解答:
(1)證明:連結(jié)CB
1,
∵P是BC
1的中點(diǎn),∴CB
1過(guò)點(diǎn)P,--(1分)
∵N為AB
1的中點(diǎn),∴PN∥AC,---------------------------(2分)
∵AC?面ABC,PN?面ABC,
∴PN∥平面ABC.--------------------------------------(4分)
(2)證法一:連結(jié)AC
1,在直角△ABC中,
∵BC=1,∠BAC=30°,
∴AC=A
1C
1=
-----------------------------------(5分)
∵
=
=
,
∴Rt△A
1C
1M~Rt△C
1CA--------------------------------------------------------(7分)
∴∠A
1MC
1=∠CAC
1,∴∠AC
1C+∠CAC
1=∠AC
1C+∠A
1MC
1=90°
∴AC
1⊥A
1M.-------------------------------------------------------------------(8分)
∵B
1C
1⊥C
1A
1,CC
1⊥B
1C
1,且C
1A
1∩CC
1=C
1∴B
1C
1⊥平面AA
1CC
1,-----------------------------------------------------------(9分)
∴B
1C
1⊥A
1M,又AC
1∩B
1C
1=C
1,故A
1M⊥平面A B
1C
1,-------------------------(11分)
證法二:連結(jié)AC
1,在直角△ABC中,∵BC=1,∠BAC=30°,
∴AC=A
1C
1=
-------------------------------------------------------------(5分)
設(shè)∠AC
1A
1=α,∠MA
1C
1=β
∵
tanαtanβ=•=•=1,------------------------------------------(7分)
∴α+β=90° 即AC
1⊥A
1M.-------------------------------------------------------(8分)
∵B
1C
1⊥C
1A
1,CC
1⊥B
1C
1,且C
1A
1∩CC
1=C
1∴B
1C
1⊥平面AA
1CC
1,-----------------------------------------------------------(9分)
∴B
1C
1⊥A
1M,又AC
1∩B
1C
1=C
1故A
1M⊥面A B
1C
1,------------------------------------------------------------(11分)】
(3)設(shè)點(diǎn)M到平面AA
1B
1的距離為h,
由(2)知B
1C
1⊥平面AA
1CC
1∵
VM-AA1B1=VB1-MAA1------------------------------------------------------------(12分)
∴
S△AA1B1•h=S△MAA1•B1C1∴
h==
=.
即點(diǎn)M到平面AA
1B
1的距離為
.----------------------------------------------(14分)