两个呈夹角 $\theta$ 的直线间的性质（B009）

问题

若直线

L_{1}

的表达式为

\frac{x - x_{1}}{l_{1}}

=

\frac{y - y_{1}}{m_{1}}

=

\frac{z - z_{1}}{n_{1}}

, 直线

L_{2}

的表达式为

\frac{x - x_{2}}{l_{2}}

=

\frac{y - y_{2}}{m_{2}}

=

\frac{z - z_{2}}{n_{2}}

. 此外，直线

L_{1}

和

L_{2}

的方向向量分别为

\vec{s_{1}}

=

(l_{1}, m_{1}, n_{1})

和

\vec{s_{2}}

=

(l_{2}, m_{2}, n_{2})

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

选项

[A].

\cos θ

=

\frac{| l_{1} l_{2} + m_{1} m_{2} + n_{1} n_{2} |}{\sqrt{l_{1}^{2} + m_{1}^{2} + n_{1}^{2}} \times \sqrt{l_{2}^{2} + m_{2}^{2} + n_{2}^{2}}}

[B].

\cos θ

=

\frac{| l_{1} l_{2} + m_{1} m_{2} + n_{1} n_{2} |}{\sqrt{l_{1} + m_{1} + n_{1}} \times \sqrt{l_{2} + m_{2} + n_{2}}}

[C].

\cos θ

=

\frac{l_{1} l_{2} + m_{1} m_{2} + n_{1} n_{2}}{(l_{1}^{2} + m_{1}^{2} + n_{1}^{2}) \times (l_{2}^{2} + m_{2}^{2} + n_{2}^{2})}

[D].

\cos θ

=

\frac{l_{1} l_{2} + m_{1} m_{2} + n_{1} n_{2}}{\sqrt{l_{1}^{2} + m_{1}^{2} + n_{1}^{2}} \times \sqrt{l_{2}^{2} + m_{2}^{2} + n_{2}^{2}}}

答案

$\cos θ$ $=$ $\frac{| \vec{s_{1}} \cdot \vec{s_{2}} |}{| \vec{s_{1}} | | \vec{s_{2}} |}$ $=$ $\frac{| l_{1} l_{2} + m_{1} m_{2} + n_{1} n_{2} |}{\sqrt{l_{1}^{2} + m_{1}^{2} + n_{1}^{2}} \times \sqrt{l_{2}^{2} + m_{2}^{2} + n_{2}^{2}}}$

问题

选项

相关文章：