This is from a section in my course book on elementary differential geometry:
Since the tangent plane $T_p S$ of a surface $S$ at a point $p \in S$ passes through the origin of $\mathbb{R}^3$, it is completely determined by giving a unit vector perpendicular to it…
There are plenty of surfaces with points whose tangent plane doesn't pass through the origin, so why does it say so here?
Best Answer
I THINK:
The author identified the tangent space $T_p\mathbb{R}^3$ and $\mathbb{R}^3$ itself. In this case $T_pS$ is a plane passing through the origin i.e. a subspace of co-dimension one and hence the normal and $p$ determine the plane.