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2022年高考数学北京17

  2022-12-16 20:10:06  

(14分)如图,在三棱柱$ABC-A_{1}B_{1}C_{1}$中,侧面$BCC_{1}B_{1}$为正方形,平面$BCC_{1}B_{1}\bot$平面$ABB_{1}A_{1}$,$AB=BC=2$,$M$,$N$分别为$A_{1}B_{1}$,$AC$的中点.
(Ⅰ)求证:$MN//$平面$BCC_{1}B_{1}$;
(Ⅱ)再从条件①、条件②这两个条件中选择一个作为已知,求直线$AB$与平面$BMN$所成角的正弦值.
条件①:$AB\bot MN$;
条件②:$BM=MN$.
注:如果选择条件①和条件②分别解答,按第一个解答计分.

分析:(1)通过证面面平证线面平行;
(2)通过证明$BC$,$BA$,$BB_{1}$两两垂直,从而建立以$B$为坐标原点,$BC$,$BA$,$BB_{1}$为坐标轴建立如图所示的空间直角坐标系,利用向量法求线面角的正弦值.
解答:解:$(I)$证明:取$AB$中点$K$,连接$NK$,$MK$,
$\because M$为$A_{1}B_{1}$的中点.$\therefore B_{1}M//BK$,且$B_{1}M=BK$,
$\therefore$四边形$BKMB_{1}$是平行四边形,故$MK//BB_{1}$,
$MK\not\subsetset$平面$BCC_{1}B_{1}$;$BB_{1}\subset$平面$BCC_{1}B_{1}$,
$\therefore MK//$平面$BCC_{1}B_{1}$,
$\because K$是$AB$中点,$N$是$AC$的点,
$\therefore NK//BC$,$\because NK\not\subsetset$平面$BCC_{1}B_{1}$;$BC\subset$平面$BCC_{1}B_{1}$,
$\therefore NK//$平面$BCC_{1}B_{1}$,又$NK\bigcap MK=K$,
$\therefore$平面$NMK//$平面$BCC_{1}B_{1}$,
又$MN\subset$平面$NMK$,$\therefore MN//$平面$BCC_{1}B_{1}$;
$(II)\because$侧面$BCC_{1}B_{1}$为正方形,平面$BCC_{1}B_{1}\bot$平面$ABB_{1}A_{1}$,平面$BCC_{1}B_{1}\bigcap$平面$ABB_{1}A_{1}=BB_{1}$,
$\therefore CB\bot$平面$ABB_{1}A_{1}$,$\therefore CB\bot AB$,又$NK//BC$,$\therefore AB\bot NK$,
若选①:$AB\bot MN$;又$MN\bigcap NK=N$,$\therefore AB\bot$平面$MNK$,
又$MK\subset$平面$MNK$,$\therefore AB\bot MK$,又$MK//BB_{1}$,
$\therefore AB\bot BB_{1}$,$\therefore BC$,$BA$,$BB_{1}$两两垂直,
若选②:$\because CB\bot$平面$ABB_{1}A_{1}$,$NK//BC$,$\therefore NK\bot$平面$ABB_{1}A_{1}$,$KM\subset$平面$ABB_{1}A_{1}$,
$\therefore MK\bot NK$,又$BM=MN$,$NK=\dfrac{1}{2}BC$,$BK=\dfrac{1}{2}AB$,
$\therefore \Delta BKM\cong \Delta NKM$,$\therefore \angle BKM=\angle NKM=90^\circ$,
$\therefore AB\bot MK$,又$MK//BB_{1}$,$\therefore AB\bot BB_{1}$,
$\therefore BC$,$BA$,$BB_{1}$两两垂直,
以$B$为坐标原点,$BC$,$BA$,$BB_{1}$为坐标轴建立如图所示的空间直角坐标系,

则$B(0$,0,$0)$,$N(1$,1,$0)$,$M(0$,1,$2)$,$A(0$,2,$0)$,
$\therefore$$\overrightarrow{BM}=(0$,1,$2)$,$\overrightarrow{BN}=(1$,1,$0)$,

设平面$BMN$的一个法向量为$\overrightarrow{n}=(x$,$y$,$z)$,
则$\left\{\begin{array}{l}{\overrightarrow{n}\cdot \overrightarrow{BM}=y+2z=0}\\ {\overrightarrow{n}\cdot \overrightarrow{BN}=x+y=0}\end{array}\right.$,令$z=1$,则$y=-2$,$x=2$,
$\therefore$平面$BMN$的一个法向量为$\overrightarrow{n}=(2$,$-2$,$1)$,
又$\overrightarrow{BA}=(0$,2,$0)$,
设直线$AB$与平面$BMN$所成角为$\theta$,
$\therefore \sin \theta =\vert \cos  < \overrightarrow{n}$,$\overrightarrow{BA} > \vert =\dfrac{\vert \overrightarrow{n}\cdot \overrightarrow{BA}\vert }{\vert \overrightarrow{n}\vert \cdot \vert \overrightarrow{BA}\vert }=\dfrac{4}{\sqrt{4+4+1}\times 2}=\dfrac{2}{3}$.
$\therefore$直线$AB$与平面$BMN$所成角的正弦值为$\dfrac{2}{3}$.
点评:本题考查线面平行的证明,线面角的求法,属中档题.

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