如图, AB 是圆的直径, PA 垂直圆所在的平面, C 是圆上的点. (1)求证:平面 PAC ⊥平面 PBC ;(2)若
如图,AB是圆的直径,PA垂直圆所在的平面,C是圆上的点.(1)求证:平面PAC⊥平面PBC;(2)若AB=2,AC=1,PA=1,求二面角C-PB-A的余弦值....
如图, AB 是圆的直径, PA 垂直圆所在的平面, C 是圆上的点. (1)求证:平面 PAC ⊥平面 PBC ;(2)若 AB =2, AC =1, PA =1,求二面角 C - PB - A 的余弦值.
展开
雯吧乣
推荐于2016-10-31
·
TA获得超过151个赞
知道答主
回答量:120
采纳率:100%
帮助的人:43.5万
关注
(1)见解析(2) |
(1)由 AB 是圆的直径,得 AC ⊥ BC , 由 PA ⊥平面 ABC , BC ?平面 ABC ,得 PA ⊥ BC . 又 PA ∩ AC = A , PA ?平面 PAC , AC ?平面 PAC , 所以 BC ⊥平面 PAC . 因为 BC ?平面 PBC , 所以平面 PBC ⊥平面 PAC . (2)过 C 作 CM ∥ AP ,则 CM ⊥平面 ABC . 如图,以点 C 为坐标原点,分别以直线 CB 、 CA 、 CM 为 x 轴, y 轴, z 轴建立空间直角坐标系. 在Rt△ ABC 中,因为 AB =2, AC =1,所以 BC = ![](https://iknow-pic.cdn.bcebos.com/b2de9c82d158ccbf62e118891ad8bc3eb1354127?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto) . 因为 PA =1,所以 A (0,1,0), B ( ![](https://iknow-pic.cdn.bcebos.com/b2de9c82d158ccbf62e118891ad8bc3eb1354127?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto) ,0,0), P (0,1,1).故 ![](https://iknow-pic.cdn.bcebos.com/f11f3a292df5e0fe647a24b25f6034a85edf7227?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto) =( ![](https://iknow-pic.cdn.bcebos.com/b2de9c82d158ccbf62e118891ad8bc3eb1354127?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto) ,0,0), ![](https://iknow-pic.cdn.bcebos.com/5d6034a85edf8db1c873d1d70a23dd54564e7427?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto) =(0,1,1). 设平面 BCP 的法向量为 n 1 =( x 1 , y 1 , z 1 ),则 ![](https://iknow-pic.cdn.bcebos.com/86d6277f9e2f0708551687b3ea24b899a901f276?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto) 所以 不妨令 y 1 =1,则 n 1 =(0,1,-1).因为 ![](https://iknow-pic.cdn.bcebos.com/060828381f30e92438b731824f086e061d95f776?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto) =(0,0,1), ![](https://iknow-pic.cdn.bcebos.com/29381f30e924b899282fd48b6d061d950a7bf676?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto) =( ![](https://iknow-pic.cdn.bcebos.com/b2de9c82d158ccbf62e118891ad8bc3eb1354127?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto) ,-1,0), 设平面 ABP 的法向量为 n 2 =( x 2 , y 2 , z 2 ),则 ![](https://iknow-pic.cdn.bcebos.com/9358d109b3de9c82dd38543c6f81800a19d84327?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto) 所以 不妨令 x 2 =1,则 n 2 =(1, ![](https://iknow-pic.cdn.bcebos.com/b2de9c82d158ccbf62e118891ad8bc3eb1354127?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto) ,0).于是cos〈 n 1 , n 2 〉= ![](https://iknow-pic.cdn.bcebos.com/810a19d8bc3eb13523c619f4a51ea8d3fd1f447f?x-bce-process=image/quality,q_85) = ![](https://iknow-pic.cdn.bcebos.com/810a19d8bc3eb135209a1a08a51ea8d3fd1f4427?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto) . 由题图可判断二面角为锐角,所以二面角 C - PB - A 的余弦值为 ![](https://iknow-pic.cdn.bcebos.com/810a19d8bc3eb135209a1a08a51ea8d3fd1f4427?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto) . |
收起
为你推荐: