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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】
展开这是一份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】,共3页。
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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