(2014?南京三模)如图,在正四棱锥P-ABCD中,PA=AB=2,点M,N分别在线段PA和BD上,BN=13BD.(1)若PM=1

2025-05-23 14:56:28
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回答1:

解答:(本小题满分10分)
(1)证明:连接AC,BD交于点O,以OA为x轴正方向,以OB为y轴正方向,
OP为z轴建立空间直角坐标系.
∵PA=AB=

2
,则A(1,0,0),B(0,1,0),
D(0,-1,0),P(0,0,1).
BN
=
1
3
BD
,得N(0,
1
3
,0),
PM
=
1
3
PA
,得M(
1
3
,0,
2
3
),
MN
=(-
1
3
1
3
,-
2
3
)
AD
=(-1,-1,0)

MN
?
AD
=0
,∴MN⊥AD.
(2)∵M在PA上,设
PM
PA
,得M(λ,0,1-λ),
BM
=(λ,-1,1-λ)
BD
=(0,-2,0)

设平面MBD的法向量
n
=(x,y,z)

n
?
BD
=0
n
?
BM
=0
,得
-2y=0
λx-y+(1-λ)z=0

取z=λ,得
n
=(λ-1,0,λ)

∵平面ABD的法向量为
OP
=(0,0,1)
,二面角M-BD-A的大小为
π
4

∴cos
π
4
=|
n
?
OP
|
n
||