1、第一章1.11.1.2 空間向量的數(shù)量積運(yùn)算A級(jí)基礎(chǔ)過關(guān)練1設(shè)a，b，c是任意的非零向量，且它們互相不共線，則下列命題：(ab)c(ca)b0；|a|b|ab|；(ab)c(ca)b不與c垂直；(3a2b)(3a2b)9|a|24|b|2其中正確的有()ABCD【答案】D【解析】對(duì)于，由于(ab)c與(ca)b是兩個(gè)向量，前者與c共線，后者與b共線，方向任意，故不正確，正確2在空間四邊形ABCD中，eq o(AB,sup6()eq o(CD,sup6()eq o(BC,sup6()eq o(AD,sup6()eq o(CA,sup6()eq o(BD,sup6()()A0B1C2D3【答案】A

2、【解析】如圖，設(shè)eq o(AB,sup6()b，eq o(AC,sup6()c，eq o(AD,sup6()d，則eq o(CD,sup6()dc，eq o(BD,sup6()db，eq o(BC,sup6()cb，所以原式b(dc)d(cb)c(db)03在正四面體PABC中，棱長為1，且D為棱AB的中點(diǎn)，則eq o(PC,sup6()eq o(PD,sup6()的值為()Aeq f(1,4)Beq f(1,4)Ceq f(1,2)Deq f(1,2)【答案】D【解析】如圖，PABC為正四面體，則APCBPCAPB60因?yàn)镈是棱AB中點(diǎn)，所以eq o(PD,sup6()eq f(1,2)(e

3、q o(PA,sup6()eq o(PB,sup6()，所以eq o(PC,sup6()eq o(PD,sup6()eq o(PC,sup6()eq f(1,2)(eq o(PA,sup6()eq o(PB,sup6()eq f(1,2)eq o(PC,sup6()eq o(PA,sup6()eq f(1,2)eq o(PC,sup6()eq o(PB,sup6()eq f(1,2)11cos60eq f(1,2)11cos60eq f(1,2)故選D4已知|a|13，|b|19，|ab|24，則|ab|等于()A22B48C28D32【答案】A【解析】因?yàn)閨ab|2a2b22ab，|ab|2

4、a2b22ab，所以|ab|22(a2b2)|ab|22(132192)242484，所以|ab|225設(shè)a，b，c為非零向量，則(ab)c()A是三個(gè)向量的數(shù)量積B是與a共線的向量C是與c共線的向量D無意義【答案】C【解析】由a，b，c為非零向量可得abeq blc|rc|(avs4alco1(a)eq blc|rc|(avs4alco1(b)cosa，b，顯然ab為數(shù)量，設(shè)為t，則(ab)ctc，即有(ab)c是與c共線的向量，故A，B，D均錯(cuò)誤，C正確故選C6已知非零向量b在非零向量a方向上的投影為零，則向量a，b的關(guān)系是()AabBabCa與b相交Da與b重合【答案】B【解析】非零向量

5、b在非零向量a方向上的投影為|b|cosa，b0，又因?yàn)閎0，故|b|0所以有cosa，b0，得a，beq f(,2)，故ab故選B7(多選)(2021年邢臺(tái)月考)下列關(guān)于數(shù)量積的運(yùn)算正確的是()A|ab|a|b|B|ab|eq r(a22abb2)C(ab)ca(bc)D(ab)cacbc【答案】BD【解析】對(duì)于A選項(xiàng)，|ab|a|b|cosa，b|，故A錯(cuò)誤；對(duì)于B選項(xiàng)，|ab|eq r(（ab）2)eq r(a22abb2)，故B正確；對(duì)于C選項(xiàng)，(ab)c與c共線，a(bc)與a共線，故C錯(cuò)誤；對(duì)于D選項(xiàng)，由數(shù)量積的運(yùn)算律知(ab)cacbc，故D正確故選BD8如圖，在ABCD中，AD

6、4，CD3，ADC60，PA平面ABCD，PA6，則線段PC的長為_【答案】7【解析】因?yàn)镻A平面ABCD，AD，DC平面ABCD，故PAAD，PADC又因?yàn)閑q o(PC,sup6()eq o(PA,sup6()eq o(AD,sup6()eq o(DC,sup6()，所以|eq o(PC,sup6()|2(eq o(PA,sup6()eq o(AD,sup6()eq o(DC,sup6()2|eq o(PA,sup6()|2|eq o(AD,sup6()|2|eq o(DC,sup6()|22eq o(PA,sup6()eq o(AD,sup6()2eq o(AD,sup6()eq o(D

7、C,sup6()2eq o(DC,sup6()eq o(PA,sup6()6242322|eq o(AD,sup6()|eq o(DC,sup6()|cos120611249，所以|eq o(PC,sup6()|7，即PC的長為79如圖，在空間四邊形ABCD中，每條邊的長度和兩條對(duì)角線的長度都等于1，M，N分別是AB，AD的中點(diǎn)，則eq o(MN,sup6()eq o(DC,sup6()_【答案】eq f(1,4)【解析】eq o(MN,sup6()eq o(DC,sup6()eq f(1,2)eq o(BD,sup6()eq o(DC,sup6()eq f(1,2)|eq o(BD,sup6

8、()|eq o(DC,sup6()|coseq o(BD,sup6()，eq o(DC,sup6()eq f(1,2)11cos120eq f(1,4)10如圖，已知正四面體OABC的棱長為1，求： (1)(eq o(OA,sup6()eq o(OB,sup6()(eq o(CA,sup6()eq o(CB,sup6()；(2)|eq o(OA,sup6()eq o(OB,sup6()eq o(OC,sup6()|解：(1)(eq o(OA,sup6()eq o(OB,sup6()(eq o(CA,sup6()eq o(CB,sup6()(eq o(OA,sup6()eq o(OB,sup6(

9、)(eq o(OA,sup6()eq o(OC,sup6()eq o(OB,sup6()eq o(OC,sup6()(eq o(OA,sup6()eq o(OB,sup6()(eq o(OA,sup6()eq o(OB,sup6()2eq o(OC,sup6()1211cos60211cos6011cos6012211cos601(2)|eq o(OA,sup6()eq o(OB,sup6()eq o(OC,sup6()|eq r(（o(OA,sup6()o(OB,sup6()o(OC,sup6()）2)eq r(o(OA,sup6()2o(OB,sup6()2o(OC,sup6()22（o(

10、OA,sup6()o(OB,sup6()o(OB,sup6()o(OC,sup6()o(OA,sup6()o(OC,sup6()）)eq r(1212122（11cos60）3)eq r(,6)B級(jí)能力提升練11已知|a|3eq r(2)，|b|4，a與b的夾角為135，mab，nab，若mn，則()A1Beq f(3,2)C2D1【答案】B【解析】mn(ab)(ab)|a|2abab|b|2183eq r(2)4cos1353eq r(2)4cos135166121664因?yàn)閙n，所以640，所以eq f(3,2)12(多選)如圖，在四面體ABCD中，各棱長均為a，E，F(xiàn)，G分別是AB，AD

11、，DC的中點(diǎn)，則下列向量的數(shù)量積等于a2的是()A2eq o(BA,sup6()eq o(AC,sup6()B2eq o(AD,sup6()eq o(BD,sup6()C2eq o(EF,sup6()eq o(CB,sup6()D2eq o(FG,sup6()eq o(AC,sup6()【答案】BD【解析】依題意，四面體ABCD是正四面體，對(duì)于A，eq o(BA,sup6()，eq o(AC,sup6()120，2eq o(BA,sup6()eq o(AC,sup6()2a2cos120a2，A不是；對(duì)于B，eq o(AD,sup6()，eq o(BD,sup6()60，2eq o(AD,su

12、p6()eq o(BD,sup6()2a2cos60a2，B是；對(duì)于C，因?yàn)镋，F(xiàn)是AB，AD的中點(diǎn)，則2eq o(EF,sup6()eq o(BD,sup6()，而eq o(BD,sup6()，eq o(CB,sup6()120，2eq o(EF,sup6()eq o(CB,sup6()eq o(BD,sup6()eq o(CB,sup6()a2cos120eq f(1,2)a2，C不是；對(duì)于D，因?yàn)镕，G是AD，DC的中點(diǎn)，則2eq o(FG,sup6()eq o(AC,sup6()，2eq o(FG,sup6()eq o(AC,sup6()eq o(AC,sup6()2a2，D是故選BD

13、13在正方體ABCDA1B1C1D1中，下面給出命題：|eq o(A1A,sup6()eq o(A1D1,sup6()eq o(A1B1,sup6()|23|eq o(A1B1,sup6()|2；eq o(A1C,sup6()(eq o(A1B1,sup6()eq o(A1A,sup6()0；eq o(AD1,sup6()與eq o(A1B,sup6()的夾角為60；此正方體體積為|eq o(AB,sup6()eq o(AA1,sup6()eq o(AD,sup6()|其中錯(cuò)誤命題的序號(hào)是_【答案】【解析】因?yàn)閨eq o(A1A,sup6()eq o(A1D1,sup6()eq o(A1B1,

14、sup6()|eq o(A1C,sup6()|eq r(3)|eq o(A1B1,sup6()|，故正確；因?yàn)閑q o(A1C,sup6()(eq o(A1B1,sup6()eq o(A1A,sup6()(eq o(A1B1,sup6()eq o(A1D1,sup6()eq o(A1A,sup6()(eq o(A1B1,sup6()eq o(A1A,sup6()eq o(A1B1,sup6()2eq o(A1B1,sup6()eq o(A1D1,sup6()eq o(A1A,sup6()eq o(A1D1,sup6()eq o(A1A,sup6()20，故正確；AD1與A1B兩異面直線的夾角為

15、60，但eq o(AD1,sup6()與eq o(A1B,sup6()的夾角為120；因?yàn)閑q o(AB,sup6()eq o(AA1,sup6()0，故錯(cuò)誤，正確的應(yīng)是|eq o(AB,sup6()|eq o(AA1,sup6()|eq o(AD,sup6()|14已知兩個(gè)單位向量a，b的夾角為60(1)若caeq blc(rc)(avs4alco1(3f(2,2)b(R)，且bc0，則的值為_；(2)向量ab在b方向上的投影數(shù)量為_【答案】(1)2或3(2)eq f(3,2)【解析】(1) bcabeq blc(rc)(avs4alco1(3f(2,2)b2cos603eq f(2,2)e

16、q f(26,2)0，所以2或3(2)向量ab在b方向上的投影數(shù)量為eq f(（ab）b,blc|rc|(avs4alco1(b)eq f(f(1,2)1,1)eq f(3,2)15如圖，在平行六面體ABCDA1B1C1D1中，底面ABCD是邊長為1的正方形，AA1eq r(2)設(shè)eq o(AB,sup6()a，eq o(AD,sup6()b，eq o(AA1,sup6()c(1)試用a，b，c表示向量eq o(AC,sup6()，eq o(BD1,sup6()；(2)若A1ADA1AB120，求直線AC與BD1所成的角解：(1)由向量的加減運(yùn)算法則知eq o(AC,sup6()eq o(AB,sup6()eq o(AD,sup6()ab，eq o(BD1,sup6()eq o(AD1,sup6()eq o(AB,sup6()bca(2)由題意知|a|b|1，|c|eq r(2)，a，b90，a，cb，c120，eq o(AC,sup6()eq o(BD1,sup6()(ab)(bca)aca2b2bc1eq r(2)cos120111eq r(2)cos120eq f(r(2),2)eq f(r(2),2)eq r(2)因?yàn)閨eq o(AC,sup6()|eq r(2)|eq o(BD1,sup6()|eq