---
title : "幾何代數 (Geometric Algebra) II"
path: "幾何代數 Geometric Algebra II"
---

{%hackmd @RintarouTW/About %}

# 幾何代數 (Geometric Algebra) II

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## Solving Linear Equation

$$
\begin{cases}
a_1x+b_1y=c_1\\
a_2x+b_2y=c_2
\end{cases}\implies
x\pmatrix{a_1\\a_2}+y\pmatrix{b_1\\b_2}=\pmatrix{c_1\\c_2}\\[4ex]
x\vec{a}+y\vec{b}=\vec{c}\\[2ex]
(x\vec{a}+y\vec{b})\wedge\vec{b}=\vec{c}\wedge\vec{b}\\[2ex]
x\vec{a}\wedge\vec{b}=\vec{c}\wedge\vec{b}\\[2ex]
\implies x={\vec{c}\wedge\vec{b} \over \vec{a}\wedge\vec{b}}\\[4ex]
\vec{a}\wedge(x\vec{a}+y\vec{b})=\vec{a}\wedge\vec{c}\\[2ex]
y\vec{a}\wedge\vec{b}=\vec{a}\wedge\vec{c}\\[2ex]
\implies y = {\vec{a}\wedge\vec{c} \over \vec{a}\wedge\vec{b}}
$$

Example:

$$
\begin{cases}
x-y=1\\
x+2y=0
\end{cases}
\implies
\begin{cases}
\vec{a}=e_1+e_2\\
\vec{b}=-e_1+2e_2\\
\vec{c}=e_1
\end{cases}
\\[4ex]
\begin{aligned}
\vec{a}\wedge\vec{b} &= (e_1+e_2)\wedge(-e_1+2e_2)\\[2ex]
&= 2e_1\wedge e_2 - e_2\wedge e_1\\[2ex]
&= 2e_1\wedge e_2 + e_1 \wedge e_2\\[2ex]
&= 3(e_1\wedge e_2)\\[2ex]
\vec{c}\wedge\vec{b} &= e_1 \wedge (-e_1+2e_2)\\[2ex]
&= 2(e_1\wedge e_2)\\[2ex]
\implies x &= {2(e_1\wedge e_2) \over 3(e_1\wedge e_2)}
= {2 \over 3}\\[2ex]
\vec{a}\wedge\vec{c} &= (e_1+e_2)\wedge e_1\\[2ex]
&= -e_1 \wedge e_2\\[2ex]
\implies y &= {{-e_1 \wedge e_2} \over {3(e_1\wedge e_2)}}
= {-1 \over 3}\\
\end{aligned}
$$

## Equation and Determinant ($\det$) - Cramer's Rule

$$
\cases{
x=\Large{\vec{c}\wedge \vec{b} \over \vec{a}\wedge \vec{b}}\\
y=\Large{\vec{a}\wedge \vec{c} \over \vec{a}\wedge \vec{b}}
} \implies {
\pmatrix{a_1 & b_1\\a_2 & b_2}\pmatrix{x\\y}=\pmatrix{c_1\\c_2
}
} \implies \cases{
\vec{a}=a_1e_1+a_2e_2\\
\vec{b}=b_1e_1+b_2e_2\\
\vec{c}=c_1e_1+c_2e_2
}\\[4ex]
\eqalign{
x &= {(c_1e_1+c_2e_2)\wedge(b_1e_1+b_2e_2) \over (a_1e_1+a_2e_2)\wedge(b_1e_1+b_2e_2)}\\[2ex]
&= {{c_1b_2(e_1\wedge e_2)-b_1c_2(e_1\wedge e_2)}
\over {a_1b_2(e_1\wedge e_2)-b_1a_2(e_1\wedge e_2)}}\\[2ex]
&= {c_1b_2 - b_1c_2 \over a_1b_2 - b_1a_2} =
{\det\pmatrix{c_1 & b_1\\c_2 & b_2} \over \det\pmatrix{a_1 & b_1\\a_2 & b_2}}\quad\checkmark\\[2ex]
y &= {(a_1e_1+a_2e_2)\wedge(c_1e_1+c_2e_2) \over (a_1e_1+a_2e_2)\wedge(b_1e_1+b_2e_2)}\\[2ex]
&= {{a_1c_2(e_1\wedge e_2)-c_1a_2(e_1\wedge e_2)}
\over {a_1b_2(e_1\wedge e_2)-b_1a_2(e_1\wedge e_2)}}\\[2ex]
&= {a_1c_2 - c_1a_2 \over a_1b_2 - b_1a_2} =
{\det\pmatrix{a_1 & c_1\\a_2 & c_2} \over \det\pmatrix{a_1 & b_1\\a_2 & b_2}}\quad\checkmark\\
}
$$

## Two Reflections = Rotation
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:::info
$$
e^{\theta I}u = ue^{-\theta I}\tag{1}\label{eq:eq1}
$$
:::

$u' := u$ reflected by $v$

$$
u' = v^{-1} u v
$$

$u'' := u'$ reflected by $w$

$$
u'' = w^{-1} v^{-1} u v w
$$

$$
\\[2ex]
\eqalign{
w^{-1}v^{-1} = (vw)^{-1}\implies
u'' &= (vw)^{-1}uvw\\[2ex]
vw &= |v||w|\cos \theta + |v||w|\sin \theta I\\[2ex]
&= |v||w|e^{\theta I}\\[2ex]
\implies (vw)^{-1} &= {e^{-\theta I} \over |v||w|}\\[2ex]
\therefore u'' &= {e^{-\theta I} \over |v||w|} u (|v||w|e^{\theta I})\\[2ex]
&= e^{-\theta I} u e^{\theta I}\\[2ex]
&= ue^{\theta I}e^{\theta I}\eqref{eq:eq1}\\[2ex]
&= ue^{2\theta I}
}
$$

:::success
This is why we need to use $\theta \over 2$ to rotate $\theta$
:::

###### tags: `math` `geometric` `algebra` `geometry` `wedge product`