📂Geometry

Intersection of a Plane and a Normal Vector

Definition ¹

Let a subset of the Euclidean space $2$ have coordinates $U \subset \mathbb{R}^{2}$ and $u_{1}$, then the directional derivatives $\mathbf{x}_{1}$ and $\mathbf{x}_{2}$ can be referred to as follows on a simple surface $\mathbf{x} : U \to \mathbb{R}^{3}$.

$$\begin{align*} \mathbf{x}_{1} := {{ \partial \mathbf{x} } \over { \partial u_{1} }} & , & \mathbf{x}_{2} := {{ \partial \mathbf{x} } \over { \partial u_{2} }} \end{align*}$$

1. The plane perpendicular to $p = \mathbf{x} (a,b)$ at point $\mathbf{x}_{1} \times \mathbf{x}_{2}$ is called the Tangent Plane at $p$ for $\mathbf{x}$.
2. The following defined $\mathbf{n}$ is called the Unit Normal at $p$. $$\mathbf{n}(a,b) := {{ \mathbf{x}_{1} \times \mathbf{x}_{2} } \over { \left| \mathbf{x}_{1} \times \mathbf{x}_{2} \right| }}$$

Explanation

Just as thinking of a tangent when speaking of a curve is natural, so is considering the tangent plane of a surface. The tangent plane at $p$ is the plane that best approximates the surface around $p$.

Since simple surfaces are defined by $\mathbf{x}_{1} \times \mathbf{x}_{2} \ne 0$, the existence of the normal $\mathbf{n}$ is always guaranteed.

From the following theorem, one can understand that the tangent plane is a set of tangent vectors and becomes a vector space. For this reason, the tangent plane is also called a tangent space. The tangent space on the surface $M$ at point $p$ is denoted as $T_{p}M$.

Theorem ²

The set of all tangent vectors at point $p = \mathbf{x}(a,b)$ of a simple surface $\mathbf{x} : U \to \mathbb{R}^{3}$ is a $2$-dimensional vector space with a basis of $\left\{ \mathbf{x}_{1}(a,b), \mathbf{x}_{2}(a,b) \right\}$. Moreover, the tangent plane at $p$ is parallel to any line passing through some origin of $\mathbb{R}^{3}$.

Proof

The tangentvector $\mathbf{x}_{1}, \mathbf{x}_{2}$ at point $p$ is linearly independent. (Since $\mathbf{x}_{1} \times \mathbf{x}_{2} \ne \mathbf{0}$) The set of all tangent vectors at $p$ is a vector space, thus it is at least a $2$-dimensional vector space. To show that this vector space is $2$-dimensional, it suffices to show that $\left\{ \mathbf{x}_{1}, \mathbf{x}_{2} \right\}$ generates it.

Let $\mathbf{X}$ be a tangent vector at point $p$. And let $\boldsymbol{\gamma}$ be a curve on $\mathbf{x}(U)$ such that $\boldsymbol{\gamma}(0) = p, \dot{\boldsymbol{\gamma}}(0) = \mathbf{X}$. And express $\boldsymbol{\gamma}(t)$ as follows.

$$\boldsymbol{\gamma}(t) = \mathbf{x}\left( \gamma^{1}(t), \gamma^{2}(t) \right)$$

Then, by the chain rule,

$$\dfrac{d \boldsymbol{\gamma}}{d t} = \dfrac{\partial \mathbf{x}}{\partial u^{1}}\dfrac{d \gamma^{1}}{d t} + \dfrac{\partial \mathbf{x}}{\partial u^{2}}\dfrac{d \gamma^{2}}{d t} = \sum_{i}\dfrac{d \gamma^{i}}{d t}\mathbf{x}_{i}$$

$$\implies \mathbf{X} = \dfrac{d \boldsymbol{\gamma}}{d t}(0) = \sum_{i}\dfrac{d \gamma^{i}}{d t}(0)\mathbf{x}_{i}(a,b)$$

As any tangent vector $\mathbf{X}$ is represented as a linear combination of $\left\{ \mathbf{x}_{i} \right\}$, $\left\{ \mathbf{x}_{i} \right\}$ generates the set of all tangent vectors at $p$. Therefore, the set of all tangent vectors at $p=\mathbf{x}(a,b)$ is a $2$-dimensional vector space with a basis of $\left\{ \mathbf{x}_{1}(a,b), \mathbf{x}_{2}(a,b) \right\}$.