如图在直三棱柱abcda1b1c1d1中.角abc=90ab=bc=1aa1=根号2如图在直三棱柱abcda1b1c1d1中.角abc=90ab=bc=1aa1=根号2求异面直线ba1与ac夹角余弦值急,

Question

如图在直三棱柱abcda1b1c1d1中.角abc=90 ab=bc=1aa1=根号2
如图在直三棱柱abcda1b1c1d1中.角abc=90 ab=bc=1 aa1=根号2 求异面直线ba1与ac夹角余弦值急,

Answer 1

作BD//=AC,连接AD,A1D
∵BD//=AC
∴四边形DACB是平行四边形
∴∠A1BD即BA1与AC的夹角
∵∠ABC=90°
AB=BC=1
∴BD=AC=√2
∵AA1⊥面ABC
∴BA1=DA1=√3
余弦定理
cos∠A1BD=(A1B²+BD²-A1D²)/(2A1B*BD)=√6/6