延續投稿在龍騰數亦優第9刊主題為 "利用正射影及外積的概念求兩歪斜線的公垂線段的距離及兩端點座標"的結論 $\hbox{"}{\overrightarrow {OB_2}}\!=\!{\overrightarrow {OA_2}} \!+\!\displaystyle\frac{{\overrightarrow {A_2A_1}} \cdot[\overrightarrow {d_1}\!\times\! (\overrightarrow {d_1} \!\times\! \overrightarrow {d_2})]}{\overrightarrow {d_2}\cdot[\overrightarrow {d_1}\!\times\! (\overrightarrow {d_1} \!\times\! \overrightarrow {d_2})]}\overrightarrow {d_2}\hbox{"}$, 利用同樣的推理

方法求得 $\hbox{"}{\overrightarrow {OB_1}}={\overrightarrow {OA_1}} +\displaystyle\frac{{\overrightarrow {A_1A_2}} \cdot[\overrightarrow {d_2}\times (\overrightarrow {d_1} \times \overrightarrow {d_2})]}{\overrightarrow {d_1}\cdot[\overrightarrow {d_2}\times (\overrightarrow {d_1} \times \overrightarrow {d_2})]}\overrightarrow {d_1}\hbox{''}$, 並簡化向量三重積 $\overrightarrow {d_1}\times (\overrightarrow {d_1} \times \overrightarrow {d_2})$ 及 $\overrightarrow {d_2}\times (\overrightarrow {d_1} \times \overrightarrow {d_2})$ 求得兩歪斜線的公垂線段的兩端點座標的公式解法。

一、 利用同樣的推理方法求得${\overrightarrow {OB_1}}={\overrightarrow {OA_1}} +\displaystyle\frac{{\overrightarrow {A_1A_2}} \cdot[\overrightarrow {d_2}\times (\overrightarrow {d_1} \times \overrightarrow {d_2})]}{\overrightarrow {d_1}\cdot[\overrightarrow {d_2}\times (\overrightarrow {d_1} \times \overrightarrow {d_2})]}\overrightarrow {d_1}$

如上圖所示，已知空間直角座標系中， $O$ 為原點，兩歪斜線 $L_1$ 與 $L_2$ 分別通過點 $A_1$、點 $A_2$， $L_1$ 與 $L_2$ 的方向向量分別為 $\overrightarrow {d_1}$ 與 $\overrightarrow {d_2}$， 四邊形 $A_2A''_2B_1B_2$ 為矩形， $\angle A_1A''_1 B_1=90^\circ$, $\overline {B_1B_2} \bot$ $L_1$ 與 $L_2$，試求點 $B_1$ 的座標。

解： $\because$ ${\overrightarrow {A_1A''_1}}\bot {\overrightarrow {B_1A_2}} \Rightarrow {\overrightarrow {A_1A''_1}}\cdot {\overrightarrow {B_1A_2}}=0$, 又 ${\overrightarrow {A_1A''_1}}\|\overrightarrow {d_2}\times (\overrightarrow {d_1} \times \overrightarrow {d_2})$ 且 ${\overrightarrow {B_1A_2}}={\overrightarrow {A_1A_2}}-{\overrightarrow {A_1B_1}} ={\overrightarrow {A_1A_2}}-t\overrightarrow d_1$, $${\overrightarrow {A_1A''_1}}\cdot {\overrightarrow {B_1A_2}}=0\Rightarrow[\overrightarrow {d_2}\times (\overrightarrow {d_1} \times \overrightarrow {d_2})]\cdot({\overrightarrow {A_1A_2}}-t\overrightarrow {d_1})=0,$$

解得 $t=\displaystyle\frac{{\overrightarrow {A_1A_2}}\cdot[\overrightarrow {d_2}\times (\overrightarrow {d_1} \times \overrightarrow {d_2})]}{\overrightarrow {d_1}\cdot[\overrightarrow {d_2}\times (\overrightarrow {d_1} \times \overrightarrow {d_2})]}$

因此 $$ {\overrightarrow {OB_1}}={\overrightarrow {OA_1}}+t\overrightarrow {d_1}={\overrightarrow {OA_1}}+\displaystyle\frac{{\overrightarrow {A_1A_2}}\cdot[\overrightarrow {d_2}\times (\overrightarrow {d_1} \times \overrightarrow {d_2})]}{\overrightarrow {d_1}\cdot[\overrightarrow {d_2}\times (\overrightarrow {d_1} \times \overrightarrow {d_2})]}\overrightarrow {d_1}\hbox{。} $$

二、已知 $\overrightarrow \alpha=(l_1,m_1,n_1)$, $\overrightarrow \beta=(l_2,m_2,n_2)$, $\overrightarrow \gamma=(l_3,m_3,n_3)$, 則 $\overrightarrow\alpha\times (\overrightarrow\beta\times \overrightarrow\gamma)=(\overrightarrow\alpha\cdot \overrightarrow\gamma)\overrightarrow\beta-(\overrightarrow\alpha\cdot \overrightarrow \beta) \overrightarrow\gamma$。

證明 : $\because\ \overrightarrow\beta\times \overrightarrow\gamma=(m_2n_3-m_3n_2,n_2l_3-n_3l_2,l_2m_3-l_3m_2)$, \begin{eqnarray*} &&\hskip -15pt \therefore\, \overrightarrow\alpha\times (\overrightarrow\beta\times \overrightarrow\gamma)\\ &=&(m_1(l_2m_3-l_3m_2)-n_1(n_2l_3-n_3l_2),n_1(m_2n_3-m_3n_2)-l_1(l_2m_3-l_3m_2),\\ &&l_1(n_2l_3-n_3l_2)-m_1(m_2n_3-m_3n_2))\\ &=&((\overrightarrow\alpha\cdot \overrightarrow\gamma)l_2\!-\!(\overrightarrow\alpha\cdot \overrightarrow\beta)l_3, (\overrightarrow\alpha\cdot \overrightarrow\gamma)m_2\!-\!(\overrightarrow\alpha\cdot \overrightarrow\beta)m_3,(\overrightarrow\alpha\cdot \overrightarrow\gamma)n_2\!-\!(\overrightarrow\alpha\cdot \overrightarrow\beta)n_3)\quad~\\ &=&(\overrightarrow\alpha\cdot \overrightarrow\gamma)(l_2,m_2,n_2)-(\overrightarrow\alpha\cdot \overrightarrow\beta)(l_3,m_3,n_3)\\ &=&(\overrightarrow\alpha\cdot \overrightarrow\gamma)\overrightarrow\beta-(\overrightarrow\alpha\cdot \overrightarrow\beta)\overrightarrow\gamma, \ \hbox{得證。} \end{eqnarray*}

三、 利用二的恆等式 $\overrightarrow\alpha\times (\overrightarrow\beta\times \overrightarrow\gamma)=(\overrightarrow\alpha\cdot \overrightarrow\gamma)\overrightarrow\beta-(\overrightarrow\alpha\cdot \overrightarrow \beta) \overrightarrow\gamma$, \begin{eqnarray*} \hbox{得}\ \overrightarrow {d_1}\times (\overrightarrow {d_1}\times \overrightarrow {d_2})&=&[(\overrightarrow {d_1}\cdot \overrightarrow {d_2})\overrightarrow {d_1}-(\overrightarrow {d_1}\cdot \overrightarrow {d_1})\overrightarrow {d_2}]=-[(\overrightarrow {d_1}\cdot \overrightarrow {d_1})\overrightarrow {d_2}-(\overrightarrow {d_1}\cdot \overrightarrow {d_2})\overrightarrow {d_1}],\\ \overrightarrow {d_2}\times (\overrightarrow {d_1}\times \overrightarrow {d_2})&=&[(\overrightarrow {d_2}\cdot \overrightarrow {d_2})\overrightarrow {d_1}-(\overrightarrow {d_2}\cdot \overrightarrow {d_1})\overrightarrow {d_2}]\hbox{。} %=[(\overrightarrow {d_2}\cdot\overrightarrow {d_2})\overrightarrow {d_1}-(\overrightarrow {d_1}\cdot \overrightarrow {d_2})\overrightarrow {d_2}] \end{eqnarray*}

四、 利用三的等式，向量三重積 $\overrightarrow {d_1}\times (\overrightarrow {d_1}\times \overrightarrow {d_2})=-[(\overrightarrow {d_1}\cdot \overrightarrow {d_1})\overrightarrow {d_2}-(\overrightarrow {d_1}\cdot \overrightarrow {d_2})\overrightarrow {d_1}]$,

將 $\displaystyle\frac{{\overrightarrow {A_2A_1}}\cdot[\overrightarrow {d_1}\times (\overrightarrow {d_1} \times \overrightarrow {d_2})]}{\overrightarrow {d_2}\cdot[\overrightarrow {d_1}\times (\overrightarrow {d_1} \times \overrightarrow {d_2})]}$ 化簡為 $$\displaystyle\frac{{\overrightarrow {A_2A_1}}\cdot-[(\overrightarrow {d_1}\cdot \overrightarrow {d_1}) \overrightarrow {d_2}-(\overrightarrow {d_1}\cdot \overrightarrow {d_2}) \overrightarrow {d_1}]}{\overrightarrow {d_2}\cdot-[(\overrightarrow {d_1}\cdot \overrightarrow {d_1}) \overrightarrow {d_2}-(\overrightarrow {d_1}\cdot \overrightarrow {d_2}) \overrightarrow {d_1}]}=\displaystyle\frac{(\overrightarrow {d_1}\cdot \overrightarrow {d_1}) (\overrightarrow {d_2}\cdot {\overrightarrow {A_2A_1}})-(\overrightarrow {d_1}\cdot \overrightarrow {d_2}) (\overrightarrow {d_1}\cdot {\overrightarrow {A_2A_1}})}{(\overrightarrow {d_1}\cdot \overrightarrow {d_1})(\overrightarrow {d_2}\cdot \overrightarrow {d_2})-(\overrightarrow {d_1}\cdot \overrightarrow {d_2})(\overrightarrow {d_1}\cdot \overrightarrow {d_2})}$$ \begin{eqnarray*} \hbox{因此}\ {\overrightarrow {OB_2}}&=&{\overrightarrow {OA_2}}+\displaystyle\frac{{\overrightarrow {A_2A_1}}\cdot[\overrightarrow {d_1}\times (\overrightarrow {d_1} \times \overrightarrow {d_2})]}{\overrightarrow {d_2}\cdot[\overrightarrow {d_1}\times (\overrightarrow {d_1} \times \overrightarrow {d_2})]}\overrightarrow {d_2}\\ &=&{\overrightarrow {OA_2}}+\displaystyle\frac{(\overrightarrow {d_1}\cdot \overrightarrow {d_1}) (\overrightarrow {d_2}\cdot {\overrightarrow {A_2A_1}})-(\overrightarrow {d_1}\cdot \overrightarrow {d_2}) (\overrightarrow {d_1}\cdot {\overrightarrow {A_2A_1}})}{(\overrightarrow {d_1}\cdot \overrightarrow {d_1})(\overrightarrow {d_2}\cdot \overrightarrow {d_2})-(\overrightarrow {d_1}\cdot \overrightarrow {d_2})(\overrightarrow {d_1}\cdot \overrightarrow {d_2})}\overrightarrow {d_2}\hbox{。}\end{eqnarray*}

五、 利用三的等式，向量三重積 $\overrightarrow {d_2}\times (\overrightarrow {d_1}\times \overrightarrow {d_2})=[(\overrightarrow {d_2}\cdot \overrightarrow {d_2})\overrightarrow {d_1}-(\overrightarrow {d_1}\cdot \overrightarrow {d_2})\overrightarrow {d_2}]$,

將 $\displaystyle\frac{{\overrightarrow {A_1A_2}}\cdot[\overrightarrow {d_2}\times (\overrightarrow {d_1} \times \overrightarrow {d_2})]}{\overrightarrow {d_1}\cdot[\overrightarrow {d_2}\times (\overrightarrow {d_1} \times \overrightarrow {d_2})]}$ 化簡為 $$\displaystyle\frac{{\overrightarrow {A_1A_2}}\cdot[(\overrightarrow {d_2}\cdot \overrightarrow {d_2}) \overrightarrow {d_1}-(\overrightarrow {d_1}\cdot \overrightarrow {d_2}) \overrightarrow {d_2}]}{\overrightarrow {d_1}\cdot[(\overrightarrow {d_2}\cdot \overrightarrow {d_2}) \overrightarrow {d_1}-(\overrightarrow {d_1}\cdot \overrightarrow {d_2}) \overrightarrow {d_2}]} =\displaystyle\frac{({\overrightarrow {A_1A_2}} \cdot \overrightarrow {d_1}) (\overrightarrow {d_2}\cdot \overrightarrow {d_2})-({\overrightarrow {A_1A_2}}\cdot \overrightarrow {d_2}) (\overrightarrow {d_1}\cdot \overrightarrow {d_2})}{(\overrightarrow {d_1}\cdot \overrightarrow {d_1})(\overrightarrow {d_2}\cdot \overrightarrow {d_2})-(\overrightarrow {d_1}\cdot \overrightarrow {d_2})(\overrightarrow {d_1}\cdot \overrightarrow {d_2})}$$ \begin{eqnarray*} \hbox{因此}\ {\overrightarrow {OB_1}}&=&{\overrightarrow {OA_1}}+\displaystyle\frac{{\overrightarrow {A_1A_2}}\cdot[\overrightarrow {d_2}\times (\overrightarrow {d_1} \times \overrightarrow {d_2})]}{\overrightarrow {d_1}\cdot[\overrightarrow {d_2}\times (\overrightarrow {d_1} \times \overrightarrow {d_2})]}\overrightarrow {d_1}\\ &=&{\overrightarrow {OA_1}}+\displaystyle\frac{({\overrightarrow {A_1A_2}}\cdot \overrightarrow {d_1}) (\overrightarrow {d_2}\cdot \overrightarrow {d_2})-({\overrightarrow {A_1A_2}}\cdot \overrightarrow {d_2}) (\overrightarrow {d_1}\cdot \overrightarrow {d_2})}{(\overrightarrow {d_1}\cdot \overrightarrow {d_1})(\overrightarrow {d_2}\cdot \overrightarrow {d_2})-(\overrightarrow {d_1}\cdot \overrightarrow {d_2})(\overrightarrow {d_1}\cdot \overrightarrow {d_2})}\overrightarrow {d_1}\hbox{。}\end{eqnarray*}

六、 結論 \begin{eqnarray*} {\overrightarrow {OB_1}}&=&{\overrightarrow {OA_1}}+\displaystyle\frac{({\overrightarrow {A_1A_2}}\cdot \overrightarrow {d_1}) (\overrightarrow {d_2}\cdot \overrightarrow {d_2})-({\overrightarrow {A_1A_2}}\cdot \overrightarrow {d_2}) (\overrightarrow {d_1}\cdot \overrightarrow {d_2})}{(\overrightarrow {d_1}\cdot \overrightarrow {d_1})(\overrightarrow {d_2}\cdot \overrightarrow {d_2})-(\overrightarrow {d_1}\cdot \overrightarrow {d_2})(\overrightarrow {d_1}\cdot \overrightarrow {d_2})}\overrightarrow {d_1}\hbox{。}\\ {\overrightarrow {OB_2}}&=&{\overrightarrow {OA_2}}+\displaystyle\frac{(\overrightarrow {d_1}\cdot \overrightarrow {d_1}) (\overrightarrow {d_2}\cdot {\overrightarrow {A_2A_1}})-(\overrightarrow {d_1}\cdot \overrightarrow {d_2}) (\overrightarrow {d_1}\cdot {\overrightarrow {A_2A_1}})}{(\overrightarrow {d_1}\cdot \overrightarrow {d_1})(\overrightarrow {d_2}\cdot \overrightarrow {d_2})-(\overrightarrow {d_1}\cdot \overrightarrow {d_2})(\overrightarrow {d_1}\cdot \overrightarrow {d_2})}\overrightarrow {d_2} \hbox{。}\end{eqnarray*}

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