研究目的: 試圖以另類的方法來探求兩歪斜線的公垂線段的兩端點座標。

研究過程:

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 $\overrightarrow{d_2}$, 試求 $L_1$ 與公垂線的交點 $B_1$ 及 $L_2$ 與公垂線的交點 $B_2$ 的座標。

一、 $L_2$ 與公垂線的交點 $B_2$ 的求法:

1. 過 $L_1$ 與 $B_2$ 的平面 $E_1$ 的法向量平行 $\overrightarrow{{n}_1}=\big|\overrightarrow{{d}_1}\big|^2 \overrightarrow{{d}_2}-(\overrightarrow{{d}_1}\cdot\overrightarrow{{d}_2})\overrightarrow{{d}_1}$。

證明: $\because ~~~\overrightarrow{{B_2}{B_1}}\cdot\overrightarrow{d}_1=0$ 且 $\overrightarrow{B_2B_1}\cdot\overrightarrow{d_2}=0$,

$\therefore ~~~\Big [\big|\overrightarrow{d_1}\big|^2 \overrightarrow{d_2}-(\overrightarrow{d_1}\cdot\overrightarrow{d_2})\overrightarrow{d_1}\Big ] \cdot\overrightarrow{{B_2}{B_1}}$

$=\big|\overrightarrow{d_1}\big|^2(\overrightarrow{d_2}\cdot\overrightarrow{B_2B_1})-(\overrightarrow{d_1}\cdot\overrightarrow{d_2}) (\overrightarrow{d_1}\cdot\overrightarrow{{B_2B_1}})=0$。

又 $\Big [\big|\overrightarrow{d_1}\big|^2 \overrightarrow{d_2}-(\overrightarrow{d_1}\cdot\overrightarrow{d_2})\overrightarrow{d_1}\Big ]\cdot\overrightarrow{d_1}$

$=\big|\overrightarrow{d_1}\big|^2(\overrightarrow{d_2}\cdot\overrightarrow{d_1})-(\overrightarrow{d_1}\cdot\overrightarrow{d_2})\big|\overrightarrow{d_1}\big|^2=0$。

因此 $\overrightarrow{n_1}=\big|\overrightarrow{d_1}\big|^2 \overrightarrow{d_2}-(\overrightarrow{d_1}\cdot\overrightarrow{d_2})\overrightarrow{d_1}$ 平行平面 $E_1$ 的法向量。

2. $\because$ $B_2$ 與 $A_1$ 在平面 $E_1$ 上 $\therefore \overrightarrow{A_1B_2}\cdot\overrightarrow{n_1}=0 \Rightarrow(\overrightarrow{A_2B_2}-\overrightarrow{A_2A_1})\cdot\overrightarrow{n_1}=0$。

設 $\overrightarrow{A_2B_2}=t_2\overrightarrow{d_2}$, 因此 $t_2\overrightarrow{d_2}\cdot\overrightarrow{n_1} =\overrightarrow{A_2A_1}\cdot\overrightarrow{n_1}$, \begin{eqnarray*} \hbox{解得 }~ t_2 &=& \frac{\overrightarrow{A_2A_1}\cdot\overrightarrow{n_1}}{\overrightarrow{d_2}\cdot\overrightarrow{n_1}} \\ &=& \frac{\big|\overrightarrow{d_1}\big|^2(\overrightarrow{d_2}\cdot\overrightarrow{A_2A_1}) -(\overrightarrow{d_1}\cdot\overrightarrow{d_2})(\overrightarrow{d_1}\cdot\overrightarrow{A_2A_1})} {\big|\overrightarrow{d_1}\big|^2 \big|\overrightarrow{d_2}\big|^2 - (\overrightarrow{d_1}\cdot\overrightarrow{d_2})^2} \\ &=& \frac{\left | \begin{array}{cc} \overrightarrow{d_1}\cdot\overrightarrow{d_1} & ~~~\overrightarrow{d_1}\cdot\overrightarrow{A_2A_1} \\ \overrightarrow{d_1}\cdot\overrightarrow{d_2} & ~~~\overrightarrow{d_2}\cdot\overrightarrow{A_2A_1} \end{array} \right |} {\left | \begin{array}{cc} \overrightarrow{d_1}\cdot\overrightarrow{d_1} & ~~~\overrightarrow{d_1}\cdot\overrightarrow{d_2} \\ \overrightarrow{d_1}\cdot\overrightarrow{d_2} & ~~~\overrightarrow{d_2}\cdot\overrightarrow{d_2} \end{array} \right |}\hbox{。} \end{eqnarray*}

3. 設 $\Delta=\left | \begin{array}{cc} \overrightarrow{d_1}\cdot\overrightarrow{d_1} & ~~~\overrightarrow{d_1}\cdot\overrightarrow{d_2} \\ \overrightarrow{d_1}\cdot\overrightarrow{d_2} & ~~~\overrightarrow{d_2}\cdot\overrightarrow{d_2} \end{array} \right |$ 且 $\Delta_{t_2}=\left | \begin{array}{cc} \overrightarrow{d_1}\cdot\overrightarrow{d_1} & ~~~\overrightarrow{d_1}\cdot\overrightarrow{A_2A_1} \\ \overrightarrow{d_1}\cdot\overrightarrow{d_2} & ~~~\overrightarrow{d_2}\cdot\overrightarrow{A_2A_1} \end{array} \right |$,

$\overrightarrow{OB_2}=\overrightarrow{OA_2}+\overrightarrow{A_2B_2}=\overrightarrow{OA_2}+t_2\overrightarrow{d_2}$,

$t_2=\frac{\left | \begin{array}{cc} \overrightarrow{d_1}\cdot\overrightarrow{d_1} & ~~~\overrightarrow{d_1}\cdot\overrightarrow{A_2A_1} \\ \overrightarrow{d_1}\cdot\overrightarrow{d_2} & ~~~\overrightarrow{d_2}\cdot\overrightarrow{A_2A_1} \end{array} \right |} {\left | \begin{array}{cc} \overrightarrow{d_1}\cdot\overrightarrow{d_1} & ~~~\overrightarrow{d_1}\cdot\overrightarrow{d_2} \\ \overrightarrow{d_1}\cdot\overrightarrow{d_2} & ~~~\overrightarrow{d_2}\cdot\overrightarrow{d_2} \end{array} \right |}=\displaystyle\frac{\Delta_{t_2}}{\Delta}$。

二、 $L_1$ 與公垂線的交點 $B_1$ 的求法:

4. 過 $L_2$ 與 $B_1$ 的平面 $E_2$ 的法向量平行 $\overrightarrow{n_2}=\big|\overrightarrow{d_2}\big|^2\overrightarrow{d_1} -(\overrightarrow{d_1}\cdot\overrightarrow{d_2})\overrightarrow{d_2}$。

證明: $\because$ $\overrightarrow{B_2B_1}\cdot\overrightarrow{d_1}=0$ 且 $\overrightarrow{B_2B_1}\cdot\overrightarrow{d_2}=0$,

$\therefore$ $\Big [\big|\overrightarrow{d_2}\big|^2\overrightarrow{d_1}-(\overrightarrow{d_1}\cdot\overrightarrow{d_2})\overrightarrow{d_2}\Big ] \cdot\overrightarrow{B_2B_1}$

$=\big|\overrightarrow{d_2}\big|^2(\overrightarrow{d_1}\cdot\overrightarrow{B_2B_1}) -(\overrightarrow{d_1}\cdot\overrightarrow{d_2})(\overrightarrow{d_2}\cdot\overrightarrow{B_2B_1})=0$。

又 $\Big [\big|\overrightarrow{d_2}\big|^2 \overrightarrow{d_1}-(\overrightarrow{d_1}\cdot\overrightarrow{d_2})\overrightarrow{d_2}\Big ]\cdot\overrightarrow{d_2}$

$=\big|\overrightarrow{d_2}\big|^2(\overrightarrow{d_1}\cdot\overrightarrow{d_2})-(\overrightarrow{d_1}\cdot\overrightarrow{d_2})\big|\overrightarrow{d_2}\big|^2=0$。

因此 $\overrightarrow{n_2}=\big|\overrightarrow{d_2}\big|^2 \overrightarrow{d_1}-(\overrightarrow{d_1}\cdot\overrightarrow{d_2})\overrightarrow{d_2}$ 平行平面 $E_2$ 的法向量。

5. $\because$ $B_1$ 與 $A_2$ 在平面 $E_2$ 上 $\therefore \overrightarrow{A_2B_1}\cdot\overrightarrow{n_2}=0 \Rightarrow(\overrightarrow{A_1B_1}-\overrightarrow{A_1A_2})\cdot\overrightarrow{n_2}=0$。

設 $\overrightarrow{A_1B_1}=t_1\overrightarrow{d_1}$, 因此 $t_1\overrightarrow{d_1}\cdot\overrightarrow{n_2} =\overrightarrow{A_1A_2}\cdot\overrightarrow{n_2}$, \begin{eqnarray*} \hbox{解得 }~ t_1 &=& \frac{\overrightarrow{A_1A_2}\cdot\overrightarrow{n_2}}{\overrightarrow{d_1}\cdot\overrightarrow{n_2}} \\ &=& \frac{\big|\overrightarrow{d_2}\big|^2(\overrightarrow{d_1}\cdot\overrightarrow{A_1A_2}) -(\overrightarrow{d_1}\cdot\overrightarrow{d_2})(\overrightarrow{d_2}\cdot\overrightarrow{A_1A_2})} {\big|\overrightarrow{d_1}\big|^2 \big|\overrightarrow{d_2}\big|^2 - (\overrightarrow{d_1}\cdot\overrightarrow{d_2})^2} \\ &=& \frac{\left | \begin{array}{cc} \overrightarrow{A_1A_2}\cdot\overrightarrow{d_1} & ~~~\overrightarrow{d_1}\cdot\overrightarrow{d_2} \\ \overrightarrow{A_1A_2}\cdot\overrightarrow{d_2} & ~~~\overrightarrow{d_2}\cdot\overrightarrow{d_2} \end{array} \right |} {\left | \begin{array}{cc} \overrightarrow{d_1}\cdot\overrightarrow{d_1} & ~~~\overrightarrow{d_1}\cdot\overrightarrow{d_2} \\ \overrightarrow{d_1}\cdot\overrightarrow{d_2} & ~~~\overrightarrow{d_2}\cdot\overrightarrow{d_2} \end{array} \right |}\hbox{。} \end{eqnarray*}

6. 設 $\Delta=\left | \begin{array}{cc} \overrightarrow{d_1}\cdot\overrightarrow{d_1} & ~~~\overrightarrow{d_1}\cdot\overrightarrow{d_2} \\ \overrightarrow{d_1}\cdot\overrightarrow{d_2} & ~~~\overrightarrow{d_2}\cdot\overrightarrow{d_2} \end{array} \right |$ 且 $\Delta_{t_1}=\left | \begin{array}{cc} \overrightarrow{A_1A_2}\cdot\overrightarrow{d_1} & ~~~\overrightarrow{d_1}\cdot\overrightarrow{d_2} \\ \overrightarrow{A_1A_2}\cdot\overrightarrow{d_2} & ~~~\overrightarrow{d_2}\cdot\overrightarrow{d_2} \end{array} \right |$,

$\overrightarrow{OB_1}=\overrightarrow{OA_1}+\overrightarrow{A_1B_1}=\overrightarrow{OA_1}+t_1\overrightarrow{d_1}$,

$t_1=\frac{\left | \begin{array}{cc} \overrightarrow{A_1A_2}\cdot\overrightarrow{d_1} & ~~~\overrightarrow{d_1}\cdot\overrightarrow{d_2} \\ \overrightarrow{A_1A_2}\cdot\overrightarrow{d_2} & ~~~\overrightarrow{d_2}\cdot\overrightarrow{d_2} \end{array} \right |} {\left | \begin{array}{cc} \overrightarrow{d_1}\cdot\overrightarrow{d_1} & ~~~\overrightarrow{d_1}\cdot\overrightarrow{d_2} \\ \overrightarrow{d_1}\cdot\overrightarrow{d_2} & ~~~\overrightarrow{d_2}\cdot\overrightarrow{d_2} \end{array} \right |}=\displaystyle\frac{\Delta_{t_1}}{\Delta}$。

三、 結論:

設 $\Delta=\left | \begin{array}{cc} \overrightarrow{d_1}\cdot\overrightarrow{d_1} & ~~~\overrightarrow{d_1}\cdot\overrightarrow{d_2} \\ \overrightarrow{d_1}\cdot\overrightarrow{d_2} & ~~~\overrightarrow{d_2}\cdot\overrightarrow{d_2} \end{array} \right |$, $\Delta_{t_1}=\left | \begin{array}{cc} \overrightarrow{A_1A_2}\cdot\overrightarrow{d_1} & ~~~\overrightarrow{d_1}\cdot\overrightarrow{d_2} \\ \overrightarrow{A_1A_2}\cdot\overrightarrow{d_2} & ~~~\overrightarrow{d_2}\cdot\overrightarrow{d_2} \end{array} \right |$,

$\Delta_{t_2}=\left | \begin{array}{cc} \overrightarrow{d_1}\cdot\overrightarrow{d_1} & ~~~\overrightarrow{d_1}\cdot\overrightarrow{A_2A_1} \\ \overrightarrow{d_1}\cdot\overrightarrow{d_2} & ~~~\overrightarrow{d_2}\cdot\overrightarrow{A_2A_1} \end{array} \right |$。

1. $\overrightarrow{OB_1}=\overrightarrow{OA_1}+\overrightarrow{A_1B_1}=\overrightarrow{OA_1}+t_1\overrightarrow{d_1} =\overrightarrow{OA_1}+\displaystyle\frac{\Delta_{t_1}}{\Delta}\overrightarrow{d_1}$,

2. $\overrightarrow{OB_2}=\overrightarrow{OA_2}+\overrightarrow{A_2B_2}=\overrightarrow{OA_2}+t_2\overrightarrow{d_2} =\overrightarrow{OA_2}+\displaystyle\frac{\Delta_{t_2}}{\Delta}\overrightarrow{d_2}$。

四、 實際應用:

空間二歪斜線: $L_1: \displaystyle\frac{x-11}{4}=\frac{y+5}{-3}=\frac{z+7}{-1}$, $L_2: \displaystyle\frac{x+5}{3}=\frac{y-4}{-4}=\frac{z-6}{-2}$, $\overrightarrow{d_1}=(4, -3-1)$, $\overrightarrow{d_2}=(3, -4, -2)$, $\overrightarrow{OA_1}=(11, -5, -7)$, $\overrightarrow{OA_2}=(-5, 4, 6)$, $O$ 為空間直角坐標系的原點, $\overrightarrow{A_1A_2}=(-16, 9, 13)$。試求:

(1) $L_1$ 與公垂線的交點 $B_1$。

(2) $L_2$ 與公垂線的交點 $B_2$。

解: $\Delta=\left | \begin{array}{cc} 26 & ~26 \\ 26 & ~29 \end{array} \right |$, $\Delta_{t_1}=\left | \begin{array}{cc} (-16, 9, 13)\cdot(4, -3, -1) & ~~~26 \\ (-16, 9, 13)\cdot(3, -4, -2) & ~~~29 \end{array} \right |$,

$\Delta_{t_2}=\left | \begin{array}{cc} 26 & ~(4, -3, -1)\cdot(16, -9, -13) \\ 26 & ~(3, -4, -2)\cdot(16, -9, -13) \end{array} \right |$

$t_1=\displaystyle\frac{\left | \begin{array}{cc} -104 & ~26 \\ -110 & ~29 \end{array} \right |} {\left | \begin{array}{cc} 26 & ~26 \\ 26 & ~29 \end{array} \right |} =\frac{\left | \begin{array}{cc} -52 & ~26 \\ -52 & ~29 \end{array} \right |} {\left | \begin{array}{cc} 26 & ~26 \\ 26 & ~29 \end{array} \right |} =-2$, $t_2=\displaystyle\frac{\left | \begin{array}{cc} 26 & ~104 \\ 26 & ~110 \end{array} \right |} {\left | \begin{array}{cc} 26 & ~26 \\ 26 & ~29 \end{array} \right |} =\frac{\left | \begin{array}{cc} 26 & ~52 \\ 26 & ~58 \end{array} \right |} {\left | \begin{array}{cc} 26 & ~26 \\ 26 & ~29 \end{array} \right |} =2$,

$\overrightarrow{OB_1}=(11, -5, -7)+ (-2)\cdot(4, -3, -1)=(3, 1, -5)$,

$\overrightarrow{OB_2}=(-5, 4, 6)+2\cdot(3, -4, -2)=(1, -4, 2)$。

---本文作者任教國立宜蘭高中---