2023版考前三个月冲刺专题练　第19练　空间向量与空间角【无答案版】

第19练　空间向量与空间角

[考情分析]　高考必考内容，常以空间几何体为载体考查空间角，是高考命题的重点，常与空间线、面关系的证明相结合，热点为平面与平面的夹角的求解，均以解答题的形式进行考查，难度主要体现在建立空间直角坐标系和准确计算上．题目难度为中档题．

一、异面直线所成的角

例1　(1)(2022·龙岩模拟)已知直三棱柱ABC－A1B1C1的所有棱长都相等，M为A1C1的中点，则AM与BC1所成角的正弦值为(　　)

A.  B.  C.  D.

(2)(2022·毕节模拟)在正四棱锥S－ABCD中，底面边长为2，侧棱长为4，点P是底面ABCD内一动点，且SP＝，则当A，P两点间距离最小时，直线BP与直线SC所成角的余弦值为(　　)

A.  B.  C.  D.

规律方法　(1)设直线l，m的方向向量分别为a＝(a1，b1，c1)，b＝(a2，b2，c2)，设l，m的夹角为θ，则cos θ＝＝.

(2)异面直线所成的角的范围为.

跟踪训练1　(1)(2022·丹东模拟)在三棱锥P－ABC中，PA⊥平面ABC，PA＝AB，△ABC是正三角形，M，N分别是AB，PC的中点，则直线MN，PB所成角的余弦值为(　　)

A.  B.  C.  D.

(2)(2022·长春模拟)在矩形ABCD中，O为BD的中点且AD＝2AB，将平面ABD沿对角线BD翻折至二面角A－BD－C的大小为90°，则直线AO与CD所成角的余弦值为(　　)

A.  B.  C.  D.

二、直线与平面所成的角

例2　(2022·全国甲卷)在四棱锥P－ABCD中，PD⊥底面ABCD，CD∥AB，AD＝DC＝CB＝1，AB＝2，DP＝.

(1)证明：BD⊥PA；

(2)求PD与平面PAB所成的角的正弦值．

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规律方法　(1)设直线l的方向向量为a＝(a1，b1，c1)，平面α的法向量为μ＝(a2，b2，c2)，设直线l与平面α的夹角为θ，则sin θ＝|cos〈a，μ〉|＝.

(2)线面角的范围为.

跟踪训练2　(2022·绍兴模拟)如图，在四棱台ABCD－A1B1C1D1中，底面ABCD为矩形，平面DCC1D1⊥平面ABCD，AD＝DD1＝D1C1＝C1C＝DC＝1.

(1)求证：BD1⊥DD1；

(2)求直线BD1和平面ABB1A1所成角的正弦值．

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三、平面与平面的夹角

例3　(2022·新高考全国Ⅰ改编)如图，直三棱柱ABC－A1B1C1的体积为4，△A1BC的面积为2.

(1)求A到平面A1BC的距离；

(2)设D为A1C的中点，AA1＝AB，平面A1BC⊥平面ABB1A1，求平面ABD与平面BCD夹角的正弦值．

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规律方法　(1)设平面α，β的法向量分别为μ＝(a3，b3，c3)，v＝(a4，b4，c4)，且平面α与平面β的夹角为θ，则cos θ＝|cos〈μ，v〉|＝.

(2)平面与平面的夹角的取值范围为.

跟踪训练3　(2022·重庆调研)如图，在四棱锥P－ABCD中，PA⊥平面ABCD，AC，BD相交于点N，DN＝2BN＝2，PA＝AC＝AD＝3，∠ADB＝30°.

(1)求证：AC⊥平面PAD；

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(2)若点M为PD的中点，求平面PAB与平面MAC夹角的正弦值．

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