \mathbb{R}^3$ be a smooth immersion and let $N: M \rightarrow S^2 \subset \mathbb{R}^3$ be the corresponding Gauss map. The normal curvature along a unit tangent direction $X \in TM$ can be expressed as

$$ df(X) \cdot dN(X), $$

where $\cdot$ is the usual Euclidean inner product on $\mathbb{R}^3$. The *principal curvature directions* $X_1, X_2 \in TM$ are the unit directions along which normal curvature is minimized and maximized, respectively. It is a well-established fact that (away from umbilic points) these directions are orthogonal in the sense that

$$ df(X_1) \cdot df(X_2) = 0. $$

**Question 1** Is there any *purely geometric* intuition for why principal curvature directions are orthogonal?

**Question 2** Can you argue that the bilinear form $df(X) \cdot dN(Y)$ is symmetric w.r.t. $X$ and $Y$ *without* resorting to the use of coordinates?

In fact, I'd be satisfied with an answer to Question 2 alone, since a self-adjoint operator has orthogonal eigenvalues. I'm really looking for **intuition** here, *not* just a formal proof -- if all I get is a bunch of $E,F,G$ and $e,f,g$ or (god forbid) $\Gamma_{ij}^k$ I just might scream! ;-)

Thanks!