扭啊扭啊的小虫子
幼苗
共回答了16个问题采纳率:75% 举报
(1)见解析(2)
![](https://img.yulucn.com/upload/c/84/c84b6c272aa96554d32d503f91b83c7e_thumb.jpg)
(3)
本题可通过建立空间坐标系求解.
如图,以点A为原点建立空间直角坐标系,依题意得A(0,0,0),B(0,0,2),C(1,0,1),B
1 (0,2,2),C
1 (1,2,1),E(0,1,0).
(1)证明:易得
![](https://img.yulucn.com/upload/1/ea/1ea1d88caa5d216bab8816c5f16db287_thumb.jpg)
=(1,0,-1),
![](https://img.yulucn.com/upload/e/3d/e3d0096a388adfc99e4e04fb9197b334_thumb.jpg)
=(-1,1,-1),于是
![](https://img.yulucn.com/upload/1/ea/1ea1d88caa5d216bab8816c5f16db287_thumb.jpg)
·
![](https://img.yulucn.com/upload/e/3d/e3d0096a388adfc99e4e04fb9197b334_thumb.jpg)
=0,∴B
1 C
1 ⊥CE.
(2)
![](https://img.yulucn.com/upload/a/3e/a3e1b533bd7d2e6a6444936c86d21a6a_thumb.jpg)
=(1,-2,-1).
设平面B
1 CE的法向量m=(x,y,z),
则
![](https://img.yulucn.com/upload/e/31/e31ed3a59985cc9cbe9b745a913c8726_thumb.jpg)
,即
消去x,得y+2z=0,不妨令z=1,可得一个法向量为m=(-3,-2,1).
由(1),B
1 C
1 ⊥CE,又CC
1 ⊥B
1 C
1 ,可得B
1 C
1 ⊥平面CEC
1 ,故
![](https://img.yulucn.com/upload/1/ea/1ea1d88caa5d216bab8816c5f16db287_thumb.jpg)
=(1,0,-1)为平面CEC
1 的一个法向量.
于是cos〈m,
![](https://img.yulucn.com/upload/1/ea/1ea1d88caa5d216bab8816c5f16db287_thumb.jpg)
〉=
![](https://img.yulucn.com/upload/2/2a/22a4d9cb7f0c6822481dc5ade9d9ac41_thumb.jpg)
=
![](https://img.yulucn.com/upload/4/1e/41ebc3632147b45dd5c875cb585bdccc_thumb.jpg)
=-
![](https://img.yulucn.com/upload/6/dd/6dd10050df81937bdb75de5f31cc0adc_thumb.jpg)
,从而sin〈m,
![](https://img.yulucn.com/upload/1/ea/1ea1d88caa5d216bab8816c5f16db287_thumb.jpg)
〉=
![](https://img.yulucn.com/upload/c/84/c84b6c272aa96554d32d503f91b83c7e_thumb.jpg)
,
故二面角B
1 -CE-C
1 的正弦值为
![](https://img.yulucn.com/upload/c/84/c84b6c272aa96554d32d503f91b83c7e_thumb.jpg)
.
(3)
![](https://img.yulucn.com/upload/c/5a/c5a80969f2bddfa97dddcd68135621e1_thumb.jpg)
=(0,1,0),
![](https://img.yulucn.com/upload/9/85/9856a2b5e6b6ce285e8d5dc9175d1c55_thumb.jpg)
=(1,1,1).
设
![](https://img.yulucn.com/upload/2/54/254c1389b7f3f2a5a51f96f42b684847_thumb.jpg)
=λ
![](https://img.yulucn.com/upload/9/85/9856a2b5e6b6ce285e8d5dc9175d1c55_thumb.jpg)
=(λ,λ,λ),0≤λ≤1,有
![](https://img.yulucn.com/upload/2/25/225998539162650ced3d764ecc93133f_thumb.jpg)
=
![](https://img.yulucn.com/upload/c/5a/c5a80969f2bddfa97dddcd68135621e1_thumb.jpg)
+
![](https://img.yulucn.com/upload/2/54/254c1389b7f3f2a5a51f96f42b684847_thumb.jpg)
=(λ,λ+1,λ).可取
![](https://img.yulucn.com/upload/4/35/43577227ebaea2d33605ec6be144e370_thumb.jpg)
=(0,0,2)为平面ADD
1 A
1 的一个法向量.
设θ为直线AM与平面ADD
1 A
1 所成的角,则
sinθ=|cos〈
![](https://img.yulucn.com/upload/2/25/225998539162650ced3d764ecc93133f_thumb.jpg)
,
![](https://img.yulucn.com/upload/4/35/43577227ebaea2d33605ec6be144e370_thumb.jpg)
〉|=
=
![](https://img.yulucn.com/upload/3/46/3469f6a42035d99b3351eba70d563994_thumb.jpg)
=
![](https://img.yulucn.com/upload/6/e0/6e0c3af559518713ec8e529d75a7cbec_thumb.jpg)
.
于是
![](https://img.yulucn.com/upload/6/e0/6e0c3af559518713ec8e529d75a7cbec_thumb.jpg)
=
![](https://img.yulucn.com/upload/9/11/911790de6c19d14f6e153d9a4609a4d1_thumb.jpg)
,解得λ=
![](https://img.yulucn.com/upload/a/4c/a4c846c81eeb4486d35fbc3c44915e40_thumb.jpg)
(λ=-
![](https://img.yulucn.com/upload/8/80/8802c38e6e44f5f0cfe1689069b16829_thumb.jpg)
舍去),
∴AM=
![](https://img.yulucn.com/upload/3/50/3502ed717965b960ddb1286002925cc2_thumb.jpg)
.
1年前
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