2020年1月9日福建省泉州市高2020届高2017级高三高中毕业班期末质量检测理科数学试题及参考答案解析

保密★启用前

泉州市2020届高中毕业班单科质量检查

理

科数学2020．1

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三、解答题：共70分．解答应写出文字说明、证明过程或演算步骤．第17~21题为必考题，每个试题考

生都必须作答．第22、23题为选考题，考生根据要求作答．

（一）必考题：共60分．

17．（12分）

如图，四棱锥ABCD P -的底面是正方形，⊥PA 平面ABCD ，AE PD ⊥．

（1）证明：AE ⊥平面PCD ；

（2）若AP AB =，求二面角D PC B --的余弦值．

【命题意图】本小题考查线面垂直的判定与性质、二面角的求解及空间向量的坐标运算等基础知识，考查

空间想象能力、逻辑推理及运算求解能力，考查化归与转化思想、函数与方程思想等，体现

基础性、综合性与应用性，导向对发展数学抽象、逻辑推理、直观想象等核心素养的关注．

【试题解析】

解法一：（1）因为PA ⊥平面ABCD ，CD ⊂平面ABCD ，

所以PA CD ⊥．·········································································································1分又底面ABCD 是正方形，所以AD CD ⊥．·····································································2分又PA AD A = ，所以CD ⊥平面PAD ．······································································3分又AE ⊂平面PAD ，所以CD AE ⊥．···········································································4分又因为AE PD ⊥，CD PD D = ，,CD PD ⊂平面PCD ，·············································5分所以AE ⊥平面PCD ．·······························································································6分

（2）因为PA ⊥平面ABCD ，底面ABCD 为正方形，

所以PA AB ⊥，PA AD ⊥，AB AD ⊥，分别以AB 、AD 、AP 所在的直线为x 轴、y 轴、z 轴建立空间直角坐标系A xyz -（如图所示）．······································································7分设1PA AB ==，则A 0,0,0（），B 1,0,0（），C 1,1,0（），D 0,1,0（），(0,0,1)P ，11

(0,,)22

E ，1,0,1PB =- （），1,1,1PC =- （），11(0,,22

AE = ．··························································8分由（1）得11(0,,)22AE = 为平面PCD 的一个法向量．·······················································9分设平面PBC 的一个法向量为111()m x ,y ,z = ．

由0,0,PB m PC m ⎧⋅=⎪⎨⋅=⎪⎩ 得11111

0,0,x z x y z -=⎧⎨+-=⎩令11x =，解得11z =，10y =．所以(1,0,1)m = ．·····································································································10分因此112

cos ,2m AE m AE m AE ⋅===

⋅ ．·······························································11分由图可知二面角B PC D --的大小为钝角．

故二面角B PC D --的余弦值为12

-．·········································································12分解法二：（1）同解法一．·····································································································6分

（2）过点B 作BF 垂直于PC 于点F ，连接DF 、BD ．

因为PB PD =，BC CD =，PC PC =，

所以PBC PDC △≌△．······························································································7分因此易得090DFC BFC ∠=∠=，BF DF =．································································8分所以BFD ∠为二面角B PC D --的平面角．···································································9分设1PA AB ==，

则BD =

3

BF DF ==．·························································10分