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2022年高考数学新高考Ⅰ-19

  2022-12-15 15:52:23  

(12分)如图,直三棱柱$ABC-A_{1}B_{1}C_{1}$的体积为4,△$A_{1}BC$的面积为$2\sqrt{2}$.
(1)求$A$到平面$A_{1}BC$的距离;
(2)设$D$为$A_{1}C$的中点,$AA_{1}=AB$,平面$A_{1}BC\bot$平面$ABB_{1}A_{1}$,求二面角$A-BD-C$的正弦值.

分析:(1)利用体积法可求点$A$到平面$A_{1}BC$的距离;
(2)以$B$为坐标原点,$BC$,$BA$,$BB_{1}$所在直线为坐标轴建立如图所示的空间直角坐标系,利用向量法可求二面角$A-BD-C$的正弦值.
解:(1)由直三棱柱$ABC-A_{1}B_{1}C_{1}$的体积为4,可得$V{}_{{A}_{1}-ABC}=\dfrac{1}{3}V{}_{{A}_{1}{B}_{1}{C}_{1}-ABC}=\dfrac{4}{3}$,
设$A$到平面$A_{1}BC$的距离为$d$,由$V{}_{{A}_{1}-ABC}=V{}_{A-{A}_{1}BC}$,
$\therefore$$\dfrac{1}{3}S{}_{\triangle {A}_{1}BC}\cdot d=\dfrac{4}{3}$,$\therefore$$\dfrac{1}{3}\times 2\sqrt{2}\cdot d=\dfrac{4}{3}$,解得$d=\sqrt{2}$.
(2)连接$AB_{1}$交$A_{1}B$于点$E$,$\because AA_{1}=AB$,$\therefore$四边形$ABB_{1}A_{1}$为正方形,
$\therefore AB_{1}\bot A_{1}B$,又$\because$平面$A_{1}BC\bot$平面$ABB_{1}A_{1}$,平面$A_{1}BC\bigcap$平面$ABB_{1}A_{1}=A_{1}B$,
$\therefore AB_{1}\bot$平面$A_{1}BC$,$\therefore AB_{1}\bot BC$,
由直三棱柱$ABC-A_{1}B_{1}C_{1}$知$BB_{1}\bot$平面$ABC$,$\therefore BB_{1}\bot BC$,又$AB_{1}\bigcap BB_{1}=B_{1}$,
$\therefore BC\bot$平面$ABB_{1}A_{1}$,$\therefore BC\bot AB$,
以$B$为坐标原点,$BC$,$BA$,$BB_{1}$所在直线为坐标轴建立如图所示的空间直角坐标系,

$\because AA_{1}=AB$,$\therefore BC\times \sqrt{2}AB\times \dfrac{1}{2}=2\sqrt{2}$,又$\dfrac{1}{2}AB\times BC\times AA_{1}=4$,解得$AB=BC=AA_{1}=2$,
则$B(0$,0,$0)$,$A(0$,2,$0)$,$C(2$,0,$0)$,$A_{1}(0$,2,$2)$,$D(1$,1,$1)$,
则$\overrightarrow{BA}=(0$,2,$0)$,$\overrightarrow{BD}=(1$,1,$1)$,$\overrightarrow{BC}=(2$,0,$0)$,
设平面$ABD$的一个法向量为$\overrightarrow{n}=(x$,$y$,$z)$,
则$\left\{\begin{array}{l}{\overrightarrow{n}\cdot \overrightarrow{BA}=2y=0}\\ {\overrightarrow{n}\cdot \overrightarrow{BD}=x+y+z=0}\end{array}\right.$,令$x=1$,则$y=0$,$z=-1$,
$\therefore$平面$ABD$的一个法向量为$\overrightarrow{n}=(1$,0,$-1)$,
设平面$BCD$的一个法向量为$\overrightarrow{m}=(a$,$b$,$c)$,
$\left\{\begin{array}{l}{\overrightarrow{m}\cdot \overrightarrow{BC}=2a=0}\\ {\overrightarrow{m}\cdot \overrightarrow{BD}=a+b+c=0}\end{array}\right.$,令$b=1$,则$a=0$,$c=-1$,
平面$BCD$的一个法向量为$\overrightarrow{m}=(0$,1,$-1)$,
$\cos  < \overrightarrow{n}$,$\overrightarrow{m} > =\dfrac{1}{\sqrt{2}\cdot \sqrt{2}}=\dfrac{1}{2}$,
二面角$A-BD-C$的正弦值为$\sqrt{1-(\dfrac{1}{2})^{2}}=\dfrac{\sqrt{3}}{2}$.
点评:本题考查求点到面的距离,求二面角的正弦值,属中档题.

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