新教材適用2024-2025學(xué)年高中數(shù)學(xué)第3章空間向量與立體幾何3空間向量基本定理及向量的直角坐標(biāo)運(yùn)算3.2空間向量運(yùn)算的坐標(biāo)表示及應(yīng)用課后訓(xùn)練北師大版選擇性必修第一冊(cè)

3.2空間向量運(yùn)算的坐標(biāo)表示及應(yīng)用A組1.已知向量a=(1,1,0),b=(-1,0,2),則|3a+b|等于().A.15 B.4 C.5 D.172.已知點(diǎn)A(3,3,3),B(6,6,6),O為原點(diǎn),則向量OA與BO的夾角是(A.0 B.π C.323.設(shè)點(diǎn)A(3,3,1),B(1,0,5),C(0,1,0),則AB的中點(diǎn)M到點(diǎn)C的距離|CM|的值為().A.534 B.532 C.534.已知A(4,1,3),B(2,-5,1),C為線段AB上一點(diǎn),且AB=3AC,則C的坐標(biāo)為().A.72,-12,52 B.83,C.103,-1,73 D.52,-725.如圖,F是正方體ABCD-A1B1C1D1的棱CD的中點(diǎn).E是BB1上一點(diǎn),若D1F⊥DE,則有().(第5題)A.B1E=EB B.B1E=2EBC.B1E=12EB D.E與B6.(多選題)已知向量a=(4,-2,-4),b=(6,-3,2),則下列結(jié)論正確的有().A.a+b=(10,-5,-2)B.a-b=(2,-1,-6)C.a·b=22D.|a|=67.已知向量a=(1,λ,2),b=(2,-1,2),且a,b夾角的余弦值為89,則實(shí)數(shù)λ等于8.如圖,在棱長(zhǎng)為a的正方體ABCD-A1B1C1D1中,以D為坐標(biāo)原點(diǎn),DA,DC,DD1所在直線分別為x軸、y軸、z軸,建立空間直角坐標(biāo)系,過(guò)點(diǎn)B作BM⊥AC1于點(diǎn)M,求點(diǎn)M的坐標(biāo).(第8題)9.已知四邊形ABCD的頂點(diǎn)分別是A(3,-1,2),B(1,2,-1),C(-1,1,-3),D(3,-5,3),求證:四邊形ABCD是一個(gè)梯形.B組1.從點(diǎn)P(1,2,3)動(dòng)身,沿著向量v=(-4,-1,8)方向取點(diǎn)Q,使|PQ|=18,則點(diǎn)Q的坐標(biāo)為().A.(-7,0,19)B.(9,4,-13)C.(-7,0,19)或(9,4,-13)D.(-1,-2,3)或(1,-2,-3)2.已知空間向量a=(1,1,0),b=(-1,0,2),則與向量a+b方向相反的單位向量e的坐標(biāo)是().A.(0,1,2) B.(0,-1,-2)C.0,15,25 D.0,-15,-3.已知點(diǎn)A(1,0,0),B(0,-1,1),O(0,0,0),向量OA+λOB與OB的夾角為120°,則實(shí)數(shù)λ的值為(A.±66 B.66 C.-64.已知O為坐標(biāo)原點(diǎn),OA=(1,2,3),OB=(2,1,2),OP=(1,1,2),點(diǎn)Q在直線OP上運(yùn)動(dòng),則當(dāng)QA·QB取得最小值時(shí),點(diǎn)Q的坐標(biāo)為(A.12,34,1C.43,43,85.若△ABC的三個(gè)頂點(diǎn)分別為A(0,0,2),B-32,12,2,C(-1,0,2),則∠6.已知點(diǎn)A(λ+1,μ-1,3),B(2λ,μ,λ-2μ),C(λ+3,μ-3,9)三點(diǎn)共線,則實(shí)數(shù)λ=,μ=.

7.如圖,在棱長(zhǎng)為1的正方體OABC-O1A1B1C1中,E,F分別是棱AB,BC上的動(dòng)點(diǎn),且AE=BF=x,其中0≤x≤1,以O(shè)為原點(diǎn)建立空間直角坐標(biāo)系O-xyz.(第7題)(1)求證:A1F⊥C1E;(2)若A1,E,F,C1四點(diǎn)共面,求證:A18.如圖,在四棱錐P-ABCD中,PA⊥底面ABCD,AB⊥AD,AB+AD=4,CD=2,∠CDA=45°.設(shè)AB=AP,在線段AD上是否存在一個(gè)點(diǎn)G,使得點(diǎn)G到點(diǎn)P,B,C,D的距離都相等?并說(shuō)明理由.(第8題)

參考答案3.2空間向量運(yùn)算的坐標(biāo)表示及應(yīng)用A組1.D3a+b=3(1,1,0)+(-1,0,2)=(3,3,0)+(-1,0,2)=(2,3,2),故|3a+b|=4+9+4=2.BOA=(3,3,3),BO=(-6,-6,-6),則OA·BO=3×3×(-6)=-54,|OA|=33,|BO|=63,所以cos<OA,BO>=OA·BO|OA3.C∵AB的中點(diǎn)為M2,32,3,∴CM=2,12,3,故|CM|=|CM|=224.C設(shè)C(x,y,z),則AC=(x-4,y-1,z-3).又AB=(-2,-6,-2),AB=3AC,∴(-2,-6,-2)=(3x-12,3y-3,3z-9).∴3x-5.A如圖,分別以DA,DC,DD1為x軸、y軸、z軸,建立空間直角坐標(biāo)系.設(shè)正方體的棱長(zhǎng)為2,則D(0,0,0),F(0,1,0),D1(0,0,2).設(shè)E(2,2,z),則D1F=(0,1,-2),DE=(2,2,z),所以D1F·DE=0×2+1×2-2z=0,所以(第5題)6.ACDa+b=(10,-5,-2),a-b=(-2,1,-6),a·b=24+6-8=22,|a|=42+(-7.-2或255cos<a,b>=a1×解得λ=-2或λ=2558.解由題意知A(a,0,0),B(a,a,0),C1(0,a,a).所以AC1=(-a,a,a設(shè)M(x,y,z),則AM=(x-a,y,z),BM=(x-a,y-a,z).因?yàn)锽M⊥AC1,所以所以-a(x-a)+a(y-a)+az=0,即x-y-z=0. ①因?yàn)锳,M,C1三點(diǎn)共線,所以可令A(yù)M=λAC所以x-a=-λa,y=λa,z=λa(λ∈R),即x=a-λa,y=λa,z=λa. ②由①②,得x=2a3,y=a3,所以點(diǎn)M的坐標(biāo)為2a3,9.證明因?yàn)锳B=(1,2,-1)-(3,-1,2)=(-2,3,-3),CD=(3,-5,3)-(-1,1,-3)=(4,-6,6),-24=3-又因?yàn)锳B與CD不共線,所以AB∥CD.所以A,B,C,D四點(diǎn)共面.因?yàn)锳D=(3,-5,3)-(3,-1,2)=(0,-4,1),BC=(-1,1,-3)-(1,2,-1)=(-2,-1,-2),0-2≠-4所以四邊形ABCD為梯形.B組1.A設(shè)Q(x0,y0,z0),則存在實(shí)數(shù)λ,使得PQ=λv,即(x0-1,y0-2,z0-3)=λ(-4,-1,8).由|PQ|=18,得(-4λ)2+(-因?yàn)镻Q與v同方向,所以λ=2.所以(x0-1,y0-2,z0-3)=2(-4,-1,8),解得x2.D∵a=(1,1,0),b=(-1,0,2),∴a+b=(0,1,2),|a+b|=5,∴與向量a+b方向相反的單位向量e的坐標(biāo)是-15(0,1,2)=0,-15,-253.C∵OA=(1,0,0),OB=(0,-1,1),∴OA+λOB=(1,-λ,λ).∴(OA+λOB)·OB=λ+λ=2λ,|OA+λOB|=1+λ2+λ2=∴cos120°=(OA+λOB)·OB|OA+λOB||OB|=2λ4.C由題意知存在實(shí)數(shù)λ,使得OQ=λOP,則QA=OA-OQ=OA-λOP=(1-λ,2-λ,3-2λ),QB=OB-OQ=OB-λOP=(2-λ,1-λ,2-2λ),所以QA·QB=(1-λ,2-λ,3-2λ)·(2-λ,1-λ,2-2λ)=2(3λ2-8λ+所以當(dāng)λ=43時(shí),QA·QB最小,此時(shí)OQ=43OP=43,45.30°由題意,知AB=-32,12,0,AC=(-1,0,0),所以|AB|=1,|AC|=所以cosA=AB·AC|AB||AC6.00AB=(λ-1,1,λ-2μ-3),AC=(2,-2,6),由A,B,C三點(diǎn)共線,得AB∥AC,即λ-12=-12=7.證明(1)由已知條件,得A1(1,0,1),F(1-x,1,0),C1(0,1,1),E(1,x,0),則A1F=(-x,1,-1),C1E=(1,所以A1F·C1E=-x+(所以A1F⊥C1E,即A(2)由(1)知,A1F=(-x,1,-1),A1C1=(-1,1,0),A1E=(0,x,-1).設(shè)A解得λ=12,μ=1所以A18.解因?yàn)镻A⊥平面ABCD,且AB?平面ABCD,AD?平面ABCD,所以PA⊥AB,PA⊥AD.又AB⊥AD,所以AP,AB,AD兩兩垂直.(第8題)如圖,以A為坐標(biāo)原點(diǎn),建立空間直角坐標(biāo)系A(chǔ)-xyz.在平面ABCD內(nèi),作CE∥AB交AD于點(diǎn)E,則CE⊥AD.在Rt△CDE中,∠CDE=45°,CD=2,則CE=DE=1.設(shè)AB=AP=t,則B(t,0,0),P(0,0,t).由AB+AD=4,得AD=4-t,所以E(0,3-t,0),C(1,3-t,0),D(0,4-t,0).假設(shè)在線段AD上存在一個(gè)點(diǎn)G,使得點(diǎn)G到點(diǎn)P,B,C,D的距離都相等.設(shè)G(0,m,0)(0≤m≤4-t)