平面束的方程不包含平面(2)的证明

设直线$L$由方程组

\[\left\{ \begin{gathered} {A_1}x + {B_1}x + {C_1}x + {D_1} = 0\begin{array}{*{20}{c}} {}&{} \end{array}\left( 1 \right) \hfill \\ {A_2}x + {B_2}x + {C_2}x + {D_2} = 0\begin{array}{*{20}{c}} {}&{} \end{array}\left( 2 \right) \hfill \\ \end{gathered} \right.\]

所确定，其中系数${A_1}$、${B_1}$、${B_1}$与${A_2}$、${B_2}$、${B_2}$不成比例.（若成比例，两平面将平行或重合，从而不会相交形成直线$L$.）建立三元一次方程

$$\eqalign{ & {A_1}x + {B_1}x + {C_1}x + {D_1} + \cr & \lambda ({A_2}x + {B_2}x + {C_2}x + {D_2}) = 0\mathop {}\limits^{} \left( 3 \right) \cr} $$

其中$\lambda $为任意常数. ${A_1}$、${B_1}$、${B_1}$与${A_2}$、${B_2}$、${B_2}$不成比例，所以对于任何一个$\lambda $，方程（3）的系数${A_1} + \lambda {A_2},{B_1} + \lambda {B_2},{C_1} + \lambda {C_2}$不全为零.成比例时，有

$$\frac{{{A_1}}}{{{A_2}}} = \frac{{{B_1}}}{{{B_2}}} = \frac{{{C_1}}}{{{C_2}}} = k$$

则

$$\eqalign{ & {A_1} = k{A_2},{B_1} = k{B_2},{C_1} = k{C_2} \cr & {A_1} + \lambda {A_2} = {B_1} + \lambda {B_2} \cr & = {C_1} + \lambda {C_2} = 0 \cr & k{A_2} + \lambda {A_2} = k{B_2} + \lambda {B_2} \cr & = k{C_2} + \lambda {C_2} = 0 \cr} $$

$k = - \lambda $时，方程（3）的系数全为零.

方程（3）的系数${A_1} + \lambda {A_2},{B_1} + \lambda {B_2},{C_1} + \lambda {C_2}$不全为零时，方程（3）表示一个平面，若一点在直线$L$上，则点的坐标必定同时满足方程（1），（2），（3），故方程（3）表示过直线$L$的平面，且对应于不同的$\lambda $值，方程（3）表示通过直线$L$的不同平面.反之，通过直线$L$的任何平面（除了平面（2）外）都包含在方程（3）表示的一族平面内.

为什么不包括平面（2）呢？

假设包含平面（2），则

\[\begin{gathered} {A_1}x + {B_1}y + {C_1}z + {D_1} \hfill \\ + \lambda \left( {{A_2}x + {B_2}y + {C_2}z + {D_2}} \right) = 0 \hfill \\ {A_1}x + {B_1}y + {C_1}z + {D_1} \hfill \\ + \lambda {A_2}x + \lambda {B_2}y + \lambda {C_2}z + \lambda {D_2} = 0 \hfill \\ \left( {{A_1} + \lambda {A_2}} \right)x + \left( {{B_1} + \lambda {B_2}} \right)y \hfill \\ + \left( {{C_1} + \lambda {C_2}} \right)z + \left( {{D_1} + \lambda {D_2}} \right) = 0 \hfill \\ \left\{ \begin{gathered} {A_1} + \lambda {A_2} = {A_2} \hfill \\ {B_1} + \lambda {B_2} = {B_2} \hfill \\ {C_1} + \lambda {C_2} = {C_2} \hfill \\ {D_1} + \lambda {D_2} = {D_2} \hfill \\ \end{gathered} \right. \hfill \\ \Rightarrow \left\{ \begin{gathered} \frac{{{A_1}}}{{{A_2}}} = 1 - \lambda \hfill \\ \frac{{{B_1}}}{{{B_2}}} = 1 - \lambda \hfill \\ \frac{{{C_1}}}{{{C_2}}} = 1 - \lambda \hfill \\ \frac{{{D_1}}}{{{D_2}}} = 1 - \lambda \hfill \\ \end{gathered} \right. \hfill \\ \Rightarrow \frac{{{A_1}}}{{{A_2}}} = \frac{{{B_1}}}{{{B_2}}} = \frac{{{C_1}}}{{{C_2}}} = \frac{{{D_1}}}{{{D_2}}} \hfill \\ = 1 - \lambda \hfill \\ \end{gathered} \]

这与系数不成比例矛盾，故平面束不包含平面（2）.