如图,已知在侧棱垂直于底面三棱柱ABC-A 1 B 1 C 1 中, AC=3,AB=5,cos∠CAB= 3 5 ,A

1个回答

  • (1)证明:在△ABC中,由余弦定理得BC=4,∴△ABC为直角三角形,∴AC⊥BC.

    又∵CC 1⊥面ABC,∴CC 1⊥AC,CC 1∩BC=C,∴AC⊥面BCC 1∴AC⊥BC 1

    (2)证明:设B 1C交BC 1于点E,则E为BC 1的中点,连接DE,则DE为△ABC 1的中位线,

    则在△ABC 1中,DE ∥ AC 1,又DE⊂面CDB 1,则AC 1∥ 面B 1CD.

    (3)在△ABC中过C作CF⊥AB垂足为F,

    由面ABB 1A 1⊥面ABC知,CF⊥面ABB 1A 1,∴ V A 1 - B 1 CD = V C- A 1 D B 1 .

    而 S △D A 1 B 1 =

    1

    2 A 1 B 1 •A A 1 =5×4×

    1

    2 =10 ,

    CF=

    AC•BC

    AB =

    3×4

    5 =

    12

    5 ,

    V A 1 - B 1 CD =

    1

    3 ×10×

    12

    5 =8 .