I think that it's well-known result that for a differential map $f : \mathbb{R}^m \to \mathbb{R}$ that describes a compact regular submanifold $M$ of dimension $(m-1)$ (hyperplane) you get that its tangent space in any $p \in M$ is $$T_p(M) = Ker(d(f)_p)$$
My question is if also $T_p(M)$ can be understood as the orthogonal vector space of the one generated by the vector $\vec{0p}$, that is, the line $span_{\mathbb{R}}\{\vec{0p}\}$.
At least I think that is true for the $1,2$-dimensional spheres and the cylinder.
But for example for the torus or an ellipse?
Thanks for the support!
No. In $\Bbb R^2$, consider the function $$ f(x, y) = (x-3)^2 + y^2 - 1 $$ The zero-set of this function is the circle $C$ of radius $1$ and center $(3, 0)$, so it contains the point $p = (3, 1)$ for instance. The tangent space at $p$ is the line through $p$ parallel to the $x$-axis, so every vector in that space is orthogonal to $[0, 1]$.
On the other hand, the vector $0p$ is $[3, 1]$, which is orthogonal only to the zero-vector in $T_p(C)$.