第七章第7课时空间向量及其运算随堂检测(含解析)1.(2012·大同调研)已知△ABC中,A(2,-5,3),AB=(4,1,2),BC=(3,-2,5).求:(1)顶点B和顶点C的坐标;(2)CA·AB.解:(1)设B(x1,y1,z1),由AB=(x1,y1,z1)-(2,-5,3)=(4,1,2),故B(6,-4,5),同理C(9,-6,10).(2)∵CA=-AC=-(AB+BC)=(-7,1,-7),∴CA·AB=(-7,1,-7)·(4,1,2)=-28+1-14=-41.2.E,F分别是正方体ABCD-A1B1C1D1中线段A1D,AC上的点,且DE=AF=AC.求证:(1)EF∥BD1;(2)EF⊥A1D.证明:(1)建立如图所示的空间直角坐标系,设AB=1,连接DF,则D(0,0,0),A(1,0,0),B(1,1,0),C(0,1,0),A1(1,0,1),D1(0,0,1),E,又DF=DA+AC,可得F.∵EF=,BD1=(-1,-1,1)=-3EF,∴BD1∥EF,又F不在BD1上,∴EF∥BD1.(2)∵A1D=(-1,0,-1),EF·A1D=·(-1,0,-1)=0,∴EF⊥A1D,即EF⊥A1D.