歡迎閱讀本文檔，希望本文檔能對(duì)您有所幫助！歡迎閱讀本文檔，希望本文檔能對(duì)您有所幫助！歡迎閱讀本文檔，希望本文檔能對(duì)您有所幫助！歡迎閱讀本文檔，希望本文檔能對(duì)您有所幫助！歡迎閱讀本文檔，希望本文檔能對(duì)您有所幫助！歡迎閱讀本文檔，希望本文檔能對(duì)您有所幫助！1.1.3空間向量的坐標(biāo)與空間直角坐標(biāo)系A(chǔ)級(jí)必備知識(shí)基礎(chǔ)練1.已知向量a=(1,-2,1),a+b=(-1,2,-1),則向量b等于()A.(2,-4,2) B.(-2,4,-2)C.(-2,0,-2) D.(2,1,-3)2.已知a=(-3,2,5),b=(1,x,-1),且a·b=2,則x的值是()A.3 B.4C.5 D.63.(多選題)已知向量a=(1,1,0),則與a共線的單位向量e=()A.-2B.(0,1,0)C.22D.(-1,-1,0)4.若△ABC的三個(gè)頂點(diǎn)坐標(biāo)分別為A(1,-2,1),B(4,2,3),C(6,-1,4),則△ABC的形狀是()A.銳角三角形 B.直角三角形C.鈍角三角形 D.等邊三角形5.(多選題)對(duì)于任意非零向量a=(x1,y1,z1),b=(x2,y2,z2),以下說法錯(cuò)誤的有()A.若a⊥b,則x1x2+y1y2+z1z2=0B.若a∥b,則xC.cos<a,b>=xD.若x1=y1=z1=1,則a為單位向量6.若a=(2,-3,5),b=(-3,1,-4),則|a-2b|=.7.已知向量a=(5,3,1),b=-2,t,-25,若a與b的夾角為鈍角,則實(shí)數(shù)t的取值范圍為.

8.已知向量a=(1,2,-2),b=(4,-2,4),c=(3,m,n).(1)求a-b;(2)若a∥c,求m,n;(3)求cos<a,b>.B級(jí)關(guān)鍵能力提升練9.(多選題)已知點(diǎn)P是△ABC所在的平面外一點(diǎn),若AB=(-2,1,4),AP=(1,-2,1),AC=(4,2,0),則()A.AP⊥AB B.AP⊥BPC.BC=53 D.AP∥BC10.設(shè)x,y∈R,向量a=(x,1,1),b=(1,y,1),c=(2,-2,2),且a⊥c,b∥c,則|a+b|=()A.22 B.3 C.5 D.411.已知點(diǎn)A(1,-1,2),B(5,-6,2),C(1,3,-1),則AB在AC上的投影的數(shù)量為12.已知空間向量a=(1,-2,3),則向量a在坐標(biāo)平面xOy上的投影向量是.

13.已知A,B,C三點(diǎn)的坐標(biāo)分別是(2,-1,2),(4,5,-1),(-2,2,3),AP=12(AB?14.如圖所示,在四棱錐P-ABCD中,底面ABCD為矩形,側(cè)棱PA⊥底面ABCD,AB=3,BC=1,PA=2,E為PD的中點(diǎn).(1)求cos<AC,PB(2)在側(cè)面PAB內(nèi)找一點(diǎn)N,使NE⊥平面PAC,求N點(diǎn)的坐標(biāo).15.已知點(diǎn)A(0,2,3),B(-2,1,6),C(1,-1,5).(1)求以AB,AC為鄰邊的平行四邊形的面積;(2)若|a|=3,且向量a分別與向量AB,AC垂直,求向量C級(jí)學(xué)科素養(yǎng)創(chuàng)新練16.P是平面ABC外的點(diǎn),四邊形ABCD是平行四邊形,AB=(2,-1,-4),AD=(4,2,0),AP=(-1,2,-1).(1)求證:PA⊥平面ABCD;(2)對(duì)于向量a=(x1,y1,z1),b=(x2,y2,z2),c=(x3,y3,z3),定義一種運(yùn)算:(a×b)·c=x1y2z3+x2y3z1+x3y1z2-x1y3z2-x2y1z3-x3y2z1,試計(jì)算(AB×AD)·AP的絕對(duì)值,說明其與幾何體P-ABCD的體積關(guān)系,并由此猜想向量這種運(yùn)算(AB×AD)

1.1.3空間向量的坐標(biāo)與空間直角坐標(biāo)系1.B2.C因?yàn)閍=(-3,2,5),b=(1,x,-1),所以a·b=-3+2x-5=2,解得x=5.故選C.3.AC對(duì)A,存在實(shí)數(shù)λ=-2,使(1,1,0)=-2-22,-22,0,且-22,-22,0=12+12=1,故A對(duì)B,不存在實(shí)數(shù)λ,使(1,1,0)=λ(0,1,0),故B錯(cuò)誤;對(duì)C,存在實(shí)數(shù)λ=2,使(1,1,0)=222,22,0,且22,22,0=12+對(duì)D,|(-1,-1,0)|=1+1=2,不是單位向量,故D錯(cuò)誤.故選4.AAB=(3,4,2),AC=(5,1,3),BC=(2,-3,1).由AB·AC>0,得A為銳角;由CA·CB>0,得C為銳角;由BA·BC>0,得B為銳角5.BD對(duì)于A選項(xiàng),因?yàn)閍⊥b,則a·b=x1x2+y1y2+z1z2=0,A選項(xiàng)正確;對(duì)于B選項(xiàng),若x2=0,且y2≠0,z2≠0,若a∥b,分式x1x2無(wú)意義對(duì)于C選項(xiàng),由空間向量數(shù)量積的坐標(biāo)運(yùn)算可知cos<a,b>=x1x2對(duì)于D選項(xiàng),若x1=y1=z1=1,則|a|=12+12+12=3,此時(shí)6.258因?yàn)閍=(2,-3,5),b=(-3,1,-4),所以a-2b=(8,-5,13),所以|a-2b|=827.-∞,-65∪-65,5215由已知得a·b=5×(-2)+3t+1×-25=3t-525,因?yàn)閍與b的夾角為鈍角,所以a·b<0,即3t-525<0,所以t<若a與b的夾角為180°,則存在λ<0,使a=λb(λ<0),即(5,3,1)=λ-2,t,-25,所以5=-2故t的取值范圍是-∞,-65∪-65,52158.解(1)∵a=(1,2,-2),b=(4,-2,4),∴a-b=(1,2,-2)-(4,-2,4)=(1-4,2-(-2),-2-4)=(-3,4,-6).(2)∵a=(1,2,-2),c=(3,m,n),且a∥c,∴31=m2=n-(3)∵a=(1,2,-2),b=(4,-2,4),∴a·b=1×4+2×(-2)+(-2)×4=-8,|a|=12+22+(-2)2=3,|b|=42+9.ACAP·AB=-2-2+4=0,∴AP⊥AB,即AP⊥AB,BP=BA+AP=(2,-1,-4)+(1,-2,1)=(3,-3,-3),BP·AP=3+6-3=6≠0,∴AP與BPBC=AC?AB=(4,2,0)-(-2,1,4)∴|BC|=62+12+假設(shè)AP=kBC(k∈R),則1=6k,-2=k,1=-4k,無(wú)解,因此假設(shè)不成立10.C∵a⊥c,∴a·c=2x-2+2=0,∴x=0.又b∥c,∴12=y-∴a=(0,1,1),b=(1,-1,1),∴a+b=(0,1,1)+(1,-1,1)=(1,0,2),∴|a+b|=1+4=5.故選11.-4∵AB=(5,-6,2)-(1,-1,2)=(4,-5,0),AC=(1,3,-1)-(1,-1,2)=(0,4,-3),∴cos<AB,AC>=0-20+042+(-5)2×42+(-3)2=-2012.(1,-2,0)13.5,12,0∵CB=(6,3,-4),設(shè)P(a則(a-2,b+1,c-2)=3,∴a=5,b=12,c=0,∴P514.解(1)由題意,建立如圖所示的空間直角坐標(biāo)系,則A(0,0,0),B(3,0,0),C(3,1,0),D(0,1,0),P(0,0,2),E0,12,1,從而AC=(3,1,0),PB=(3,0,-2).則cos<AC,PB>=(2)由于N點(diǎn)在側(cè)面PAB內(nèi),故可設(shè)N點(diǎn)坐標(biāo)為(x,0,z),則NE=-x,12,1-z,由NE⊥平面PAC可得,NE化簡(jiǎn)得z即N點(diǎn)的坐標(biāo)為36,0,1.15.解(1)AB=(-2,-1,3),AC=(1,-3,2),設(shè)θ為AB,AC則cosθ=AB·∴sinθ=32.∴以AB,AC為鄰邊的平行四邊形的面積S=|AB||AC|sinθ=73(2)設(shè)a=(x,y/r/