三维直角坐标系上点到直线的距离(B009)

问题

在三维直角坐标系中,已知点 $M_{1}$ $(x_{1}, y_{1}, z_{1})$ 为直线 $Ax$ $+$ $By$ $+$ $Cz$ $+$ $D$ $=$ $0$ 上的任意一点,向量 $\vec{s}$ $=$ $(l, m, n)$ 为该直线的方向向量,则点 $M_{0}$ $(x_{0}, y_{0}, z_{0})$ 到该直线的距离 $d$ $=$ $?$

选项

[A].   $d$ $=$ $\frac{\left(x_{1}-x_{0}, y_{1}-y_{0}, z_{1}-z_{0}\right) \times(l, m, n)}{\sqrt{l^{2}+m^{2}+n^{2}}}$

[B].   $d$ $=$ $\frac{\left|\left(x_{1}-x_{0}, y_{1}-y_{0}, z_{1}-z_{0}\right) \times(l, m, n)\right|}{\sqrt{l^{2}+m^{2}+n^{2}}}$

[C].   $d$ $=$ $\frac{\left|\left(x_{1}-x_{0}, y_{1}-y_{0}, z_{1}-z_{0}\right) \times(l, m, n)\right|}{\sqrt{l+m+n}}$

[D].   $d$ $=$ $\frac{\left|\left(x_{1}-x_{0}, y_{1}-y_{0}, z_{1}-z_{0}\right) \times(l, m, n)\right|}{l^{2}+m^{2}+n^{2}}$



显示答案

$d$ $=$ $\frac{\left|\overrightarrow{M_{\textcolor{orange}{0}} M_{\textcolor{cyan}{1}}} \times \textcolor{red}{\vec{s}} \right|}{|\textcolor{red}{\vec{s}}|}$ $=$ $\frac{\left|\left(x_{\textcolor{cyan}{1}}-x_{\textcolor{orange}{0}}, y_{\textcolor{cyan}{1}}-y_{\textcolor{orange}{0}}, z_{\textcolor{cyan}{1}}-z_{\textcolor{orange}{0}}\right) \times(\textcolor{red}{l}, \textcolor{red}{m}, \textcolor{red}{n})\right|}{\sqrt{\textcolor{red}{l}^{2}+\textcolor{red}{m}^{2}+\textcolor{red}{n}^{2}}}$