直纹面

证明单叶双曲面的同族的两条直母线异面，且同族的任意三条直母线都不平行于同一平面

第一次玩这种东东，完全不熟行的说，看着显然，就是不会证……

搞第一个先，单叶双曲面为
\[\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}-\frac{z^{2}}{c^{2}}=1,\]
其中 $a,b,c>0$。设其上某直母线 $L_1$ 为
\[\left\{\begin{aligned}
\lambda _{1}\left( \frac{x}{a}+\frac{z}{c} \right)&=\lambda _{2}\left( 1+\frac{y}{b} \right), \\
\lambda _{2}\left( \frac{x}{a}-\frac{z}{c} \right)&=\lambda _{1}\left( 1-\frac{y}{b} \right),
\end{aligned}\right.\]
其中 $\lambda_1,\lambda_2$ 不同为 0。不妨设 $\lambda_2\ne0$ 时另一同族直母线 $L_2$ 为
\[\left\{\begin{aligned}
\lambda _{3}\left( \frac{x}{a}+\frac{z}{c} \right)&=\lambda _{2}\left( 1+\frac{y}{b} \right), \\
\lambda _{2}\left( \frac{x}{a}-\frac{z}{c} \right)&=\lambda _{3}\left( 1-\frac{y}{b} \right),
\end{aligned}\right.\]
其中 $\lambda_3\ne\lambda_1$。
令 $z=0$ 可以求出 $L_1$ 过点 $\displaystyle P_1\left(\frac{2 a \lambda_1\lambda_2}{\lambda_1^2+\lambda_2^2},\frac{b(\lambda_1^2-\lambda_2^2)}{\lambda_1^2+\lambda_2^2},0\right)$，$L_2$ 过点 $\displaystyle P_2\left(\frac{2 a \lambda_3\lambda_2}{\lambda_3^2+\lambda_2^2},\frac{b(\lambda_3^2-\lambda_2^2)}{\lambda_3^2+\lambda_2^2},0\right)$，又 $L_1, L_2$ 的方向向量分别为
\begin{align*}
\overrightarrow{n_{1}}&=\left( \frac{\lambda _{1}}{a},-\frac{\lambda _{2}}{b},\frac{\lambda _{1}}{c} \right)\times \left( \frac{\lambda _{2}}{a},\frac{\lambda _{1}}{b},-\frac{\lambda _{2}}{c} \right),\\
\overrightarrow{n_{2}}&=\left( \frac{\lambda _{3}}{a},-\frac{\lambda _{2}}{b},\frac{\lambda _{3}}{c} \right)\times \left( \frac{\lambda _{2}}{a},\frac{\lambda _{3}}{b},-\frac{\lambda _{2}}{c} \right),
\end{align*}
而直线异面的充要条件是 $\overrightarrow{P_{1}P_{2}}\cdot \left( \overrightarrow{n_{1}}\times \overrightarrow{n_{2}} \right)\ne 0$，代入上面的数据可知
\[\overrightarrow{P_{1}P_{2}}\cdot \left( \overrightarrow{n_{1}}\times \overrightarrow{n_{2}} \right)=-\frac{4\lambda_2^2(\lambda _1-\lambda _3)^2}{abc},\]
的确不为 0。

续第二个

再设有另一直线 $L_3$ 为
\[\left\{\begin{aligned}
\lambda _{4}\left( \frac{x}{a}+\frac{z}{c} \right)&=\lambda _{2}\left( 1+\frac{y}{b} \right), \\
\lambda _{2}\left( \frac{x}{a}-\frac{z}{c} \right)&=\lambda _{4}\left( 1-\frac{y}{b} \right),
\end{aligned}\right.\]
其中 $\lambda_1,\lambda_3,\lambda_4$ 互不相等。
这样，三条直线的方向向量分别为
\begin{align*}
\overrightarrow{n_{1}}&=\left( \frac{\lambda _{1}}{a},-\frac{\lambda _{2}}{b},\frac{\lambda _{1}}{c} \right)\times \left( \frac{\lambda _{2}}{a},\frac{\lambda _{1}}{b},-\frac{\lambda _{2}}{c} \right),\\
\overrightarrow{n_{2}}&=\left( \frac{\lambda _{3}}{a},-\frac{\lambda _{2}}{b},\frac{\lambda _{3}}{c} \right)\times \left( \frac{\lambda _{2}}{a},\frac{\lambda _{3}}{b},-\frac{\lambda _{2}}{c} \right), \\
\overrightarrow{n_{3}}&=\left( \frac{\lambda _{4}}{a},-\frac{\lambda _{2}}{b},\frac{\lambda _{4}}{c} \right)\times \left( \frac{\lambda _{2}}{a},\frac{\lambda _{4}}{b},-\frac{\lambda _{2}}{c} \right),
\end{align*}
而 $L_1,L_2,L_3$ 不平行于同一平面等价于 $(\overrightarrow{n_1}\times\overrightarrow{n_2})\times(\overrightarrow{n_1}\times\overrightarrow{n_3})\ne \vec0$，代入以上数据，为
\[(\overrightarrow{n_1}\times\overrightarrow{n_2})\times(\overrightarrow{n_1}\times\overrightarrow{n_3})=\left( \frac{4\lambda _{2}^{3}(\lambda _{1}^{2}-\lambda _{2}^{2})M}{a^{2}b^{3}c^{3}},-\frac{8\lambda _{1}\lambda _{2}^{4}M}{a^{3}b^{2}c^{3}},-\frac{4\lambda _{2}^{3}(\lambda _{1}^{2}+\lambda _{2}^{2})M}{a^{3}b^{3}c^{2}} \right),\]
其中 $M=(\lambda _{1}-\lambda _{3})(\lambda _{1}-\lambda _{4})(\lambda _{3}-\lambda _{4})\ne 0$，可见显然不为 $\vec0$。