Question

如图,在直三棱柱ABC-A1B1C1中,AB=1,AC=AA=根号3角ABC=60度,求证:AB垂直A1C,求二面角A-A1C-B的大小

Answer 1

解:
∵AB=1,AC=根号3,角ABC=60°
∴BC=2
∴角BAC=90°
∴AB⊥AC
∵AB⊥AA1
且AC与AA1共面
∴AB⊥面ACA1
∵A1C在面ACA1上
∴AB⊥A1C
2)设A(0,0,0)D为A1C中点,连接AD,BD
∵AC=AA1且角A1AC=90°
∴AD⊥A1C
∵A1B=2=BC
∴BD⊥A1C
∴∠ADB即为A-A1C-B的二面角
则D(0,根号3/2,根号3/2)
B(-2,0,0)C(0,根号3,0)A1(0,0,根号3)
∴AB=(-2,0,0)
DB=(2,根号3/2,根号3/2)
AD=(0,根号3/2,根号3/2)
∴cos∠ADB=AD*AB/(2|AD||AB|)=0
则∠ADB为直角