【文档说明】2024-2025学年精品同步试题 数学（选择性必修第一册 人教A版2019） 第1章 1-1-2 空间向量的数量积运算 Word版含解析.docx，共(8)页，310.825 KB，由小赞的店铺上传

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1.1.2空间向量的数量积运算课后训练巩固提升A组1.若a,b均为非零向量,则a·b=|a||b|是a与b共线的()A.充分不必要条件B.必要不充分条件C.充分必要条件D.既不充分又不必要条件解析:a·b=

|a||b|⇒cos<a,b>=1⇒<a,b>=0°,即a与b共线.反之不成立,因为当a与b反向共线时,a·b=-|a||b|.答案:A2.已知a+b+c=0,|a|=2,|b|=3,|c|=4,则a与b的夹角<a,b>的余弦值为()A.12B.√22C.34D.

14解析:∵a+b+c=0,∴a+b=-c.∴(a+b)2=|a|2+|b|2+2a·b=|c|2.∴a·b=32.∴cos<a,b>=𝑎·𝑏|𝑎||𝑏|=14.答案:D3.已知空间四边形ABCD的每条边和对角线的长都等于a,点E,F分别是BC,

AD的中点,则𝐴𝐸⃗⃗⃗⃗⃗·𝐴𝐹⃗⃗⃗⃗⃗的值为()A.a2B.12a2C.14a2D.√34a2解析:𝐴𝐸⃗⃗⃗⃗⃗·𝐴𝐹⃗⃗⃗⃗⃗=12(𝐴𝐵⃗⃗⃗⃗⃗+𝐴𝐶⃗⃗⃗⃗⃗)·12𝐴𝐷⃗⃗⃗⃗⃗=14

(𝐴𝐵⃗⃗⃗⃗⃗·𝐴𝐷⃗⃗⃗⃗⃗+𝐴𝐶⃗⃗⃗⃗⃗·𝐴𝐷⃗⃗⃗⃗⃗)=14(𝑎×𝑎×12+𝑎×𝑎×12)=14a2.答案:C4.设A,B,C,D是空间中不共面的四点,且满足𝐴𝐵⃗⃗⃗⃗⃗·𝐴𝐶⃗⃗⃗⃗⃗=0,𝐴𝐶⃗⃗⃗⃗⃗·

𝐴𝐷⃗⃗⃗⃗⃗=0,𝐴𝐵⃗⃗⃗⃗⃗·𝐴𝐷⃗⃗⃗⃗⃗=0,则△BCD是()A.钝角三角形B.锐角三角形C.直角三角形D.不确定解析:𝐵𝐶⃗⃗⃗⃗⃗·𝐵𝐷⃗⃗⃗⃗⃗⃗=(𝐴𝐶⃗⃗⃗⃗⃗−𝐴𝐵⃗⃗⃗⃗⃗)·(

𝐴𝐷⃗⃗⃗⃗⃗−𝐴𝐵⃗⃗⃗⃗⃗)=𝐴𝐶⃗⃗⃗⃗⃗·𝐴𝐷⃗⃗⃗⃗⃗−𝐴𝐶⃗⃗⃗⃗⃗·𝐴𝐵⃗⃗⃗⃗⃗−𝐴𝐵⃗⃗⃗⃗⃗·𝐴𝐷⃗⃗⃗⃗⃗+|𝐴𝐵⃗⃗⃗⃗⃗|2=|𝐴𝐵⃗⃗⃗⃗⃗|2>0,同理可证𝐶𝐵⃗⃗⃗

⃗⃗·𝐶𝐷⃗⃗⃗⃗⃗>0,𝐷𝐵⃗⃗⃗⃗⃗⃗·𝐷𝐶⃗⃗⃗⃗⃗>0.所以△BCD的每个内角均为锐角.故△BCD是锐角三角形.答案:B5.(多选题)已知四边形ABCD为矩形,PA⊥平面ABCD,连接AC,BD,PB,PC,PD,则下列各组向量中,数量积一定为零的是()A.

𝑃𝐶⃗⃗⃗⃗⃗与𝐵𝐷⃗⃗⃗⃗⃗⃗B.𝐷𝐴⃗⃗⃗⃗⃗与𝑃𝐵⃗⃗⃗⃗⃗C.𝑃𝐷⃗⃗⃗⃗⃗与𝐴𝐵⃗⃗⃗⃗⃗D.𝑃𝐴⃗⃗⃗⃗⃗与𝐶𝐷⃗⃗⃗⃗⃗解析:因为PA⊥平面ABCD,且CD⊂平面ABCD,所以PA⊥CD.故𝑃𝐴⃗⃗⃗⃗⃗·𝐶𝐷⃗⃗⃗⃗⃗=0.因为

AD⊥AB,PA⊥AD,且PA∩AB=A,所以AD⊥平面PAB.因为PB⊂平面PAB,所以AD⊥PB.故𝐷𝐴⃗⃗⃗⃗⃗·𝑃𝐵⃗⃗⃗⃗⃗=0.同理,𝑃𝐷⃗⃗⃗⃗⃗·𝐴𝐵⃗⃗⃗⃗⃗=0,𝑃𝐴⃗⃗⃗⃗⃗·𝐶𝐷⃗⃗⃗⃗⃗=0.因为PA⊥平面ABCD,BD⊂

平面ABCD,所以PA⊥BD.所以𝑃𝐶⃗⃗⃗⃗⃗·𝐵𝐷⃗⃗⃗⃗⃗⃗=(𝑃𝐴⃗⃗⃗⃗⃗+𝐴𝐶⃗⃗⃗⃗⃗)·𝐵𝐷⃗⃗⃗⃗⃗⃗=𝑃𝐴⃗⃗⃗⃗⃗·𝐵𝐷⃗⃗⃗⃗⃗⃗+𝐴𝐶⃗⃗⃗⃗⃗·𝐵𝐷⃗⃗⃗⃗⃗⃗=�

�𝐶⃗⃗⃗⃗⃗·𝐵𝐷⃗⃗⃗⃗⃗⃗.因为四边形ABCD为矩形,所以BD不一定与AC垂直.所以𝑃𝐶⃗⃗⃗⃗⃗与𝐵𝐷⃗⃗⃗⃗⃗⃗的数量积不一定为0.故选BCD,排除A.答案:BCD6.已知空间向量a,b,|a|

=3√2,|b|=5,m=a+b,n=a+λb,<a,b>=135°,若m⊥n,则λ的值为.解析:∵m⊥n,∴(a+b)·(a+λb)=0.∴m·n=0,即a2+λb2+(1+λ)a·b=0,即18+25λ+(1+λ)×3√2×5×cos135°=0,解得λ=-310.答案:-3

107.如图所示,在直三棱柱ABC-A1B1C1中,CA=CB=1,∠BCA=90°,AA1=2,则cos<𝐵𝐴1⃗⃗⃗⃗⃗⃗⃗⃗,𝐶𝐵1⃗⃗⃗⃗⃗⃗⃗>=.解析:𝐵𝐴1⃗⃗⃗⃗⃗⃗⃗⃗·𝐶𝐵1⃗⃗⃗⃗⃗⃗⃗=(𝐵𝐴⃗⃗⃗⃗⃗+𝐵𝐵1⃗⃗⃗⃗⃗

⃗⃗)·(𝐶𝐵⃗⃗⃗⃗⃗+𝐵𝐵1⃗⃗⃗⃗⃗⃗⃗)=𝐵𝐴⃗⃗⃗⃗⃗·𝐶𝐵⃗⃗⃗⃗⃗+𝐵𝐴⃗⃗⃗⃗⃗·𝐵𝐵1⃗⃗⃗⃗⃗⃗⃗+𝐵𝐵1⃗⃗⃗⃗⃗⃗⃗·𝐶𝐵⃗⃗⃗⃗⃗+𝐵𝐵1⃗

⃗⃗⃗⃗⃗⃗·𝐵𝐵1⃗⃗⃗⃗⃗⃗⃗,∵BB1⊥BA1,BB1⊥BC,∴𝐵𝐴⃗⃗⃗⃗⃗·𝐵𝐵1⃗⃗⃗⃗⃗⃗⃗=0,𝐵𝐵1⃗⃗⃗⃗⃗⃗⃗·𝐶𝐵⃗⃗⃗⃗⃗=0,又𝐵𝐴⃗⃗⃗⃗⃗·𝐶𝐵⃗⃗⃗⃗⃗=|𝐵𝐴⃗⃗⃗⃗⃗||𝐶𝐵⃗⃗⃗⃗⃗|·c

os(180°-∠ABC)=√2×1×cos135°=-1,𝐵𝐵1⃗⃗⃗⃗⃗⃗⃗·𝐵𝐵1⃗⃗⃗⃗⃗⃗⃗=4,∴𝐵𝐴1⃗⃗⃗⃗⃗⃗⃗⃗·𝐶𝐵1⃗⃗⃗⃗⃗⃗⃗=-1+0+0+4=3,又|𝐵𝐴1⃗⃗⃗⃗⃗⃗⃗⃗|·|𝐶𝐵1⃗⃗⃗⃗⃗⃗⃗|=√

6×√5=√30,∴cos<𝐵𝐴1⃗⃗⃗⃗⃗⃗⃗⃗,𝐶𝐵1⃗⃗⃗⃗⃗⃗⃗>=3√30=√3010.答案:√30108.如图,在平行六面体ABCD-A1B1C1D1中,AB=4,AD=3,AA1=5,∠BAD=90°,∠BAA1=∠DAA1=60°,则体对角线AC1的长度等于.解

析:|𝐴𝐶1⃗⃗⃗⃗⃗⃗⃗|2=(𝐴𝐵⃗⃗⃗⃗⃗+𝐴𝐷⃗⃗⃗⃗⃗+𝐴𝐴1⃗⃗⃗⃗⃗⃗⃗)2=|𝐴𝐵⃗⃗⃗⃗⃗|2+|𝐴𝐷⃗⃗⃗⃗⃗|2+|𝐴𝐴1⃗⃗⃗⃗⃗⃗⃗|2+2𝐴�

�⃗⃗⃗⃗⃗·𝐴𝐷⃗⃗⃗⃗⃗+2𝐴𝐵⃗⃗⃗⃗⃗·𝐴𝐴1⃗⃗⃗⃗⃗⃗⃗+2𝐴𝐷⃗⃗⃗⃗⃗·𝐴𝐴1⃗⃗⃗⃗⃗⃗⃗=16+9+25+2×4×3×cos90°+2×4×5×cos60°+2×3×5×cos60°=50+20+15=85,则|𝐴𝐶1⃗⃗⃗⃗⃗⃗⃗|=√85.答案:√

859.如图,在四棱锥P-ABCD中,底面ABCD为菱形,侧棱PA⊥底面ABCD,证明:PC⊥BD.证明:∵𝑃𝐶⃗⃗⃗⃗⃗=𝐴𝐵⃗⃗⃗⃗⃗+𝐴𝐷⃗⃗⃗⃗⃗−𝐴𝑃⃗⃗⃗⃗⃗,𝐵𝐷⃗⃗⃗⃗

⃗⃗=𝐴𝐷⃗⃗⃗⃗⃗−𝐴𝐵⃗⃗⃗⃗⃗,∴𝑃𝐶⃗⃗⃗⃗⃗·𝐵𝐷⃗⃗⃗⃗⃗⃗=(𝐴𝐵⃗⃗⃗⃗⃗+𝐴𝐷⃗⃗⃗⃗⃗−𝐴𝑃⃗⃗⃗⃗⃗)·(𝐴𝐷⃗⃗⃗⃗⃗−𝐴𝐵⃗⃗⃗⃗⃗)=(𝐴𝐵⃗⃗⃗⃗⃗+�

�𝐷⃗⃗⃗⃗⃗)·(𝐴𝐷⃗⃗⃗⃗⃗−𝐴𝐵⃗⃗⃗⃗⃗)-𝐴𝑃⃗⃗⃗⃗⃗·(𝐴𝐷⃗⃗⃗⃗⃗−𝐴𝐵⃗⃗⃗⃗⃗)=𝐴𝐷⃗⃗⃗⃗⃗2−𝐴𝐵⃗⃗⃗⃗⃗2−𝐴𝑃⃗⃗⃗⃗⃗·𝐴𝐷⃗⃗⃗⃗⃗+𝐴𝑃⃗⃗⃗⃗⃗·𝐴𝐵⃗⃗⃗⃗⃗.∵底面ABCD为菱形

,∴AD=AB,∴𝐴𝐷⃗⃗⃗⃗⃗2−𝐴𝐵⃗⃗⃗⃗⃗2=0.∵侧棱PA⊥底面ABCD,∴PA⊥AB,PA⊥AD,∴𝐴𝑃⃗⃗⃗⃗⃗·𝐴𝐷⃗⃗⃗⃗⃗=𝐴𝑃⃗⃗⃗⃗⃗·𝐴𝐵⃗⃗⃗⃗⃗=0,∴𝑃𝐶⃗⃗⃗⃗⃗·𝐵𝐷⃗⃗⃗⃗⃗⃗=𝐴𝐷⃗

⃗⃗⃗⃗2−𝐴𝐵⃗⃗⃗⃗⃗2−𝐴𝑃⃗⃗⃗⃗⃗·𝐴𝐷⃗⃗⃗⃗⃗+𝐴𝑃⃗⃗⃗⃗⃗·𝐴𝐵⃗⃗⃗⃗⃗=0,∴PC⊥BD.10.如图,已知线段AB在平面α内,线段AC⊥α于点A,线段BD⊥AB

于点B,线段DD'⊥α于点D',如果∠DBD'=30°,AB=a,AC=BD=b,求点C,D间的距离.解:|𝐶𝐷⃗⃗⃗⃗⃗|2=(𝐶𝐴⃗⃗⃗⃗⃗+𝐴𝐵⃗⃗⃗⃗⃗+𝐵𝐷⃗⃗⃗⃗⃗⃗)2=|𝐶𝐴⃗⃗⃗⃗⃗

|2+|𝐴𝐵⃗⃗⃗⃗⃗|2+|𝐵𝐷⃗⃗⃗⃗⃗⃗|2+2(𝐶𝐴⃗⃗⃗⃗⃗·𝐴𝐵⃗⃗⃗⃗⃗+𝐶𝐴⃗⃗⃗⃗⃗·𝐵𝐷⃗⃗⃗⃗⃗⃗+𝐴𝐵⃗⃗⃗⃗⃗·𝐵𝐷⃗⃗⃗⃗⃗⃗).∵AC⊥α,且AB⊂α,∴AC⊥AB.∴𝐶𝐴⃗⃗⃗⃗⃗·

𝐴𝐵⃗⃗⃗⃗⃗=0.又∠DBD'=30°,AC⊥α,DD'⊥α,∴<𝐶𝐴⃗⃗⃗⃗⃗,𝐵𝐷⃗⃗⃗⃗⃗⃗>=60°.又BD⊥AB,∴𝐴𝐵⃗⃗⃗⃗⃗·𝐵𝐷⃗⃗⃗⃗⃗⃗=0.∴|𝐶𝐷⃗⃗⃗⃗⃗|2=b2+a2+b2+2(0+b2cos60°

+0)=a2+3b2.∴|𝐶𝐷⃗⃗⃗⃗⃗|=√𝑎2+3𝑏2,即点C,D间的距离为√𝑎2+3𝑏2.B组1.已知棱长为1的正方体ABCD-A1B1C1D1的上底面A1B1C1D1的中心为O1,则𝐴𝑂1⃗⃗⃗⃗⃗⃗⃗·𝐴𝐶⃗⃗

⃗⃗⃗的值为()A.-1B.0C.1D.2解析:∵𝐴𝑂1⃗⃗⃗⃗⃗⃗⃗=𝐴𝐴1⃗⃗⃗⃗⃗⃗⃗+𝐴1𝑂1⃗⃗⃗⃗⃗⃗⃗⃗⃗=𝐴𝐴1⃗⃗⃗⃗⃗⃗⃗+12𝐴𝐶⃗⃗⃗⃗⃗,∴𝐴𝑂1⃗⃗⃗⃗⃗⃗⃗·𝐴𝐶⃗⃗⃗⃗⃗=(𝐴𝐴1⃗⃗⃗⃗⃗⃗⃗+12�

�𝐶⃗⃗⃗⃗⃗)·𝐴𝐶⃗⃗⃗⃗⃗=𝐴𝐴1⃗⃗⃗⃗⃗⃗⃗·𝐴𝐶⃗⃗⃗⃗⃗+12|𝐴𝐶⃗⃗⃗⃗⃗|2=1.故选C.答案:C2.如图,在正四面体ABCD中,E是BC的中点,那么()A.𝐴𝐸⃗⃗⃗⃗⃗·𝐵𝐶⃗⃗⃗⃗⃗<𝐴𝐸⃗⃗⃗⃗⃗·𝐶𝐷⃗⃗

⃗⃗⃗B.𝐴𝐸⃗⃗⃗⃗⃗·𝐵𝐶⃗⃗⃗⃗⃗=𝐴𝐸⃗⃗⃗⃗⃗·𝐶𝐷⃗⃗⃗⃗⃗C.𝐴𝐸⃗⃗⃗⃗⃗·𝐵𝐶⃗⃗⃗⃗⃗>𝐴𝐸⃗⃗⃗⃗⃗·𝐶𝐷⃗⃗⃗⃗⃗D.𝐴𝐸⃗⃗⃗⃗⃗·𝐵𝐶⃗⃗⃗⃗⃗与𝐴�

�⃗⃗⃗⃗⃗·𝐶𝐷⃗⃗⃗⃗⃗不能比较大小解析:∵𝐴𝐸⃗⃗⃗⃗⃗·𝐵𝐶⃗⃗⃗⃗⃗=12(𝐴𝐵⃗⃗⃗⃗⃗+𝐴𝐶⃗⃗⃗⃗⃗)·(𝐴𝐶⃗⃗⃗⃗⃗−𝐴𝐵⃗⃗⃗⃗⃗)=12(|𝐴𝐶⃗⃗⃗⃗⃗|2

-|𝐴𝐵⃗⃗⃗⃗⃗|2)=0,𝐴𝐸⃗⃗⃗⃗⃗·𝐶𝐷⃗⃗⃗⃗⃗=12(𝐴𝐵⃗⃗⃗⃗⃗+𝐴𝐶⃗⃗⃗⃗⃗)·(𝐴𝐷⃗⃗⃗⃗⃗−𝐴𝐶⃗⃗⃗⃗⃗)=12𝐴𝐵⃗⃗⃗⃗⃗·𝐴𝐷⃗⃗⃗⃗⃗−12𝐴𝐵⃗⃗⃗⃗

⃗·𝐴𝐶⃗⃗⃗⃗⃗+12𝐴𝐶⃗⃗⃗⃗⃗·𝐴𝐷⃗⃗⃗⃗⃗−12|𝐴𝐶⃗⃗⃗⃗⃗|2=12|𝐴𝐶⃗⃗⃗⃗⃗|2(cos60°-1)<0.∴𝐴𝐸⃗⃗⃗⃗⃗·𝐵𝐶⃗⃗⃗⃗⃗>𝐴𝐸⃗⃗⃗⃗⃗·𝐶𝐷⃗⃗⃗⃗⃗.答案:C3.(多选题)已知正方体ABCD-A1B1

C1D1,下列四个结论中,正确的是()A.(𝐴𝐴1⃗⃗⃗⃗⃗⃗⃗+𝐴𝐷⃗⃗⃗⃗⃗+𝐴𝐵⃗⃗⃗⃗⃗)2=3|𝐴𝐵⃗⃗⃗⃗⃗|2B.𝐴1𝐶⃗⃗⃗⃗⃗⃗⃗·(𝐴1𝐵1⃗⃗⃗⃗⃗⃗⃗⃗⃗−𝐴1

𝐴⃗⃗⃗⃗⃗⃗⃗)=0C.𝐴𝐷1⃗⃗⃗⃗⃗⃗⃗与𝐴1𝐵⃗⃗⃗⃗⃗⃗⃗的夹角为60°D.正方体的体积为|𝐴𝐵⃗⃗⃗⃗⃗·𝐴𝐴1⃗⃗⃗⃗⃗⃗⃗·𝐴𝐷⃗⃗⃗⃗⃗|解析:如图所示,(𝐴𝐴1⃗⃗⃗⃗⃗⃗⃗+𝐴𝐷⃗⃗

⃗⃗⃗+𝐴𝐵⃗⃗⃗⃗⃗)2=(𝐴𝐴1⃗⃗⃗⃗⃗⃗⃗+𝐴1𝐷1⃗⃗⃗⃗⃗⃗⃗⃗⃗+𝐷1𝐶1⃗⃗⃗⃗⃗⃗⃗⃗⃗)2=|𝐴𝐶1⃗⃗⃗⃗⃗⃗⃗|2=3|𝐴𝐵⃗⃗⃗⃗⃗|2,故A正确;𝐴1𝐶⃗⃗⃗⃗⃗⃗⃗·(𝐴1𝐵1⃗⃗⃗⃗⃗⃗⃗⃗⃗−

𝐴1𝐴⃗⃗⃗⃗⃗⃗⃗)=𝐴1𝐶⃗⃗⃗⃗⃗⃗⃗·𝐴𝐵1⃗⃗⃗⃗⃗⃗⃗.因为AB1⊥平面A1BC,A1C⊂平面A1BC,所以AB1⊥A1C,所以𝐴1𝐶⃗⃗⃗⃗⃗⃗⃗·𝐴𝐵1⃗⃗⃗⃗⃗⃗⃗=0,故B正确;𝐴𝐷1⃗⃗⃗⃗⃗⃗⃗与𝐴1𝐵⃗⃗⃗⃗⃗⃗⃗的夹角是𝐷1

𝐶⃗⃗⃗⃗⃗⃗⃗与𝐷1𝐴⃗⃗⃗⃗⃗⃗⃗夹角的补角,而𝐷1𝐶⃗⃗⃗⃗⃗⃗⃗与𝐷1𝐴⃗⃗⃗⃗⃗⃗⃗的夹角为60°,故𝐴𝐷1⃗⃗⃗⃗⃗⃗⃗与𝐴1𝐵⃗⃗⃗⃗⃗⃗⃗的夹角为120°,故C错误;正方体的体积为|𝐴𝐵⃗⃗⃗⃗⃗||𝐴𝐴1⃗⃗⃗

⃗⃗⃗⃗||𝐴𝐷⃗⃗⃗⃗⃗|,故D错误.答案:AB4.已知|a|=2,|b|=1,<a,b>=60°,则使向量a+λb与λa-2b的夹角为钝角的实数λ的取值范围是.解析:由题意知{(𝑎+𝜆𝑏)·(𝜆𝑎-2𝑏)<0,①cos<𝑎+𝜆𝑏,𝜆𝑎-2𝑏>≠-1.②由①得

λ2+2λ-2<0,解得-1-√3<λ<-1+√3.当a+λb与λa-2b反向共线时,存在实数k<0,使a+λb=k(λa-2b),即{𝑘𝜆=1,𝜆=-2𝑘,无解.所以不存在a+λb与λa-2b反向共线的情况,②始终成立.故实数λ的

取值范围为(-1-√3,-1+√3).答案:(-1-√3,-1+√3)5.在四面体OABC中,棱OA,OB,OC两两垂直,且OA=1,OB=2,OC=3,G为△ABC的重心,则𝑂𝐺⃗⃗⃗⃗⃗·(𝑂𝐴⃗⃗⃗⃗⃗+𝑂𝐵⃗⃗⃗⃗⃗+𝑂𝐶⃗⃗⃗

⃗⃗)=.解析:由已知得𝑂𝐴⃗⃗⃗⃗⃗·𝑂𝐵⃗⃗⃗⃗⃗=𝑂𝐴⃗⃗⃗⃗⃗·𝑂𝐶⃗⃗⃗⃗⃗=𝑂𝐵⃗⃗⃗⃗⃗·𝑂𝐶⃗⃗⃗⃗⃗=0.如图,取BC的中点D,连接OD,AD,则AD过点G,且AG=23AD.𝑂𝐺⃗⃗⃗⃗⃗=𝑂𝐴⃗⃗⃗⃗⃗+𝐴𝐺

⃗⃗⃗⃗⃗=𝑂𝐴⃗⃗⃗⃗⃗+23𝐴𝐷⃗⃗⃗⃗⃗=𝑂𝐴⃗⃗⃗⃗⃗+23(𝑂𝐷⃗⃗⃗⃗⃗⃗−𝑂𝐴⃗⃗⃗⃗⃗)=13𝑂𝐴⃗⃗⃗⃗⃗+23×12(𝑂𝐵⃗⃗⃗⃗⃗+𝑂𝐶⃗⃗⃗⃗⃗)=13𝑂𝐴⃗⃗⃗⃗⃗+13𝑂𝐵⃗⃗⃗⃗⃗+13

𝑂𝐶⃗⃗⃗⃗⃗.𝑂𝐺⃗⃗⃗⃗⃗·(𝑂𝐴⃗⃗⃗⃗⃗+𝑂𝐵⃗⃗⃗⃗⃗+𝑂𝐶⃗⃗⃗⃗⃗)=13(𝑂𝐴⃗⃗⃗⃗⃗+𝑂𝐵⃗⃗⃗⃗⃗+𝑂𝐶⃗⃗⃗⃗⃗)2=13(|𝑂𝐴⃗⃗⃗⃗⃗|2+|𝑂𝐵⃗⃗⃗⃗⃗|2+|𝑂𝐶⃗

⃗⃗⃗⃗|2)=13×(1+4+9)=143.答案:1436.已知正三棱柱ABC-DEF的侧棱长为2,底面边长为1,M是BC的中点,若CF上有一点N,使MN⊥AE,则𝐶𝑁𝐶𝐹=.解析:如图,设𝐶𝑁𝐶𝐹=m.∵𝐴𝐸⃗⃗⃗⃗⃗=𝐴𝐵⃗⃗⃗⃗⃗+𝐵𝐸⃗⃗⃗⃗

⃗,𝑀𝑁⃗⃗⃗⃗⃗⃗⃗=12𝐵𝐶⃗⃗⃗⃗⃗+m𝐴𝐷⃗⃗⃗⃗⃗,∴𝐴𝐸⃗⃗⃗⃗⃗·𝑀𝑁⃗⃗⃗⃗⃗⃗⃗=(𝐴𝐵⃗⃗⃗⃗⃗+𝐵𝐸⃗⃗⃗⃗⃗)·(12𝐵𝐶⃗⃗⃗⃗⃗+𝑚𝐴𝐷⃗⃗⃗⃗⃗)=12×1×1×(-12)+4m=0.

∴m=116.答案:1167.如图,在直三棱柱ABC-A1B1C1中,∠ABC=90°,AB=BC=1,AA1=√2,求𝐵𝐴1⃗⃗⃗⃗⃗⃗⃗⃗与𝐴𝐶⃗⃗⃗⃗⃗的夹角的余弦值.解:∵𝐵𝐴1⃗⃗⃗⃗⃗⃗⃗⃗=𝐵𝐴⃗⃗⃗⃗⃗+𝐵𝐵1⃗⃗⃗⃗⃗⃗⃗,𝐴𝐶⃗⃗⃗⃗⃗=�

�𝐶⃗⃗⃗⃗⃗−𝐵𝐴⃗⃗⃗⃗⃗,且𝐵𝐴⃗⃗⃗⃗⃗·𝐵𝐶⃗⃗⃗⃗⃗=𝐵𝐵1⃗⃗⃗⃗⃗⃗⃗·𝐵𝐴⃗⃗⃗⃗⃗=𝐵𝐵1⃗⃗⃗⃗⃗⃗⃗·𝐵𝐶⃗⃗⃗⃗⃗=0,∴𝐵𝐴1⃗⃗⃗⃗⃗⃗⃗⃗·𝐴𝐶⃗⃗⃗⃗⃗=-|𝐵𝐴⃗⃗⃗⃗⃗|2=-1.又|�

�𝐶⃗⃗⃗⃗⃗|=√2,|𝐵𝐴1⃗⃗⃗⃗⃗⃗⃗⃗|=√1+2=√3,∴cos<𝐵𝐴1⃗⃗⃗⃗⃗⃗⃗⃗,𝐴𝐶⃗⃗⃗⃗⃗>=𝐵𝐴1⃗⃗⃗⃗⃗⃗⃗⃗·𝐴𝐶⃗⃗⃗⃗⃗|𝐵𝐴1⃗⃗⃗⃗⃗⃗⃗⃗||𝐴𝐶⃗⃗⃗⃗⃗|=-1√6=-√66.故𝐵𝐴1⃗⃗⃗⃗⃗⃗⃗⃗与𝐴𝐶

⃗⃗⃗⃗⃗的夹角的余弦值为-√66.8.如图所示,在正三棱柱ABC-A1B1C1中,底面边长为√2.(1)设侧棱长为1,求证:AB1⊥BC1;(2)设AB1与BC1的夹角为π3,求侧棱长.答案：(1)证明:𝐴𝐵1⃗⃗⃗⃗⃗⃗⃗=𝐴𝐵⃗⃗⃗⃗⃗+𝐵𝐵1⃗⃗⃗⃗⃗⃗⃗,𝐵𝐶1⃗⃗⃗⃗⃗

⃗⃗=𝐵𝐵1⃗⃗⃗⃗⃗⃗⃗+𝐵𝐶⃗⃗⃗⃗⃗.∵BB1⊥平面ABC,∴BB1⊥AB,BB1⊥BC.∴𝐵𝐵1⃗⃗⃗⃗⃗⃗⃗·𝐴𝐵⃗⃗⃗⃗⃗=0,𝐵𝐵1⃗⃗⃗⃗⃗⃗⃗·𝐵𝐶⃗⃗⃗⃗⃗=0.又△ABC为正三角形,

∴<𝐴𝐵⃗⃗⃗⃗⃗,𝐵𝐶⃗⃗⃗⃗⃗>=π-<𝐵𝐴⃗⃗⃗⃗⃗,𝐵𝐶⃗⃗⃗⃗⃗>=π-π3=2π3.∵𝐴𝐵1⃗⃗⃗⃗⃗⃗⃗·𝐵𝐶1⃗⃗⃗⃗⃗⃗⃗=(𝐴𝐵⃗⃗⃗⃗⃗+𝐵𝐵1⃗⃗⃗⃗⃗⃗⃗)·(𝐵𝐵1⃗⃗⃗⃗⃗⃗⃗+𝐵𝐶⃗⃗⃗⃗⃗)=𝐴𝐵⃗⃗⃗⃗⃗·𝐵

𝐵1⃗⃗⃗⃗⃗⃗⃗+𝐴𝐵⃗⃗⃗⃗⃗·𝐵𝐶⃗⃗⃗⃗⃗+|𝐵𝐵1⃗⃗⃗⃗⃗⃗⃗|2+𝐵𝐵1⃗⃗⃗⃗⃗⃗⃗·𝐵𝐶⃗⃗⃗⃗⃗=|𝐴𝐵⃗⃗⃗⃗⃗||𝐵𝐶⃗⃗⃗⃗⃗|·cos<𝐴𝐵⃗⃗⃗⃗⃗,𝐵𝐶⃗⃗⃗⃗⃗>+|𝐵𝐵1⃗⃗⃗⃗⃗⃗⃗|2=-1+1=0,∴�

�𝐵1⃗⃗⃗⃗⃗⃗⃗⊥𝐵𝐶1⃗⃗⃗⃗⃗⃗⃗.∴AB1⊥BC1.(2)解由(1)知𝐴𝐵1⃗⃗⃗⃗⃗⃗⃗·𝐵𝐶1⃗⃗⃗⃗⃗⃗⃗=|𝐴𝐵⃗⃗⃗⃗⃗||𝐵𝐶⃗⃗⃗⃗⃗|cos<𝐴𝐵⃗⃗⃗

⃗⃗,𝐵𝐶⃗⃗⃗⃗⃗>+|𝐵𝐵1⃗⃗⃗⃗⃗⃗⃗|2=|𝐵𝐵1⃗⃗⃗⃗⃗⃗⃗|2-1.又|𝐴𝐵1⃗⃗⃗⃗⃗⃗⃗|=√(𝐴𝐵⃗⃗⃗⃗⃗+𝐵𝐵1⃗⃗⃗⃗⃗⃗⃗)2=√|AB⃗⃗⃗⃗⃗|2+|BB1⃗⃗⃗⃗⃗⃗⃗|2=√2+|𝐵𝐵1

⃗⃗⃗⃗⃗⃗⃗|2,|𝐵𝐶1⃗⃗⃗⃗⃗⃗⃗|=√2+|𝐵𝐵1⃗⃗⃗⃗⃗⃗⃗|2,∴cos<𝐴𝐵1⃗⃗⃗⃗⃗⃗⃗,𝐵𝐶1⃗⃗⃗⃗⃗⃗⃗>=|𝐵𝐵1⃗⃗⃗⃗⃗⃗⃗⃗|2-12+|𝐵𝐵1⃗⃗⃗⃗⃗⃗⃗⃗

|2=cosπ3=12.∴|𝐵𝐵1⃗⃗⃗⃗⃗⃗⃗|=2,即侧棱长为2.