2.如图,在棱长为1的正方体 -(A)_(1)(B)_(1)(C)_(1)(D)_(1) 中,E为线段DD1的中点,F为线段BB1的中点.-|||-(1)求点A1到直线B1E的距离;-|||-(2)求直线FC1到直线AE的距离;-|||-(3)求点A1到平面AB1E的距离;-|||-(4)求直线FC1到平面AB1E的距离.2.如图,在棱长为1的正方体 -(A)_(1)(B)_(1)(C)_(1)(D)_(1) 中,E为线段DD1的中点,F为线段BB1的中点.-|||-(1)求点A1到直线B1E的距离;-|||-(2)求直线FC1到直线AE的距离;-|||-(3)求点A1到平面AB1E的距离;-|||-(4)求直线FC1到平面AB1E的距离.

【答案】

$\left(1\right)$$d=\dfrac{\sqrt{5}}{3}$

$\left(2\right)$$d=\dfrac{\sqrt{5}}{2}$

$\left(3\right)$$d=\dfrac{2}{3}$

$\left(4\right)$$d=\dfrac{1}{3}$

【解析】

$\left(1\right)$连接${A}_{1}E$,

在$\triangle {A}_{1}{B}_{1}E$中,${A}_{1}E=\sqrt{{1}^{2}+{\left(\frac{1}{2}\right)}^{2}}=\dfrac{\sqrt{5}}{2}$

$E{B}_{1}=\sqrt{{1}^{2}+{1}^{2}+{\left(\dfrac{1}{2}\right)}^{2}}=\dfrac{3}{2}$,${A}_{1}{B}_{1}=1$

$\because {A}_{1}{{B}_{1}}^{2}+{A}_{1}{E}^{2}=E{{B}_{1}}^{2}$,$\therefore \angle E{A}_{1}{B}_{1}={90}^{\circ }$.

设${A}_{1}$到${B}_{1}E$的距离为$d$.$\therefore {S}_{\triangle {A}_{1}{B}_{1}E}=\dfrac{1}{2}\cdot {A}_{1}E\cdot {A}_{1}{B}_{1}=\dfrac{1}{2}\cdot E{B}_{1}\cdot d$

$\therefore \dfrac{\sqrt{5}}{2}\times 1=\dfrac{3}{2}d$,解得$d=\dfrac{\sqrt{5}}{3}$

$\left(2\right)$$\because F{C}_{1}\ykparallel AE$,$\therefore F{C}_{1}$到$AE$的距离为$AF$.

$AF=\sqrt{{1}^{2}+{\left(\dfrac{1}{2}\right)}^{2}}=\dfrac{\sqrt{5}}{2}$,$\therefore F{C}_{1}$到$AE$的距离为$\dfrac{\sqrt{5}}{2}$

$\left(3\right)$以$DA$为$x$轴,$DC$为$y$轴,$D{D}_{1}$为$z$轴建立空间直角坐标系.

$\therefore {A}_{1}\left(1,0,1\right)$,$A\left(1,0,0\right)$,${B}_{1}\left(1,1,1\right)$,$E\left(0,0,\dfrac{1}{2}\right)$,${C}_{1}\left(0,1,1\right)$

$\overrightarrow{A{B}_{1}}=\left(0,1,1\right)$,$\overrightarrow{AE}=\left(-1,0,\dfrac{1}{2}\right)$,$\overrightarrow{A{A}_{1}}=\left(0,0,1\right)$$\overrightarrow{{C}_{1}A}=\left(1,-1,-1\right)$

设$\overrightarrow{m}=\left(x,y,z\right)$为平面$\overrightarrow{A{B}_{1}E}$的一个法向量.

$\left\{\begin{array}{l}y+z=0\\ -x+\dfrac{1}{2}z=0\end{array}\right.$,解得$\overrightarrow{m}=\left(1,-2,2\right)$

$\therefore {A}_{1}$到平面$A{B}_{1}E$的距离$d=\dfrac{\overrightarrow{A{A}_{1}}\cdot \overrightarrow{m}}{\overrightarrow{\left|m\right|}}=\dfrac{1\times 2}{\sqrt{1+4+4}}=\dfrac{2}{3}$

$\left(4\right)$$\because F{C}_{1}\ykparallel AE$,$\therefore F{C}_{1}\ykparallel $平面$A{B}_{1}E$.

$\therefore F{C}_{1}$到平面$A{B}_{1}E$的距离即为$A{C}_{1}$到平面$A{B}_{1}E$的距离.

$\therefore d=\dfrac{\overrightarrow{{C}_{1}A}\cdot \overrightarrow{m}}{\overrightarrow{m}}=\dfrac{1\times 1+\left(-2\right)\times \left(-1\right)+2\times \left(-1\right)}{\sqrt{1+4+4}}=\dfrac{1}{3}$