91学 首页 > 数学 > 高考题 > 2022 > 2022年新高考1 > 正文 返回 打印

2022年高考数学新高考Ⅰ-9

  2022-12-15 15:51:03  

(5分)已知正方体$ABCD-A_{1}B_{1}C_{1}D_{1}$,则(  )
A.直线$BC_{1}$与$DA_{1}$所成的角为$90^\circ$              
B.直线$BC_{1}$与$CA_{1}$所成的角为$90^\circ$              
C.直线$BC_{1}$与平面$BB_{1}D_{1}D$所成的角为$45^\circ$              
D.直线$BC_{1}$与平面$ABCD$所成的角为$45^\circ$
分析:求出异面直线所成角判断$A$;证明线面垂直,结合线面垂直的性质判断$B$;分别求出线面角判断$C$与$D$.
解:如图,

连接$B_{1}C$,由$A_{1}B_{1}//DC$,$A_{1}B_{1}=DC$,得四边形$DA_{1}B_{1}C$为平行四边形,
可得$DA_{1}//B_{1}C$,$\because BC_{1}\bot B_{1}C$,$\therefore$直线$BC_{1}$与$DA_{1}$所成的角为$90^\circ$,故$A$正确;
$\because A_{1}B_{1}\bot BC_{1}$,$BC_{1}\bot B_{1}C$,$A_{1}B_{1}\bigcap B_{1}C=B_{1}$,$\therefore BC_{1}\bot$平面$DA_{1}B_{1}C$,而$CA_{1}\subset$平面$DA_{1}B_{1}C$,
$\therefore BC_{1}\bot CA_{1}$,即直线$BC_{1}$与$CA_{1}$所成的角为$90^\circ$,故$B$正确;
设$A_{1}C_{1}\bigcap B_{1}D_{1}=O$,连接$BO$,可得$C_{1}O\bot$平面$BB_{1}D_{1}D$,即$\angle C_{1}BO$为直线$BC_{1}$与平面$BB_{1}D_{1}D$所成的角,
$\because \sin \angle C_{1}BO=\dfrac{O{C}_{1}}{B{C}_{1}}=\dfrac{1}{2}$,$\therefore$直线$BC_{1}$与平面$BB_{1}D_{1}D$所成的角为$30^\circ$,故$C$错误;
$\because CC_{1}\bot$底面$ABCD$,$\therefore \angle C_{1}BC$为直线$BC_{1}$与平面$ABCD$所成的角为$45^\circ$,故$D$正确.
故选:$ABD$.
点评:本题考查空间中异面直线所成角与线面角的求法,考查空间想象能力与思维能力,考查运算求解能力,是基础题.

http://x.91apu.com//shuxue/gkt/2022/2022xgk1/2022-12-15/33386.html