2d:null 2e:["$","title",null,{"children":"「向量比」的運算性質 | lochiwei | Observable"}] 33:T5dd,md`

${mathCSS.tag.property()}

若 ${tex`\vec{v} = {\color{blue}t} \; \vec{u}`}（${tex`\vec{u}\neq\vec{0}`}），則： ${tex`\dfrac{{\color{red}\vec{v}}}{\vec{u}}={\color{blue}t}`}

若： ${tex` \left\{ \begin{aligned} \vec{c} &= {\color{blue}r} \; \vec{a} + {\color{blue}s} \; \vec{b} \\ \vec{d} &= {\color{blue}t} \; \vec{a} + {\color{blue}u} \; \vec{b} \end{aligned} \right. `} ，且 ${tex`\vec{a}\times\vec{b}\neq 0`}，則： ${tex`\dfrac{{\color{red}\vec{c}}\times{\color{red}\vec{d}}}{\vec{a}\times\vec{b}}= \begin{vmatrix} r & s \\ t & u \end{vmatrix} `}

如果上個性質中，若 ${tex`\vec{c}\parallel\vec{a}`}、 ${tex`\vec{d}\parallel\vec{b}`}，則： ${tex`\dfrac{{\color{red}\vec{c}}\times{\color{red}\vec{d}}}{\vec{a}\times\vec{b}}= \left(\dfrac{{\color{red}\vec{c}}}{\vec{a}}\right) \left(\dfrac{{\color{red}\vec{d}}}{\vec{b}}\right) `}

${tex` \dfrac{{\color{blue}{r}}{\color{red}\vec{v}}}{{\color{blue}{s}}\vec{u}}= \left(\dfrac{{\color{blue}{r}}}{{\color{blue}{s}}}\right) \left(\dfrac{{\color{red}\vec{v}}}{\vec{u}}\right) `}，其中 ${tex`{\color{blue}{r}}, {\color{blue}{s}} \in \R`}，${tex`s\neq 0`}，${tex`\vec{u}\neq \vec{0}`}

${tex` \dfrac{\vec{u}}{\vec{u}}= {\color{blue}{1}} `}，其中 ${tex`\vec{u}\neq \vec{0}`}

`34:T95e,md` 1. 此性質的證明，請參閱「[向量比](https://observablehq.com/@lochiwei/vector-ratio)」。 2. 先計算 ${tex`\vec{c}\times\vec{d}`}（⚠️ 注意：「平面外積」只是一個「實數」，並非二維向量❗）
${tex.block` \begin{aligned} {\color{red}\vec{c}}\times{\color{red}\vec{d}} &= ({\color{blue}r} \; \vec{a} + {\color{blue}s} \; \vec{b}) \times ({\color{blue}t} \; \vec{a} + {\color{blue}u} \; \vec{b}) \\ &= {\color{blue}ru} \; (\vec{a} \times \vec{b}) + {\color{blue}st} \; (\vec{b} \times \vec{a}) \\ &= ({\color{blue}ru-st})\; (\vec{a} \times \vec{b}) \end{aligned} `}
再將 ${tex`\vec{a} \times \vec{b}`} 除到左邊，即可得：
${tex.block`\dfrac{{\color{red}\vec{c}}\times{\color{red}\vec{d}}}{\vec{a}\times\vec{b}}= \begin{vmatrix} r & s \\ t & u \end{vmatrix} `} 3. 若 ${tex`\vec{c}\parallel\vec{a}`}、 ${tex`\vec{d}\parallel\vec{b}`}，則：${tex`\vec{c}={\color{blue}{r}} \; \vec{a}`}、${tex`\vec{d}={\color{blue}{u}} \; \vec{b} \quad (r, u \in \R)`}，根據「性質 1」： ${tex.block`\dfrac{{\color{red}\vec{c}}}{\vec{a}}={\color{blue}r}, \; \dfrac{{\color{red}\vec{d}}}{\vec{b}}={\color{blue}u}`} 根據「性質 2」： ${tex.block`\dfrac{{\color{red}\vec{c}}\times{\color{red}\vec{d}}}{\vec{a}\times\vec{b}}= \begin{vmatrix} r & 0 \\ 0 & u \end{vmatrix} =ru `} 故得證。 4. 根據「[向量比的定義](https://observablehq.com/@lochiwei/vector-ratio)」：${tex.block` \dfrac{{\color{blue}{r}}{\color{red}\vec{v}}}{{\color{blue}{s}}\vec{u}} \overset{\Delta}{=} \dfrac{ {\color{blue}{r}}{\color{red}\vec{v}} \cdot {\color{blue}{s}}\vec{u} }{ {\color{blue}{s}}\vec{u} \cdot {\color{blue}{s}}\vec{u} } = \dfrac{ {\color{blue}{r\cancel{s}}} ( {\color{red}\vec{v}} \cdot \vec{u} ) }{ {\color{blue}{s\cancel{s}}} ( \vec{u} \cdot \vec{u} ) } = \left(\dfrac{ {\color{blue}{r}} }{ {\color{blue}{s}} }\right) \left(\dfrac{ {\color{red}\vec{v}} \cdot \vec{u} }{ \vec{u} \cdot \vec{u} }\right) \overset{\Delta}{=} \left(\dfrac{{\color{blue}{r}}}{{\color{blue}{s}}}\right) \left(\dfrac{{\color{red}\vec{v}}}{\vec{u}}\right) `} 5. 因為 ${tex`\vec{u} = {\color{blue}{1}} \vec{u}`}，再根據「性質 1」即可得證。 `35:[] 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