2023版考前三个月冲刺专题练　第20练　空间向量与距离、探究性问题【无答案版】

第20练　空间向量与距离、探究性问题

[考情分析]　空间向量与距离、探究性问题在高考试题中出现较少，一般以解答题的形式考查，难度在中档以上．

一、空间距离

例1　(2022·吉林模拟)如图，在三棱柱ABC－A1B1C1中，侧棱AA1⊥底面A1B1C1，∠BAC＝90°，AB＝4，AC＝2，M是AB的中点，N是A1B1的中点，P是BC1与B1C的交点，点Q在线段C1N上．

(1)求证：PQ∥平面A1CM；

(2)若平面A1CM与平面ACM的夹角的余弦值是，求点B到平面A1CM的距离．

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规律方法　(1)点到直线的距离

直线l的单位方向向量为u，A是直线l上的任一点，P为直线l外一点，设＝a，则点P到直线l的距离d＝.

(2)点到平面的距离

平面α的法向量为n，A是平面α内任一点，P为平面α外一点，则点P到平面α的距离为d＝.

跟踪训练1　(2022·镇江模拟)如图所示，在三棱柱ABC－A1B1C1中，AB＝BC，点A1在平面ABC上的射影为线段AC的中点D，侧面AA1C1C是边长为2的菱形．

(1)若△ABC是正三角形，求异面直线DB1与BC所成角的余弦值；

(2)当直线CB1与平面ABB1A1所成角的正弦值为时，求线段BD的长．

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二、探究性问题

例2　(2021·全国甲卷改编)已知直三棱柱ABC－A1B1C1中，侧面AA1B1B为正方形，AB＝BC＝2，E，F分别为AC和CC1的中点，D为棱A1B1上的点，BF⊥A1B1.

(1)证明：BF⊥DE；

(2)当B1D为何值时，平面BB1C1C与平面DFE的夹角的正弦值最小？

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规律方法　空间向量求解探究性问题：

(1)假设题中的数学对象存在(或结论成立)或暂且认可其中的一部分结论；

(2)在这个前提下进行逻辑推理，把要成立的结论当作条件，据此列方程或方程组，把“是否存在”问题转化为“点的坐标(或参数)是否有解、是否有规定范围内的解”等．若由此推导出矛盾，则否定假设；否则，给出肯定结论．

跟踪训练2　(2022·北京丰台模拟)如图，在直角梯形ABCD中，AB∥CD，∠DAB＝90°，AD＝DC＝AB.以直线AB为轴，将直角梯形ABCD旋转得到直角梯形ABEF，且AF⊥AD.

(1)求证：DF∥平面BCE；

(2)在线段DF上是否存在点P，使得直线AE和平面BCP所成角的正弦值为？若存在，求出的值；若不存在，请说明理由．

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