33405 利用平面的法向量來求兩歪斜線的公垂線段的兩端點座標

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### 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 $\overrightarrow{{n}_1}=\big|\overrightarrow{{d}_1}\big|^2 \overrightarrow{{d}_2}-(\overrightarrow{{d}_1}\cdot\overrightarrow{{d}_2})\overrightarrow{{d}_1}$。

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

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

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

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

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 $\overrightarrow{n_2}=\big|\overrightarrow{d_2}\big|^2\overrightarrow{d_1} -(\overrightarrow{d_1}\cdot\overrightarrow{d_2})\overrightarrow{d_2}$。

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

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

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

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

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:

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

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:

(1) $L_1$ 與公垂線的交點 $B_1$。

(2) $L_2$ 與公垂線的交點 $B_2$。

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

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

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

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

---本文作者任教國立宜蘭高中---

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