∵BC=λAD,∴|BC|=2|λ|,BC∥AD
又∵|AB|=向量|AD|=2,向量|CB-CD|=|DB|=2√3
∴角ABD=角ADB=∠DBC=30°
(1)①若∠BCD=90°,则COS30°=BC/BD,得λ=2分之3
②若∠BDC=90°,则COS30°=BD/BC,得λ=2
(2)①若∠BCD=90°,向量CB•向量BA =3乘以2乘以COS120°=-3
若∠BDC=90°,向量CB•向量BA=4乘以2乘以COS120°=-4
∵BC=λAD,∴|BC|=2|λ|,BC∥AD
又∵|AB|=向量|AD|=2,向量|CB-CD|=|DB|=2√3
∴角ABD=角ADB=∠DBC=30°
(1)①若∠BCD=90°,则COS30°=BC/BD,得λ=2分之3
②若∠BDC=90°,则COS30°=BD/BC,得λ=2
(2)①若∠BCD=90°,向量CB•向量BA =3乘以2乘以COS120°=-3
若∠BDC=90°,向量CB•向量BA=4乘以2乘以COS120°=-4