直三棱柱ABC-A1B1C1的底面中,AB⊥AC,AB=AC=a,D为CC1的中点,CC1AC=λ(1)λ为何值时,A1D⊥平面ABD;(2)当A1D⊥平面ABD时,求C1到平面ABD的距离;(3)当

题目简介

直三棱柱ABC-A1B1C1的底面中,AB⊥AC,AB=AC=a,D为CC1的中点,CC1AC=λ(1)λ为何值时,A1D⊥平面ABD;(2)当A1D⊥平面ABD时,求C1到平面ABD的距离;(3)当

题目详情

直三棱柱ABC-A1B1C1的底面中,AB⊥AC,AB=AC=a,D为CC1的中点,
CC1
AC

(1)λ为何值时,A1D⊥平面ABD;
(2)当A1D⊥平面ABD时,求C1到平面ABD的距离;
(3)当二面角A-BD-C为60°时,求λ的值.
题型:解答题难度:中档来源:不详

答案

AB
AC
AA1
为正交基底建立空间直角坐标系,
A(0,0,0),B(a,0,0),C(0,a,0),C1(0,a,λa),D(0,a,class="stub"1
2
λa),A1
(0,0,λa)
(1)
A1D
=(0,a,-class="stub"λa
2
),
AD
=(0,a,class="stub"λa
2
)

∵A1D⊥平面ABD∴A1D⊥AD
∴0+a2-
λ2a2
4
=0有λ=2
(2)λ=2时,
C1D
=(0,0,-a),
A1D
=(0,a,-a)
C1到平面ABD的距离d=|
C1D
A1D
|
A1D
|
|=
2
2
a
(3)取BC中点E,连接AE,则AE⊥BC,又BB1⊥AE∴AE⊥平面BCD
AE
=(class="stub"a
2
,class="stub"a
2
,0),设
m
=(x,y,z)为平面ABD的一个法向量

m
AB
=0
m
AD
=0
x=0
y=-class="stub"λz
2

取z=1得
m
=(0,-class="stub"λ
2
,1),由cos60°=|
m
AE
|
m
|•|
AE
|
|得λ=2

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