空间直线方程的一般式/交面式（B009）

问题

若两个平面的法向量分别为

\vec{n_{1}}

=

(A_{1}, B_{1}, C_{1})

\vec{n_{2}}

=

(A_{2}, B_{2}, C_{2})

, 此外还有常数

D_{1}

和

D_{2}

, 则这两个平面相交所形成的直线如何表示？

[A].

{\begin{matrix} \frac{A_{1}}{x} + \frac{B_{1}}{y} + \frac{C_{1}}{z} + D_{1} = 0, \\ \frac{A_{2}}{x} + \frac{B_{2}}{y} + \frac{C_{2}}{z} + D_{2} = 0. \end{matrix}

[B].

{\begin{matrix} \frac{1}{A_{1}} x + \frac{1}{B_{1}} y + \frac{1}{C_{1}} z + \frac{1}{D_{1}} = 0, \\ \frac{1}{A_{2}} x + \frac{1}{B_{2}} y + \frac{1}{C_{2}} z + \frac{1}{D_{2}} = 0. \end{matrix}

[C].

{\begin{matrix} A_{1} x + B_{1} y + C_{1} z = 0, \\ A_{2} x + B_{2} y + C_{2} z = 0. \end{matrix}

[D].

{\begin{matrix} A_{1} x + B_{1} y + C_{1} z + D_{1} = 0, \\ A_{2} x + B_{2} y + C_{2} z + D_{2} = 0. \end{matrix}

${\begin{matrix} A_{1} x + B_{1} y + C_{1} z + D_{1} = 0, \\ A_{2} x + B_{2} y + C_{2} z + D_{2} = 0. \end{matrix}$