《2025年成才之路高中新课程学习指导高中数学选择性必修第一册北师大版》

已知正方体$ABCD - A_1B_1C_1D_1$棱长为$2$，$E$，$F$分别是$C_1C$，$D_1A_1$的中点，求点$A$到$EF$的距离.

答案：
对点训练2:以$D$点为原点,$DA,DC,DD_{1}$所在直线分别为$x$轴，$y$轴，$z$轴建立空间直角坐标系如图所示,
则$A(2,0,0),E(0,2,1),F(1,0,2)$,则$\overrightarrow{EF} = (1,-2,1),\overrightarrow{FA} = (1,0,-2)$.
$|\overrightarrow{EF}| = \sqrt{1^{2} + (-2)^{2} + 1^{2}} = \sqrt{6}$，
$|\overrightarrow{FA}| = \sqrt{1^{2} + 0^{2} + (-2)^{2}} = \sqrt{5}$，
$\overrightarrow{FA}·\overrightarrow{EF} = 1×1 + 0×(-2) + (-2)×1 = -1$，
$\overrightarrow{FA}$在$\overrightarrow{EF}$上的投影为$\frac{\overrightarrow{FA}·\overrightarrow{EF}}{|\overrightarrow{EF}|} = \frac{-1}{\sqrt{6}}$，
所以点$A$到$EF$的距离
$d = \sqrt{|\overrightarrow{FA}|^{2} - (\frac{-1}{\sqrt{6}})^{2}} = \sqrt{\frac{29}{6}} = \frac{\sqrt{174}}{6}$

例3. 如图，已知正方形$ABCD$的边长为$1$，$PD \perp$平面$ABCD$，且$PD = 1$，$E$，$F$分别为$AB$，$BC$的中点.
(1) 求点$D$到平面$PEF$的距离；
(2) 求直线$AC$到平面$PEF$的距离.

答案：
例3:
(1)建立如图所示的空间直角坐标系,则$D(0,0,0),P(0,0,1),A(1,0,0),C(0,1,0),E(\frac{1}{2},0),F(\frac{1}{2},1,0),\overrightarrow{DE}=(1,\frac{1}{2},0),\overrightarrow{DF}=(\frac{1}{2},1,0),\overrightarrow{DP}=(0,0,1)$

设$DH\perp$平面$PEF$,垂足为$H$,则
$\overrightarrow{DH}=x\overrightarrow{DE}+y\overrightarrow{DF}+z\overrightarrow{DP}=(x + \frac{1}{2}y,\frac{1}{2}x + y,z),(x + y + z = 1)$
$\overrightarrow{PE} = (1,\frac{1}{2},-1),\overrightarrow{PF} = (\frac{1}{2},1,-1)$
所以$\overrightarrow{DH}·\overrightarrow{PE} = x + \frac{1}{2}y + \frac{1}{2}(\frac{1}{2}x + y) - z = \frac{5}{4}x + y - z = 0$.
同理，$\overrightarrow{DH}·\overrightarrow{PF} = x + \frac{5}{4}y - z = 0$
又$x + y + z = 1$，所以解得$x = y = \frac{4}{17},z = \frac{9}{17}$
所以$\overrightarrow{DH} = \frac{3}{17}(2,2,3)$，所以$|\overrightarrow{DH}| = \frac{3\sqrt{17}}{17}$
因此,点$D$到平面$PEF$的距离为$\frac{3\sqrt{17}}{17}$
(2)连接$AC$，则$AC// EF$，直线$AC$到平面$PEF$的距离即为点$A$到平面$PEF$的距离，
平面$PEF$的一个法向量为$\boldsymbol{n} = (2,2,3),\overrightarrow{AE} = (0,\frac{1}{2},0)$
所求距离为$\frac{|\overrightarrow{AE}·\boldsymbol{n}|}{|\boldsymbol{n}|} = \frac{1}{\sqrt{17}} = \frac{\sqrt{17}}{17}$

如图，已知正方体$ABCD - A_1B_1C_1D_1$的棱长为$a$，求点$A$到平面$A_1BD$的距离.

答案： 对点训练3:设点$A$到平面$A_{1}BD$的距离为$h$,则
$V_{B - AA_{1}D} = \frac{1}{3}× a×\frac{1}{2}× a× a = \frac{1}{6}a^{3}$
$V_{A - A_{1}BD} = \frac{1}{3}× h×\frac{\sqrt{3}}{4}×(\sqrt{2}a)^{2} = \frac{\sqrt{3}}{6}a^{2}h$
$\because V_{A - A_{1}BD} = V_{B - AA_{1}D}$，$\therefore h = \frac{\sqrt{3}}{3}a$
$\therefore$点$A$到平面$A_{1}BD$的距离为$\frac{\sqrt{3}}{3}a$