Tangent plane and normal vector of a surface

Definition

Let $r \colon D \to \mathbb{R}^3$ be a parametrised surface that is regular at the point $(u_0, v_0)$ of $D$, and let $p = r(u_0, v_0)$. The tangent plane at $p$ is the plane $\{ p + s \, r_u(u_0, v_0) + t \, r_v(u_0, v_0) : s, t \in \mathbb{R} \}$. The vector $N(u_0, v_0) = r_u(u_0, v_0) \times r_v(u_0, v_0)$ is the normal vector of $r$ at $(u_0, v_0)$, and $n = N / \|N\|$ is the unit normal vector.

Remarks

The vector $N$ is perpendicular to $r_u$ and $r_v$, so it is perpendicular to the tangent plane. The other unit normal at $p$ is $-n$. For the graph $z = f(x, y)$, the normal is $(-f_x, -f_y, 1)$, which points upward. If the surface is the level set $g(x, y, z) = c$ of a $C^1$ function $g$ with $\nabla g(p) \ne 0$, then the gradient $\nabla g(p)$ is a normal vector at $p$. For the unit sphere with $r(\varphi, \theta) = (\sin \varphi \cos \theta, \sin \varphi \sin \theta, \cos \varphi)$, one gets $r_\varphi \times r_\theta = \sin \varphi \, r(\varphi, \theta)$, which points outward for $0 < \varphi < \pi$.

Explanations

L1 human

The tangent plane is the flat plane that touches the surface at a point, and the normal vector sticks straight out of it. The cross product of the two coordinate tangent vectors gives a normal at once.