Tangent plane at a point on a surface

Question

In Pressley’s book, he writes since the tangent plane at $P$ in $S$ passes through the origin of $\mathbb{R}^3$, it is completely determined by giving a unit vector perpendicular to it called a unit normal to $S$ at $P$. Why must the tangent plane pass through the origin? Because it is a vector space spanned by the linearly independent partial derivative of a surface patch? Even if so why ‘passing through the origin’ plays a role in the plane being completely determined by a unit vector? Will it be impossible to do so if the plane didn’t pass through the origin?

Answer 1

He said this to just identify the word "at $p$". Now the tangent plane is passing through the origin of $R^3$ which is $p$ itself. Thus, $R^3$ is a vector space with the origin at $p$ where the tangent space is a vector subspace generated by two tangent vectors to the surface at $p$. The unit normal is also at $p$.

You are dealing with two different but isometric 3-dimensional vector spaces $R^3$. The first is the one you are talking about and the second is the vector space generated by the tangent vectors and the normal vector. The tangent vectors should pass through the origin of the second which is p itself.
I copied this image from the internet. The points $p$ and $q$ are two different origins of two different spaces