User:小躍/數學

二元一次方程式

1. 當${\begin{cases}ax+by=c\\cx+dy=e\end{cases}}$ 時，$x$或$y$就只有一組解。
2. 當${\begin{cases}ax+by=c\\ax+by=e\end{cases}}\Rightarrow (c=e)$ 時，$x$或$y$就有無限多組解。
3. 當${\begin{cases}ax+by=c\\ax+by=e\end{cases}}\Rightarrow (c\neq e)$ 時，$x$或$y$就無解。

平面向量

平面向量的表示法

1. 設兩點坐標$A(x_{1},y_{1}),B(x_{2},y_{2})$，則${\color {Red}{\overrightarrow {AB}}=(x_{2}-x_{1},y_{2}-y_{1})}$
2. 兩向量平行：當${\vec {a}}(x_{1},y_{1}),{\vec {b}}(x_{2},y_{2})$ ，且${\vec {a}}//{\vec {b}}$ 時，則${\color {Red}{\frac {x_{1}}{x_{2}}}={\frac {y_{1}}{y_{2}}}}$
3. 分點公式：$A$、$P$、$B$

   內分點： $P$ 介於$A$、$B$ 之間，${\overline {AP}}:{\overline {PB}}=m:n$ (內分)，

   則${\color {Blue}{\overrightarrow {OP}}={\frac {n}{m+n}}{\overrightarrow {OA}}+{\frac {m}{m+n}}{\overrightarrow {OB}}}$

   外分點： $P$ 介於$A$、$B$ 之外，${\overline {AP}}:{\overline {PB}}=m:n$ (外分)，

   則${\color {Blue}{\overrightarrow {OP}}={\frac {-n}{m-n}}{\overrightarrow {OA}}+{\frac {m}{m-n}}{\overrightarrow {OB}}}$

4. 直線的參數式：過$A(x_{0},y_{0})$，向量${\vec {v}}=(a,b)$平行的直線上點$P(x,y)$可表示為

   ${\color {Blue}{\begin{cases}x=x_{0}+at\\y=y_{0}+bt\end{cases}}t\in \mathbb {R} }$

平面向量的內積

1. 設${\vec {a}}(a_{1},a_{2})$ 和${\vec {b}}(b_{1},b_{2})$，則${\color {Blue}{\vec {a}}\cdot {\vec {b}}=\left|{\vec {a}}\right|\left|{\vec {b}}\right|\cdot \cos \theta }={\color {Red}a_{1}b_{1}+a_{2}b_{2}}$
2. 當${\vec {a}}//{\vec {b}}$，設$a_{1},a_{2},b_{1},b_{2}$，則符合柯西不等式為：${\color {Red}(\left[a_{1}\right]^{2}\left[a_{2}\right]^{2})+(\left[b_{1}\right]^{2}\left[b_{2}\right]^{2})\geq (a_{1}b_{1}+a_{1}b_{2})^{2}}$
3. 正射影公式：
   1. 設${\vec {a}}$ 對${\vec {b}}$之正射影${\vec {c}}$，則${\color {Purple}{\vec {c}}={\Bigg (}{\frac {{\vec {a}}\cdot {\vec {b}}}{\left|{\vec {b}}\right|^{2}}}{\Bigg )}{\vec {b}}}$
   2. 設${\vec {b}}$ 對${\vec {a}}$之正射影${\vec {c}}$，則${\color {Purple}{\vec {c}}={\Bigg (}{\frac {{\vec {a}}\cdot {\vec {b}}}{\left|{\vec {a}}\right|^{2}}}{\Bigg )}{\vec {a}}}$
4. 距離公式：
   1. 設點$P(x_{0},y_{0})$到直線$L:ax+by+c=0$ 的距離為 $d(P,L)={\frac {\left|ax_{0}+by_{0}+c\right|}{\sqrt {a^{2}+b^{2}}}}$
   2. 設直線${\begin{cases}L_{1}:ax+by+c_{1}=0\\L_{2}:ax+by+c_{2}=0\end{cases}}$ 的距離為 $d(L_{1},L_{2})={\frac {\left|c_{1}c_{2}\right|}{\sqrt {a^{2}+b^{2}}}}$

二階行列式

1. 公式：${\begin{vmatrix}a&b\\c&d\end{vmatrix}}=ad-bc$
2. 當解析失败 (SVG（MathML可通过浏览器插件启用）：从服务器“http://localhost:6011/zh.wikipedia.org/v1/”返回无效的响应（“Math extension cannot connect to Restbase.”）：): {\displaystyle \begin{cases} {\color{Red}a_1}x+{\color{Blue}b_1}y={\color{Olive Green}c_1} \\ {\color{Orange Red}a_2}x+{\color{Navy Blue}b_2}y={\color{Emerald}c_2} \end{cases} } 時，

   ∵解析失败 (未知函数“\begin{vmatrix}”): {\displaystyle \vartriangle= \begin{vmatrix} {\color{Red}a_1} & {\color{Blue}b_1} \\ {\color{Orange Red}a_2} & {\color{Navy Blue}b_2} \end{vmatrix} ={\color{Red}a_1} \cdot {\color{Navy Blue}b_2}-{\color{Orange Red}a_2} \cdot {\color{Blue}b_1}}

   解析失败 (未知函数“\begin{vmatrix}”): {\displaystyle \vartriangle_x= \begin{vmatrix} {\color{Olive Green}c_1} & {\color{Blue}b_1} \\ {\color{Emerald}c_2} & {\color{Navy Blue}b_2} \end{vmatrix} ={\color{Olive Green}c_1} \cdot {\color{Navy Blue}b_2}-{\color{Emerald}c_2} \cdot {\color{Blue}b_1}}

   解析失败 (未知函数“\begin{vmatrix}”): {\displaystyle \vartriangle_y= \begin{vmatrix} {\color{Red}a_1} & {\color{Olive Green}c_1} \\ {\color{Orange Red}a_2} & {\color{Emerald}c_2} \end{vmatrix} ={\color{Red}a_1} \cdot {\color{Emerald}c_2}-{\color{Orange Red}a_2} \cdot {\color{Olive Green}c_1}}

   ∴$x={\frac {\vartriangle _{x}}{\vartriangle }},y={\frac {\vartriangle _{y}}{\vartriangle }}$