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Last edited by hbghlyj at 2023-3-3 03:33:00 例题3.3 (点到平面的距离) 点$P(x_0,y_0,z_0)$到平面$\pi:Ax+By+Cz+D=0$的距离可表为$$\frac{\left|A x_{0}+B y_{0}+C z_{0}+D\right|}{\sqrt{A^{2}+B^{2}+C^{2}}}$$
[证] 设过点$P$垂直于平面$\pi$的直线与平面的交点为$Q(x_1,y_1,z_1)$.与平面的法向量$(A,B,C)$同方向的单位向量为$\boldsymbol e$时,则有$$\boldsymbol{e}=\frac{1}{\sqrt{A^{2}+B^{2}+C^{2}}}(A, B, C)$$令$\overrightarrow{QP}=t\boldsymbol e$,则$|t|$为所求距离.由于\begin{array}{l}x_{0}-x_{1}=\frac{t A}{\sqrt{A^{2}+B^{2}+C^{2}}} \\ y_{0}-y_{1}=\frac{t B}{\sqrt{A^{2}+B^{2}+C^{2}}} \\ z_{0}-z_{1}=\frac{t C}{\sqrt{A^{2}+B^{2}+C^{2}}} \\ A x_{1}+B y_{1}+C z_{1}+D=0\end{array}从而得$$t=\frac{A x_{0}+B y_{0}+C z_{0}+D}{\sqrt{A^{2}+B^{2}+C^{{2}}}}$$
例题3.4 (二直线的最短距离) 设二不平行直线$g_1,g_2$的方程分别为\begin{gathered}\frac{x-x_{1}}{L_{1}}=\frac{y-y_{1}}{M_{1}}=\frac{z-z_{1}}{N_{1}} \\ \frac{x-x_{2}}{L_{2}}=\frac{y-y_{2}}{M_{2}}=\frac{z-z_{2}}{N_{2}}\end{gathered}则$g_1$与$g_2$的最短距离等于$$\left|\begin{array}{ccc}x_{2}-x_{1} & y_{2}-y_{1} & z_{2}-z_{1} \\ L_{1} & M_{1} & N_{1} \\ L_{2} & M_{2} & N_{2}\end{array}\right|\over\sqrt{\left|\begin{array}{ll}M_{1} & N_{1} \\ M_{2} & N_{2}\end{array}\right|^{2}+\left|\begin{array}{ll}N_{1} & L_{1} \\ N_{2} & L_{2}\end{array}\right|^{2}+\left|\begin{array}{ll}L_1 & M_{1} \\ L_{2} & M_{2}\end{array}\right|^{2}}$$的绝对值.
[证] 含有$g_1$而平行于$g_2$的平面$π$的方程是$$\left|\begin{array}{ccc}x-x_{1} & y-y_{1} & z-z_{1} \\ L_{1} & M_{1} & N_{1} \\ L_{2} & M_{2} & N_{2}\end{array}\right|=0$$$g_1$与$g_2$间的最短距离等于$g_2$上的点到平面$π$的距离. 由例题3.3可得到上面的结果.
又,设$P_1(x_1,y_1,z_1),$$P_2(x_2,y_2,z_2),$${\boldsymbol a}_1=(L_1,M_1,N_1),$${\boldsymbol a}_2=(L_2,M_2,N_2)$,则$g_1$与$g_2$的最短距离的表达式等于$\overrightarrow{P_1P_2}·\frac{{\boldsymbol a}_1×{\boldsymbol a}_2}{|{\boldsymbol a}_1×{\boldsymbol a}_2|}$的绝对值.
如果从点$P(x_1,y_1,z_1)$向直线$$\frac{x-x_{0}}{L}=\frac{y-y_{0}}{M}=\frac{z-z_{0}}{N}$$作垂线,其垂线长为$h$,试证,$$h={
\sqrt{\left|\begin{array}{cc}y_{1}-y_{0} & z_{1}-z_{0} \\ M & N\end{array}\right|^{2}+\left|\begin{array}{cc}z_{1}-z_{0} & x_{1}-x_{0} \\ N & L\end{array}\right|^{2}+\left|\begin{array}{cc}x_{1}-x_{0} & y_{1}-y_{0} \\ L & M\end{array}\right|^{2}}\over\sqrt{L^{2}+M^{2}+N^{2}}}$$
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