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# 平行四邊形在 n 維的面積

- [n 維向量內積 (Dot Product)](/XNUf1M5zQW-hLRDJZC7SWg)
- [n 維向量外積 (Cross Product)](/LGyZNOCKTWqf_Dap0yh6uw)

了解如何計算平行四邊形在多維度的計算過程，即可發現其中包括了行列式的規則由來，由此進入 Wedge Product (楔積) 就理所當然了。

## 2D

$$
\cases{
\vec{u}=\pmatrix{u_1\\u_2}\\
\vec{v}=\pmatrix{v_1\\v_2}\\
}\implies |\vec{u}||\vec{v}||\sin\theta|=|u_1v_2-v_1u_2|
$$

### 面積與內積

$$
\cases{
\vec{u_\perp}=\pmatrix{-u_2\\u_1}\\
\vec{v_\perp}=\pmatrix{-v_2\\v_1}
}\implies
|\vec{u}||\vec{v}||\sin\theta| = |\vec{v}\cdot\vec{u_\perp}| = |\vec{u}\cdot\vec{v_\perp}|
$$

## 3D

$$
\cases{
\vec{u}=\pmatrix{u_1\\u_2\\u_3}\\
\vec{v}=\pmatrix{v_1\\v_3\\v_3}\\
}\implies |\vec{u}||\vec{v}||\sin\theta| = ?
$$

$$
\cases{
\vec{u}\cdot\vec{v}=|\vec{u}||\vec{v}|\cos\theta\\
\cos^2\theta+\sin^2\theta=1
}\\
\implies\cos\theta=\frac{\vec{u}\cdot\vec{v}}{|\vec{u}||\vec{v}|}\\
(|\vec{u}||\vec{v}|\sin\theta)^2=|\vec{u}|^2|\vec{v}|^2-(\vec{u}\cdot\vec{v})^2
$$

$$
\begin{array}l
(|\vec{u}||\vec{v}|\sin\theta)^2
&=|\vec{u}|^2|\vec{v}|^2-(\vec{u}\cdot\vec{v})^2\\
&=(u_1^2+u_2^2+u_3^2)(v_1^2+v_2^2+v_3^2)-(u_1v_1+u_2v_2+u_3v_3)^2\\
&=(u_1v_2-v_1u_2)^2+(u_2v_3-v_2u_3)^2+(u_3v_1-v_3u_1)^2\\
&=|\vec{u}\times\vec{v}|^2
\end{array}\\
|\vec{u}||\vec{v}||\sin\theta|=\sqrt{(|\vec{u}||\vec{v}|\sin\theta)^2}=\sqrt{|\vec{u}\times\vec{v}|^2}=|\vec{u}\times\vec{v}|
$$

## n-D

$$
\cases{
\vec{u}=\pmatrix{u_1\\u_2\\u_3\\\vdots\\u_n}\\
\vec{v}=\pmatrix{v_1\\v_3\\v_3\\\vdots\\v_n}\\
}\implies |\vec{u}||\vec{v}||\sin\theta|=?
$$

$$
\cases{
\vec{u}\cdot\vec{v}=|\vec{u}||\vec{v}|\cos\theta\\
\cos^2\theta+\sin^2\theta=1
}\\
\implies\cos\theta=\frac{\vec{u}\cdot\vec{v}}{|\vec{u}||\vec{v}|}\\
(|\vec{u}||\vec{v}|\sin\theta)^2=|\vec{u}|^2|\vec{v}|^2-(\vec{u}\cdot\vec{v})^2
$$

:::info
[拉格朗日恆等式 (Lagrange's Identity)](/u9RoqmLDQX2FMcFKmRizLg)
$$
(\sum_{i=1}^n u_i^2)(\sum_{i=1}^n v_i^2)-(\sum_{i=1}^n u_iv_i)^2=\sum_{i=1}^{n-1}\sum_{j=i+1}^n (u_iv_j-v_iu_j)^2
$$
:::

$$
\begin{array}l
(|\vec{u}||\vec{v}|\sin\theta)^2
&=|\vec{u}|^2|\vec{v}|^2-(\vec{u}\cdot\vec{v})^2\\
&=(u_1^2+u_2^2+\cdots+u_n^2)(v_1^2+v_2^2+\cdots+v_n^2)-(u_1v_1+u_2v_2+\cdots+u_nv_n)^2\\
&=(\sum_{i=1}^n u_i^2)(\sum_{i=1}^n v_i^2)-(\sum_{i=1}^n u_iv_i)^2\\
&=\sum_{i=1}^{n-1}\sum_{j=i+1}^n (u_iv_j-v_iu_j)^2\\
&=|\vec{u}\times\vec{v}|^2
\end{array}\\
\therefore|\vec{u}||\vec{v}||\sin\theta|=\sqrt{(|\vec{u}||\vec{v}|\sin\theta)^2}=\sqrt{|\vec{u}\times\vec{v}|^2}=|\vec{u}\times\vec{v}|
$$

###### tags: `math` `lagrange` `parallelogram`