1.1.2　空間向量的數(shù)量積運(yùn)算 練習(xí)冊(cè)答案

1.1.2空間向量的數(shù)量積運(yùn)算1.B[解析]當(dāng)<a,b>=π時(shí),a·b<0,但<a,b>不是鈍角,即由“a·b<0”不能推出“<a,b>為鈍角”,又當(dāng)<a,b>為鈍角時(shí),a·b<0,所以“a·b<0”是“<a,b>為鈍角”的必要不充分條件.故選B.2.B[解析]AC·BD=(AB+BD+DC)·BD=AB·BD+BD2+DC·BD.因?yàn)锳B⊥BD,CD⊥BD,所以AB·BD=0,DC·BD=0,所以AC·BD=0+12+0=1,故選B3.B[解析]由正方體的性質(zhì)可得,AB⊥AD,AB⊥AA1,故AB·AD=0,AB·AA1=0,∵AB=a,AD=b,AA1=c,∴a·(b+c)=AB·(AD+AA1)=AB·AD+AB4.D[解析]如圖,連接BD,A'D,則B'D'=BD,∴<A'B,B'D'>是∠DBA'的補(bǔ)角.∵A'D=A'B=BD,∴∠DBA'=60°,∴<A'5.C[解析]由已知得PC=PA+AB+BC,所以|PC|2=(PA+AB+BC)2=PA2+AB2+BC2+2PA·AB+2PA·BC+2AB·BC=22+12+32+2×2×1×-12+2×2×3×-12+2×1×3×-12=3,所以|PC|=6.D[解析]不妨設(shè)平行六面體ABCD-A1B1C1D1的各條棱長(zhǎng)均為1,則由題意得AB·AD=12,AB·AA1=12,AD·AA1=12,BD1=BA+AD+DD1=-AB+AD+AA1,AC=AB+AD,所以BD1·AC=(-AB+AD+AA1)·(AB+AD)=-AB2+AD2-AB·AD+AB·AD+AB·AAB21+1+1-1-1+1=2,|AC|=(AB+AD)2=AB2+AD2+2AB·AD=1+1+1=3,所以cos<BD1,7.A[解析]由題意可知柏拉圖多面體EABCDF的側(cè)面均為等邊三角形,四邊形ABCD為邊長(zhǎng)為1的菱形,如圖,連接AC,BD,又△AEC≌△BED,所以AC=BD,故四邊形ABCD為正方形,同理可得四邊形BEDF也為正方形.取AE的中點(diǎn)K,連接PK,KQ,則PQ=PK+KQ=12DA+12EB,同理可得MN=12DF+12AB,則PQ·MN=12DA+12EB·12DF+12AB=14DA·DF+14DA·AB+14EB·DF+14EB·AB=148.BC[解析]由題意得a⊥b,a⊥c,b⊥c,∴a·b=0,a·c=0,b·c=0.對(duì)于選項(xiàng)A,a·(b+c)=a·b+a·c=0,故A錯(cuò)誤;對(duì)于選項(xiàng)B,a·(a+b+c)=|a|2+a·b+a·c=1,故B正確;對(duì)于選項(xiàng)C,(a+b)·(b+c)=a·b+a·c+|b|2+b·c=1,故C正確;對(duì)于選項(xiàng)D,(a+b)·c=a·c+b·c=0,故D錯(cuò)誤.故選BC.9.ABD[解析]對(duì)于選項(xiàng)A,AB·AD=|AB|·|AD|cos∠BAD=5×4×12=10,故A正確;對(duì)于選項(xiàng)B,AB·AC'=AB·(AB+AD+AA')=|AB|2+AB·AD+AB·AA'=|AB|2+|AB|·|AD|cos∠BAD+|AB|·|AA'|cos∠BAA'=52+5×4×12+5×2×12=40,故B正確;對(duì)于選項(xiàng)C,BD'=BA+AA'+A'D'=|BA5223,故C錯(cuò)誤;對(duì)于選項(xiàng)D,cos∠ACC'=CA·CC'|CA-961122<0,所以∠ACC'為鈍角,則△ACC'為鈍角三角形,故D正確.10.120°[解析]由題意得c=-(a+b),則c2=(a+b)2=a2+2a·b+b2=4+2×2×3×cos<a,b>+9=7,所以cos<a,b>=-12,又<a,b>∈[0,180°],所以<a,b>=120°11.143[解析]由題意得OA·OB=OA·OC=OB·OC=0,且OG=OA+OB+OC3,故OG·(OA+OB+OC)=13(OA+OB+OC)2=13(|OA|2+|OB|2+|OC|2)=1312.4或2[解析]由題意知,FE=FA+AA'+A'E,所以FE2=(FA+AA'+A'E)2=FA2+AA'2+A'E2+2FA·AA'+2FA·A'E+2AA'·A'E,∵異面直線a,b所成的角為60°,A'E=2,AF=3,EF=23,∴23=9+AA'2+413.解:(1)因?yàn)镸為BD的中點(diǎn),P為BB1的中點(diǎn),AB=a,AD=b,AA1=c,所以PM=PB+BM=12B1B+12BD=-12BB1+12(AD-AB)=12AD-12AB(2)因?yàn)樵谄叫辛骟wABCD-A1B1C1D1中,底面ABCD是邊長(zhǎng)為2的正方形,側(cè)棱AA1=4,且∠A1AD=∠A1AB=60°,所以|a|=|b|=2,|c|=4,a·b=0,a·c=b·c=2×4×12=所以|PM|2=14(b-a-c)2=14(b2+a2+c2-2a·b-2b·c+2a·c)=14×(4+4+16-0-8+所以|PM|=6,即線段PM的長(zhǎng)為6.14.解:設(shè)VA=a,VB=b,VC=c,正四面體V-ABC的棱長(zhǎng)為1,則|a|=|b|=|c|=1,a·b=b·c=a·c=12(1)證明:由題意知VD=13(a+b+c),則AO=AV+VO=AV+12VD=-a+16(a+b+c)=16同理可得BO=16(a+c-5b),CO=16(a+b-5所以AO·BO=136(b+c-5a)·(a+c-5b)=136×18×12-9=0,所以AO⊥同理可得AO⊥CO,BO⊥CO,所以AO,BO,CO兩兩垂直.(2)DM=DV+VM=-13(a+b+c)+12c=16(-2a-所以|DM|=16(-2又|AO|=16(b+c-5a)2=22,DM·AO=16(-2a-2b+c)·16(b+c-5a)=136(-2a·b-2a·c+10a2-2b2-2b·c+10a所以cos<DM,AO>=1412×22=22,故15.ACD[解析]設(shè)正方體ABCD-A1B1C1D1的棱長(zhǎng)為a,如圖.對(duì)于A,連接B1C,易知△AB1C為等邊三角形,所以|AB1×AC|=2a×2a×sinπ3=3a2,連接B1D1,易知DB=D1B1,△AB1D1為等邊三角形,所以|AD1×DB|=|AD1×D1B1|=2a×2a×sin2π3=3a2,故A正確.對(duì)于B,根據(jù)定義,AB×AD=-AD×AB,故B錯(cuò)誤.對(duì)于C,6|BC×AC|=6×a×2a×22=6a2=S,故C正確.對(duì)于D,因?yàn)锽1D1⊥A1C1,DD1⊥A1C1,B1D1∩DD1=D1,所以A1C1⊥平面BB1D1D,又BD1?平面BB1D1D,所以A1C1⊥BD1.因?yàn)锳D1⊥A1D,AB⊥A1D,AD1∩AB=A,所以A1D⊥平面ABD1,又BD1?平面ABD1,所以A1D⊥BD1,結(jié)合外積的定義可知16.解:(1)連接AM,則MN=AN-AM=AN-(AD+DM)=λ(AB+AF)-[AD+λ(AB-AD)]=λ(a+c)-[b+λ(a-b)]=(λ-1)b+λc,當(dāng)λ=12時(shí),MN=-12b+12c,所以|MN|=-12又AE=AB+AF=a+c,所以MN·AE=12(c-b)·(a+c)=12(a·c+c2-b·a-b·c)=12×9又易知|AE|=5,所以cos<MN,AE>=MN·AE|MN||AE|=9432(2)