# 问题及解答

## 三维空间中两条异面直线, 求其公垂线

Posted by haifeng on 2016-09-01 10:38:48 last update 2016-09-01 10:38:48 | Edit | Answers (1)

(1) 证明: 存在唯一的点 $P\in L_1$ 和 $Q\in L_2$, 使得两点连线 $PQ$ 同时垂直于 $L_1$ 和 $L_2$.

(2) 求 $P$ 和 $Q$ 的坐标(用 $a,b,v,w$ 表示).

1

Posted by haifeng on 2016-09-07 09:44:50

$L_1$ 与 $L_2$ 的方程如下

$L_1:\ \{a+vt\mid t\in\mathbb{R}\},\quad L_2:\ \{b+ws\mid s\in\mathbb{R}\}.$

$\begin{cases} \langle(b-a)+ws-vt,\ v\rangle=0,\\ \langle(b-a)+ws-vt,\ w\rangle=0.\\ \end{cases}$

$\begin{cases} \langle w, v\rangle s-|v|^2 t=\langle(a-b),\ v\rangle,\\ |w|^2 s-t\langle v, w\rangle=\langle(a-b),\ w\rangle.\\ \end{cases}$

$D=\begin{vmatrix} \langle w, v\rangle & -|v|^2\\ |w|^2 & -\langle v, w\rangle \end{vmatrix}=-\langle v, w\rangle^2+|v|^2\cdot |w|^2$

$s=\frac{1}{D}\begin{vmatrix} \langle (a-b),v\rangle & -|v|^2\\ \langle (a-b),w\rangle & -\langle v,w\rangle\\ \end{vmatrix}=\frac{|v|^2\langle a-b,w\rangle-\langle v,w\rangle\langle a-b,v\rangle}{|v|^2 |w|^2 -\langle v,w\rangle^2}.$

$t=\frac{1}{D}\begin{vmatrix} \langle v,w\rangle & \langle (a-b),v\rangle \\ |w|^2 & \langle (a-b),w\rangle \\ \end{vmatrix}=\frac{\langle v, w\rangle\langle a-b,w\rangle-|w|^2 \langle a-b,v\rangle}{|v|^2 |w|^2 -\langle v,w\rangle^2}.$

\begin{aligned} s=\frac{\langle a-b,w\rangle-\langle v,w\rangle\langle a-b,v\rangle}{1-\langle v,w\rangle^2},\\ t=\frac{\langle v,w\rangle\langle a-b,w\rangle-\langle a-b,v\rangle}{1-\langle v,w\rangle^2}.\\ \end{aligned}

\begin{aligned} P=a+vt=a+\frac{\langle b-a,\ v-\langle v,w\rangle w\rangle}{1-\langle v,w\rangle^2}v,\\ Q=b+ws=b+\frac{\langle a-b,\ w-\langle v,w\rangle v\rangle}{1-\langle v,w\rangle^2}w. \end{aligned}