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2022年高考数学天津17

  2022-12-16 20:38:08  

(15分)直三棱柱$ABC-A_{1}B_{1}C_{1}$中,$AA_{1}=AB=AC=2$,$AA_{1}\bot AB$,$AC\bot AB$,$D$为$A_{1}B_{1}$中点,$E$为$AA_{1}$中点,$F$为$CD$中点.
(1)求证:$EF//$平面$ABC$;
(2)求直线$BE$与平面$CC_{1}D$的正弦值;
(3)求平面$A_{1}CD$与平面$CC_{1}D$夹角的余弦值.

分析:利用中位线可证(1),建立空间直角坐标系设$\overrightarrow{n}=(x$,$y$,$z)$是平面$CC_{1}D$的法向量,平面$A_{1}CD$的法向量为$\overrightarrow{m}=(x$,$y$,$z)$,可解.
解:(1)证明:取$BB_{1}$的中点$G$,连接$FG$,$EG$,连接$AD$交$EG$于$K$,
再连接$FK$,
$\because EK//A_{1}B_{1}$,且$E$是$AA_{1}$的中点,则$K$是$AD$的中点,
$\therefore FK//AC$,$EG//AB$,
又$FK\not\subset$平面$ABC$,$AC\subset$平面$ABC$,
$\therefore FK//$平面$ABC$,
同理可得,$EG//$平面$ABC$,
又$FK\bigcap EG=K$,
$\therefore$平面$EFG//$平面$ABC$,
$\therefore EF//$平面$ABC$,
(2)在直三棱柱$ABC-A_{1}B_{1}C_{1}$中,$AC\bot AB$,则可建立如图所示的空间直角坐标系,

又$AA_{1}=AB=AC=2$,$D$为$A_{1}B_{1}$中点,$E$为$AA_{1}$中点,$F$为$CD$中点.
故$B(2$,2,$0)$,$E(1$,0,$0)$,$C(2$,0,$2)$,$C_{1}(0$,0,$2)$,$D(0$,1,$0)$,
则$\overrightarrow{BE}=(-1$,$-2$,$0)$,$\overrightarrow{C{C}_{1}}=(-2$,0,$0)$,$\overrightarrow{CD}=(-2$,1,$-2)$,
设$\overrightarrow{n}=(x$,$y$,$z)$是平面$CC_{1}D$的法向量,则有:$\overrightarrow{n}\cdot \overrightarrow{C{C}_{1}}=0$,$\overrightarrow{n}\cdot \overrightarrow{CD}=0$,即$\left\{\begin{array}{l}{-2x=0}\\ {-2x+y-2z=0}\end{array}\right.$,令$z=1$,则$x=0$,$y=2$,
所以$\overrightarrow{n}=(0,2,1)$,
设直线$BE$与平面$CC_{1}D$的夹角为$\theta$,则$\sin \theta =\vert \cos  < \overrightarrow{BE},\overrightarrow{n} > \vert =\vert \dfrac{-2\times 2}{\sqrt{5}\times \sqrt{5}}\vert =\dfrac{4}{5}$,
(3)$\because A_{1}(0$,0,$0)$,则$\overrightarrow{{A}_{1}C}=(2$,0,$2)$,$\overrightarrow{{A}_{1}D}=(0$,1,$0)$,
设平面$A_{1}CD$的法向量为$\overrightarrow{m}=(x$,$y$,$z)$,则有$\overrightarrow{m}\cdot \overrightarrow{{A}_{1}C}=0$,$\overrightarrow{m}\cdot \overrightarrow{{A}_{1}D}=0$,
即$\left\{\begin{array}{l}{2x+2z=0}\\ {y=0}\end{array}\right.$,令$x=1$,则$y=0$,$z=-1$,故$\overrightarrow{m}=(1,0,-1)$,
设平面$A_{1}CD$与平面$CC_{1}D$的夹角为$\beta$,
所以$\cos \beta =\vert \cos  < \overrightarrow{n},\overrightarrow{m} > \vert =\vert \dfrac{-1\times 1}{\sqrt{5}\times \sqrt{2}}\vert =\dfrac{\sqrt{10}}{10}$.
点评:本题考查了利用空间向量求线面角以及二面角的大小,属于较难题.

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