两个呈夹角 $\theta$ 的平面间的性质（B009）

问题

若平面

π_{1}

的表达式为

A_{1} x

+

B_{1} y

+

C_{1} z

+

D_{1}

=

0

, 平面

π_{2}

的表达式为

A_{2} x

+

B_{2} y

+

C_{2} z

+

D_{2}

=

0

. 此外，平面

π_{1}

和

π_{2}

的法向量分别为

\vec{n_{1}}

=

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

和

\vec{n_{2}}

=

(A_{1}, B_{2}, C_{2})

那么，若 $π_{1}$ 与 $π_{2}$ 之间的夹角为 $θ$ $(0 ⩽ θ ⩽ \frac{π}{2})$ , 则 $\cos θ$ $=$ $?$

选项

[A].

\cos θ

=

\frac{| A_{1} A_{2} + B_{1} B_{2} + C_{1} C_{2} |}{(A_{1}^{2} + B_{1}^{2} + C_{1}^{2}) \times (A_{2}^{2} + B_{2}^{2} + C_{2}^{2})}

[B].

\cos θ

=

\frac{A_{1} A_{2} + B_{1} B_{2} + C_{1} C_{2}}{\sqrt{A_{1}^{2} + B_{1}^{2} + C_{1}^{2}} \times \sqrt{A_{2}^{2} + B_{2}^{2} + C_{2}^{2}}}

[C].

\cos θ

=

\frac{| A_{1} A_{2} + B_{1} B_{2} + C_{1} C_{2} |}{\sqrt{A_{1}^{2} + B_{1}^{2} + C_{1}^{2}} \times \sqrt{A_{2}^{2} + B_{2}^{2} + C_{2}^{2}}}

[D].

\cos θ

=

\frac{| A_{1} A_{2} - B_{1} B_{2} - C_{1} C_{2} |}{(A_{1}^{2} + B_{1}^{2} + C_{1}^{2}) \times (A_{2}^{2} + B_{2}^{2} + C_{2}^{2})}

答案

$\cos θ$ $=$ $\frac{| \vec{n_{1}} \cdot \vec{n_{2}} |}{| \vec{n_{1}} | | \vec{n_{2}} |}$ $=$ $\frac{| A_{1} A_{2} + B_{1} B_{2} + C_{1} C_{2} |}{\sqrt{A_{1}^{2} + B_{1}^{2} + C_{1}^{2}} \times \sqrt{A_{2}^{2} + B_{2}^{2} + C_{2}^{2}}}$

问题

选项

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