【高考必備】2017年高考數(shù)學(xué)（文）熱點(diǎn)題型和提分秘籍專題33空間向量及其運(yùn)算word版含解析

1、PAGE 1.了解空間直角坐標(biāo)系，會用空間直角坐標(biāo)表示點(diǎn)的位置2.會簡單應(yīng)用空間兩點(diǎn)間的距離公式3.了解空間向量的概念，了解空間向量的基本定理及其意義，掌握空間向量的正交分解及其坐標(biāo)表示4.掌握空間向量的線性運(yùn)算及其坐標(biāo)表示5.掌握空間向量的數(shù)量積及其坐標(biāo)表示，能運(yùn)用向量的數(shù)量積判斷向量的共線與垂直熱點(diǎn)題型一 空間向量的運(yùn)算例1、如圖，在長方體ABCDA1B1C1D1中，O為AC的中點(diǎn)。(1)化簡：eq o(A1O,sup10()eq f(1,2)eq o(AB,sup10()eq f(1,2)eq o(AD,sup10()；(2)設(shè)E是棱DD1上的點(diǎn)，且eq o(DE,sup10()eq f

2、(2,3)eq o(DD1,sup10()，若eq o(EO,sup10()xeq o(AB,sup10()yeq o(AD,sup10()zeq o(AA1,sup10()，試求x、y、z的值。eq f(2,3)eq o(D1D,sup10()eq f(1,2)(eq o(DA,sup10()eq o(AB,sup10()eq f(2,3)eq o(A1A,sup10()eq f(1,2)eq o(DA,sup10()eq f(1,2)eq o(AB,sup10()eq f(1,2)eq o(AB,sup10()eq f(1,2)eq o(AD,sup10()eq f(2,3)eq o(AA

3、1,sup10()，xeq f(1,2)，yeq f(1,2)，zeq f(2,3)。【提分秘籍】空間向量的表示方法用已知不共面的向量表示某一向量時，應(yīng)結(jié)合圖形，將已知向量和未知向量轉(zhuǎn)化至三角形或平行四邊形中，然后利用三角形法則或平行四邊形法則，把所求向量用已知向量表示出來。【舉一反三】 如圖，在平行六面體ABCDA1B1C1D1中，M為AC與BD的交點(diǎn)，若eq o(A1B1,sup10()a，eq o(A1D1,sup10()b，eq o(A1A,sup10()c，則下列向量中與eq o(B1M,sup10()相等的向量是()Aeq f(1,2)aeq f(1,2)bcB.eq f(1,2)

4、aeq f(1,2)bcC.eq f(1,2)aeq f(1,2)bcDeq f(1,2)aeq f(1,2)bc【答案】A熱點(diǎn)題型二 共線、共面向量定理的應(yīng)用 例2、已知E、F、G、H分別是空間四邊形ABCD的邊AB、BC、CD、DA的中點(diǎn)，(1)求證：E、F、G、H四點(diǎn)共面；(2)求證：BD平面EFGH。由共面向量定理的推論知：E、F、G、H四點(diǎn)共面。(2)因為eq o(EH,sup10()eq o(AH,sup10()eq o(AE,sup10()eq f(1,2)eq o(AD,sup10()eq f(1,2)eq o(AB,sup10()eq f(1,2)(eq o(AD,sup10

5、()eq o(AB,sup10()eq f(1,2)eq o(BD,sup10()，所以EHBD。又EH平面EFGH，BD平面EFGH，所以BD平面EFGH。【提分秘籍】 應(yīng)用共線向量定理、共面向量定理證明點(diǎn)共線、點(diǎn)共面的方法比較：三點(diǎn)(P，A，B)共線空間四點(diǎn)(M，P，A，B)共面eq o(PA,sup10()eq o(PB,sup10()eq o(MP,sup10()xeq o(MA,sup10()yeq o(MB,sup10()對空間任一點(diǎn)O，eq o(OP,sup10()eq o(OA,sup10()teq o(AB,sup10()(t為參數(shù))對空間任一點(diǎn)O，eq o(OP,sup10

6、()eq o(OM,sup10()xeq o(MA,sup10()yeq o(MB,sup10()對空間任一點(diǎn)O，eq o(OP,sup10()(1t)eq o(OA,sup10()teq o(OB,sup10()(t為參數(shù))對空間任一點(diǎn)O，eq o(OP,sup10()(1xy)eq o(OM,sup10()xeq o(OA,sup10()yeq o(OB,sup10() 【舉一反三】 已知A、B、C三點(diǎn)不共線，對平面ABC外的任一點(diǎn)O，若點(diǎn)M滿足eq f(1,3)(eq o(OA,sup10()eq o(OB,sup10()eq o(OC,sup10()。(1)判斷eq o(MA,sup1

7、0()，eq o(MB,sup10()，eq o(MC,sup10()三個向量是否共面；(2)判斷點(diǎn)M是否在平面ABC內(nèi)。 (2)由(1)知eq o(MA,sup10()，eq o(MB,sup10()，eq o(MC,sup10()共面，且共過同一點(diǎn)M，四點(diǎn)M，A，B，C共面。從而點(diǎn)M在平面ABC內(nèi)。熱點(diǎn)題型三 空間向量數(shù)量積的運(yùn)算例3如圖所示，已知空間四邊形ABCD的每條邊和對角線長都等于1，點(diǎn)E，F(xiàn)，G分別是AB，AD，CD的中點(diǎn)，計算： (1)eq o(EF,sup10()eq o(BA,sup10()。(2)eq o(EG,sup10()eq o(BD,sup10()。 (2)eq

8、o(EG,sup10()eq o(EB,sup10()eq o(BC,sup10()eq o(CG,sup10()eq f(1,2)eq o(AB,sup10()(eq o(AC,sup10()eq o(AB,sup10()eq f(1,2)(eq o(AD,sup10()eq o(AC,sup10()eq f(1,2)eq o(AB,sup10()eq f(1,2)eq o(AC,sup10()eq f(1,2)eq o(AD,sup10()eq f(1,2)aeq f(1,2)beq f(1,2)c，eq o(BD,sup10()eq o(AD,sup10()eq o(AB,sup10()

9、ca。所以eq o(EG,sup10()eq o(BD,sup10()eq blc(rc)(avs4alco1(f(1,2)af(1,2)bf(1,2)c)(ca)eq f(1,2)a2eq f(1,2)abeq f(1,2)bceq f(1,2)c2aceq f(1,2)eq f(1,4)eq f(1,4)eq f(1,2)eq f(1,2)eq f(1,2)?！咎岱置丶?空間向量數(shù)量積計算的兩種方法(1)基向量法：ab|a|b|cosa，b。(2)坐標(biāo)法：設(shè)a(x1，y1，z1)，b(x2，y2，z2)，則abx1x2y1y2z1z2。2利用數(shù)量積可解決有關(guān)垂直、夾角、長度問題(1)a0

10、，b0，abab0。(2)|a|eq r(a2)。(3)cosa，beq f(ab,|a|b|)?！九e一反三】 已知A(1,0,0)，B(0，1,1)，eq o(OA,sup10()eq o(OB,sup10()與eq o(OB,sup10()的夾角為120，則的值為()Aeq f(r(6),6) B.eq f(r(6),6) Ceq f(r(6),6) Deq r(6)【解析】eq o(OA,sup10()eq o(OB,sup10()(1，)，cos120eq f(,r(122)r(2)eq f(1,2)，得eq f(r(6),6)。經(jīng)檢驗eq f(r(6),6)不合題意，舍去，eq f(

11、r(6),6)。【答案】C 1.【2016高考浙江文數(shù)】如圖，已知平面四邊形ABCD，AB=BC=3，CD=1，AD=，ADC=90沿直線AC將ACD翻折成，直線AC與所成角的余弦的最大值是_【答案】則，與平行的單位向量為，所以，所以時，取得最大值，為1若a，b，c為空間的一組基底，則下列各項中，能構(gòu)成基底的一組向量是()Aa，ab，ab Bb，ab，abCc，ab，ab Dab，ab，a2b【答案】C2在正方體ABCDA1B1C1D1中，給出以下向量表達(dá)式：(eq o(A1D1,sup10()eq o(A1A,sup10()eq o(AB,sup10()； (eq o(BC,sup10()e

12、q o(BB1,sup10()eq o(D1C1,sup10()；(eq o(AD,sup10()eq o(AB,sup10()2eq o(DD1,sup10()； (eq o(B1D1,sup10()eq o(A1A,sup10()eq o(DD1,sup10()。其中能夠化簡為向量eq o(BD1,sup10()的是()A B C D【解析】(eq o(A1D1,sup10()eq o(A1A,sup10()eq o(AB,sup10()eq o(AD1,sup10()eq o(AB,sup10()eq o(BD1,sup10()；(eq o(BC,sup10()eq o(BB1,sup1

13、0()eq o(D1C1,sup10()eq o(BC1,sup10()eq o(D1C1,sup10()eq o(BD1,sup10()；(eq o(AD,sup10()eq o(AB,sup10()2eq o(DD1,sup10()eq o(BD,sup10()2eq o(DD1,sup10()eq o(BD1,sup10()；(eq o(B1D1,sup10()eq o(A1A,sup10()eq o(DD1,sup10()eq o(B1D,sup10()eq o(DD1,sup10()eq o(B1D1,sup10()eq o(BD1,sup10()，所以選A。 【答案】A3若向量a(

14、1，2)，b(2，1,2)且a與b的夾角的余弦值為eq f(8,9)，則等于()A2 B2C2或eq f(2,55) D2或eq f(2,55)【答案】C4平行六面體ABCDABCD中，若eq o(AC,sup10()xeq o(AB,sup10()2yeq o(BC,sup10()3zeq o(CC,sup10()，則xyz()A1 B.eq f(7,6)C.eq f(5,6) D.eq f(2,3)【解析】eq o(AC,sup10()eq o(AC,sup10()eq o(CC,sup10()eq o(AD,sup10()eq o(AB,sup10()eq o(CC,sup10()eq

15、o(AB,sup10()eq o(BC,sup10()eq o(CC,sup10()xeq o(AB,sup10()2yeq o(BC,sup10()3zeq o(CC,sup10()，故x1，yeq f(1,2)，zeq f(1,3)，xyz1eq f(1,2)eq f(1,3)eq f(7,6)。 【答案】B5已知直線AB、CD是異面直線，ACCD，BDCD，且AB2，CD1，則異面直線AB與CD夾角的大小為()A30 B45C60 D75【解析】coseq o(AB,sup10()，eq o(CD,sup10()eq f(o(AB,sup10()o(CD,sup10(),|o(AB,su

16、p10()|o(CD,sup10()|)eq f(o(AC,sup10()o(CD,sup10()o(DB,sup10()o(CD,sup10(),21)eq f(o(CD,sup10()2,2)eq f(1,2)，eq o(AB,sup10()與eq o(CD,sup10()所成角為60。 【答案】C6正方體ABCDA1B1C1D1的棱長為a，點(diǎn)M在eq o(AC1,sup10()上且eq o(AM,sup10()eq f(1,2)eq o(MC1,sup10()，N為B1B的中點(diǎn)，則|eq o(MN,sup10()|為()A.eq f(r(21),6)a B.eq f(r(6),6)aC.

17、eq f(r(15),6)a D.eq f(r(15),3)a點(diǎn)M在eq o(AC1,sup10()上且eq o(AM,sup10()eq f(1,2)eq o(MC1,sup10()，(xa，y，z)eq f(1,2)(x，ay，az)，xeq f(2,3)a，yeq f(a,3)，zeq f(a,3)，得Meq blc(rc)(avs4alco1(f(2a,3)，f(a,3)，f(a,3)，|eq o(MN,sup10()|eq r(blc(rc)(avs4alco1(af(2,3)a)2blc(rc)(avs4alco1(af(a,3)2blc(rc)(avs4alco1(f(a,2)f

18、(a,3)2)eq f(r(21),6)a。 【答案】A7如圖所示，已知空間四邊形ABCD，F(xiàn)為BC的中點(diǎn)，E為AD的中點(diǎn)，若eq o(EF,sup10()(eq o(AB,sup10()eq o(DC,sup10()，則_?！窘馕觥咳鐖D所示，取AC的中點(diǎn)G，連接EG、GF，則eq o(EF,sup10()eq o(EG,sup10()eq o(GF,sup10()eq f(1,2)(eq o(AB,sup10()eq o(DC,sup10()eq f(1,2)。 【答案】eq f(1,2)8已知O(0,0,0)，A(1,2,3)，B(2,1,2)，P(1,1,2)，點(diǎn)Q在直線OP上運(yùn)動，當(dāng)e

19、q o(QA,sup10()eq o(QB,sup10()最小時，點(diǎn)Q的坐標(biāo)是_。【答案】eq blc(rc)(avs4alco1(f(4,3)，f(4,3)，f(8,3)9在正方體ABCDA1B1C1D1中，下面給出四個命題：(eq o(A1A,sup10()eq o(A1D1,sup10()eq o(A1B1,sup10()23(eq o(A1B1,sup10()2；eq o(A1C,sup10()(eq o(A1B1,sup10()eq o(A1A,sup10()0；eq o(AD1,sup10()與eq o(A1B,sup10()的夾角為60；此正方體的體積為|eq o(AB,sup1

20、0()eq o(AA1,sup10()eq o(AD,sup10()|。則正確命題的序號是_(填寫所有正確命題的序號)。【解析】|eq o(A1A,sup10()eq o(A1D1,sup10()eq o(A1B1,sup10()|eq o(A1C,sup10()|eq r(3)|eq o(A1B1,sup10()|，正確；eq o(A1C,sup10()(eq o(A1B1,sup10()eq o(A1A,sup10()eq o(A1C,sup10()eq o(A1B1,sup10()eq o(A1C,sup10()eq o(A1A,sup10()；(eq o(A1C,sup10()，eq

21、o(A1B1,sup10()(eq o(A1C,sup10()，eq o(A1A,sup10()，|eq o(A1B1,sup10()|eq o(A1A,sup10()|eq o(A1C,sup10()eq o(A1B1,sup10()eq o(A1C,sup10()eq o(A1A,sup10()0.正確；AD1與A1B兩異面直線的夾角為60，但eq o(AD1,sup10()與eq o(A1B,sup10()的夾角為120，eq o(A1B,sup10()eq o(D1C,sup10()，注意方向，eq o(AB,sup10()eq o(AA1,sup10()0，正確的應(yīng)是|eq o(AB

22、,sup10()|eq o(AA1,sup10()|eq o(AD,sup10()|。 【答案】10如圖所示，在平行六面體ABCDA1B1C1D1中，設(shè)eq o(AA1,sup10()a，eq o(AB,sup10()b，eq o(AD,sup10()c，M、N、P分別是AA1、BC、C1D1的中點(diǎn)，試用a，b，c表示以下各向量：(1)eq o(AP,sup10()；(2)eq o(A1N,sup10()；(3)eq o(MP,sup10()eq o(NC1,sup10()。 (3)M是AA1的中點(diǎn)，eq o(MP,sup10()eq o(MA,sup10()eq o(AP,sup10()eq f(1,2)eq o(A1A,sup10()eq o(AP,sup10()eq f(1,2)aeq blc(rc)(avs4alco1(acf(1,2)b)eq f(1,2)aeq f(1,2)bc。又eq o(NC1,sup10()eq o(NC,sup10()eq o(CC1,sup10()eq f(1,2)eq o(BC,sup10()eq o(AA1,sup10()eq f(1,2)eq o(AD,sup10()eq o(AA1,sup10()eq f(1,2)ca，eq o(MP,sup10()eq o(NC1,sup10()eq blc(rc)(avs