空间中某点关于直线的垂足

青青子衿

空间中的点\(J\,(\,\xi,\,\eta,\,\zeta\,)\,\)关于直线\(\,\,l_1\colon\dfrac{x-x_{\overset{\,}1}}{X}=\dfrac{y-y_{\overset{\,}1}}{Y}=\dfrac{z-z_{\overset{\,}1}}{Z}\,\,\)的垂足\(K\,(\,\xi',\,\eta',\,\zeta'\,)\,\)
没有想到，表达形式可以怎么简洁：
\begin{cases}
\xi'=x_{\overset{\,}1}+X\cdot\dfrac{X\cdot(\xi-x_{\overset{\,}1})+Y\cdot(\eta-y_{\overset{\,}1})+Z\cdot(\zeta-z_{\overset{\,}1})}{X^2+Y^2+Z^2}\\
\eta'=y_{\overset{\,}1}+Y\cdot\dfrac{X\cdot(\xi-x_{\overset{\,}1})+Y\cdot(\eta-y_{\overset{\,}1})+Z\cdot(\zeta-z_{\overset{\,}1})}{X^2+Y^2+Z^2}\\
\zeta'=z_{\overset{\,}1}+Z\cdot\dfrac{X\cdot(\xi-x_{\overset{\,}1})+Y\cdot(\eta-y_{\overset{\,}1})+Z\cdot(\zeta-z_{\overset{\,}1})}{X^2+Y^2+Z^2}\\
\end{cases}
...
\begin{cases}
X\cdot(\xi'-\xi)+Y\cdot(\eta'-\eta)+Z\cdot(\zeta'-\zeta)=0\\
\\
\dfrac{\xi'-x_{\overset{\,}1}}{X}=\dfrac{\eta'-y_{\overset{\,}1}}{Y}\\
\\
\dfrac{\eta'-y_{\overset{\,}1}}{Y}=\dfrac{\zeta'-z_{\overset{\,}1}}{Z}
\end{cases}