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如圖多面體$PABCDA_{1}B_{1}C_{1}D_{1}$由正方體$ABCD-A_{1}B_{1}C_{1}D_{1}$和四棱錐$P-ABCD$組成.正方體$ABCD-A_{1}B_{1}C_{1}D_{1}$棱長(zhǎng)為$2$四棱錐$P-ABCD$側(cè)棱長(zhǎng)都相等高為$1$.（Ⅰ）求證$B_{1}C\bot($平面$PCD$；（Ⅱ）求二面角$B-PB_{1}-C$的余弦值.","title_text":"如圖多面體$PABCDA_{1}B_{1}C_{1}D_{1}$由正方體$ABCD-A_{1}B_{1}

2022-08-04 01:21:09 游戲 來源：
導(dǎo)讀 想必現(xiàn)在有很多小伙伴對(duì)于如圖，多面體$PABCDA_{1}B_{1}C_{1}D_{1}$由正方體$ABCD-A_{1}B_{1}C_{1}D_{1}$和四棱錐$P-ABCD$組成 正方體$ABC

想必現(xiàn)在有很多小伙伴對(duì)于如圖，多面體$PABCDA_{1}B_{1}C_{1}D_{1}$由正方體$ABCD-A_{1}B_{1}C_{1}D_{1}$和四棱錐$P-ABCD$組成.正方體$ABCD-A_{1}B_{1}C_{1}D_{1}$棱長(zhǎng)為$2$，四棱錐$P-ABCD$側(cè)棱長(zhǎng)都相等，高為$1$.（Ⅰ）求證：$B_{1}C\bot $平面$PCD$；（Ⅱ）求二面角$B-PB_{1}-C$的余弦值.","title_text":"如圖，多面體$PABCDA_{1}B_{1}C_{1}D_{1}$由正方體$ABCD-A_{1}B_{1}C_{1}D_{1}$和四棱錐$P-ABCD$組成.正方體$ABCD-A_{1}B_{1}C_{1}D_{1}$棱長(zhǎng)為$2$，四棱錐$P-ABCD$側(cè)棱長(zhǎng)都相等，高為$1$.（Ⅰ）求證：$B_{1}C\bot $平面$PCD$；（Ⅱ）求二面角$B-PB_{1}-C$的余弦值.方面的知識(shí)都比較想要了解，那么今天小好小編就為大家收集了一些關(guān)于如圖，多面體$PABCDA_{1}B_{1}C_{1}D_{1}$由正方體$ABCD-A_{1}B_{1}C_{1}D_{1}$和四棱錐$P-ABCD$組成.正方體$ABCD-A_{1}B_{1}C_{1}D_{1}$棱長(zhǎng)為$2$，四棱錐$P-ABCD$側(cè)棱長(zhǎng)都相等，高為$1$.（Ⅰ）求證：$B_{1}C\bot $平面$PCD$；（Ⅱ）求二面角$B-PB_{1}-C$的余弦值.","title_text":"如圖，多面體$PABCDA_{1}B_{1}C_{1}D_{1}$由正方體$ABCD-A_{1}B_{1}C_{1}D_{1}$和四棱錐$P-ABCD$組成.正方體$ABCD-A_{1}B_{1}C_{1}D_{1}$棱長(zhǎng)為$2$，四棱錐$P-ABCD$側(cè)棱長(zhǎng)都相等，高為$1$.（Ⅰ）求證：$B_{1}C\bot $平面$PCD$；（Ⅱ）求二面角$B-PB_{1}-C$的余弦值.方面的知識(shí)分享給大家，希望大家會(huì)喜歡哦。

1、（Ⅰ）證明：以$D_{1}?$為坐標(biāo)原點(diǎn)，分別以$D_{1}A_{1}$，$D_{1}C_{1}$，$D_{1}D$所在直線為$x$，$y$，$z$軸建立空間直角坐標(biāo)系$,$

2、由已知可得：$B_{1}left(2,2,0right)$，$Cleft(0,2,2right)$，$Pleft(1,1,3right)$，$Dleft(0,0,2right)$，$Bleft(2,2,2right)$，

3、則$overrightarrow {B_{1}C}=left(-2,0,2right),overrightarrow {CD}=left(0,-2,0right),overrightarrow {CP}=left(1,-1,1right)$，

4、$because overrightarrow {B_{1}C}cdot overrightarrow {CD}=0,overrightarrow {B_{1}C}cdot overrightarrow {CP}=0,therefore B_{1}Cbot CD,B_{1}Cbot CP$，

5、又$CDcap CP=C$，$therefore B_{1}Cbot $平面$PCD$；

6、（Ⅱ$) $設(shè)平面$CPB_{1}?$的一個(gè)法向量為$overrightarrow {n}=left(x,y,zright)$.

7、由$left{begin{array}{l}overrightarrow {n}cdot overrightarrow {B_{1}C}=-2x+2z=0overrightarrow {n}cdot overrightarrow {CP}=x-y+z=0end{array}right.$，取$z=1$，可得$overrightarrow {n}=left(1,2,1right)$.

8、又平面$BPB_{1}?$的一個(gè)法向量$overrightarrow {m}=left(2,-2,0right)=2left(1,-1,0right)$，

9、$therefore cos lt overrightarrow {m}$，$overrightarrow {n} gt =xlongequal[|overrightarrow {m}|cdot |overrightarrow {n}|]{overrightarrow {m}cdot overrightarrow {n}}=dfrac{1-2}{sqrt {2}times sqrt {6}}=-dfrac{sqrt {3}}{6}$.

10、由圖可知，二面角$B-PB_{1}-C$為鈍二面角，

11、$therefore $二面角$B-PB_{1}-C$的余弦值為$-dfrac{sqrt {3}}{6}$.

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