人教A版選修2-1第三章3.1.3空間向量的數(shù)量積運(yùn)算達(dá)標(biāo)過(guò)關(guān)訓(xùn)練

1、3.1.3空間向量的數(shù)量積運(yùn)算一、選擇題1若a，b均為非零向量，則ab|a|b|是a與b共線的()A充分不必要條件B必要不充分條件C充要條件D既不充分也不必要條件解析：若ab|a|b|cosa，b|a|b|，則cosa，b1，a與b共線，反之，若a，b共線也可能ab|a|b|.答案：A2下列命題中，正確的命題個(gè)數(shù)為()eq r(aa)|a|；m(a)b(m)ab(m，R)；a(bc)(bc)a；(ab)2a22abb2.A1 B2C3 D4解析：由向量的數(shù)量積及其運(yùn)算律可知，四個(gè)命題都正確答案：D3(2019蚌埠第二中學(xué)高二月考)已知空間四邊形ABCD中，ACDBDC90，且AB2，CD1，則

2、AB與CD所成的角是()A30 B45C60 D90解析：根據(jù)已知ACDBDC90，得eq o(AC,sup6()eq o(CD,sup6()eq o(DB,sup6()eq o(CD,sup6()0，eq o(AB,sup6()eq o(CD,sup6()(eq o(AC,sup6()eq o(CD,sup6()eq o(DB,sup6()eq o(CD,sup6()eq o(AC,sup6()eq o(CD,sup6()|eq o(CD,sup6()|2eq o(DB,sup6()eq o(CD,sup6()|eq o(CD,sup6()|21，coseq o(AB,sup6()，eq o

3、(CD,sup6()eq f(o(AB,sup6()o(CD,sup6(),|o(AB,sup6()|o(CD,sup6()|)eq f(1,2)，AB與CD所成的角為60.答案：C4正三棱柱ABCA1B1C1的各棱長(zhǎng)均為2，E，F(xiàn)分別是AB，A1C1的中點(diǎn)，則EF的長(zhǎng)為()A2 Beq r(3)Ceq r(5) Deq r(7)解析：如圖所示，設(shè)eq o(AB,sup6()a，eq o(AC,sup6()b，eq o(AA1,sup6()c，則|a|b|c|2，且a，b60，a，c90，b，c90.eq o(EF,sup6()eq o(EA,sup6()eq o(AA1,sup6()eq o

4、(A1F,sup6()eq f(1,2)eq o(AB,sup6()AA1eq f(1,2)eq o(A1C1,sup6()eq f(1,2)aceq f(1,2)b，|eq o(EF,sup6()|2eq f(1,4)a2c2eq f(1,4)b22eq f(1,2)aceq f(1,2)bceq f(1,4)abeq f(1,4)44eq f(1,4)42eq blc(rc)(avs4alco1(00f(1,4)22f(1,2)14115，|eq o(EF,sup6()|eq r(5)，故EF的長(zhǎng)為eq r(5).答案：C5如圖，在直三棱柱ABCA1B1C1中，ABBCAA1，ABC90，

5、點(diǎn)E，F(xiàn)分別是棱AB，BB1的中點(diǎn)，則直線EF和BC1所成的角是()A45 B60C90 D120解析：eq o(EF,sup6()eq o(BC1,sup6()eq f(1,2)eq o(AB1,sup6()(eq o(BC,sup6()eq o(BB1,sup6()eq f(1,2)(eq o(AB,sup6()eq o(BB1,sup6()(eq o(BC,sup6()eq o(BB1,sup6()eq f(1,2)(eq o(AB,sup6()eq o(BC,sup6()eq o(BB1,sup6()eq o(BC,sup6()eq o(AB,sup6()eq o(BB1,sup6()

6、eq o(BB1,sup6()eq o(BB1,sup6()eq f(1,2)(000eq o(BB1,sup6()2)eq f(1,2)|eq o(BB1,sup6()|2.設(shè)|eq o(BB1,sup6()|1，則eq o(EF,sup6()eq o(BC1,sup6()eq f(1,2)，|eq o(EF,sup6()|eq f(r(2),2)，|eq o(BC1,sup6()|eq r(2)，coseq o(EF,sup6()，eq o(BC1,sup6()eq f(o(EF,sup6()o(BC1,sup6(),avs4al(|o(EF,sup6()|o(BC1,sup6()|)eq

7、 f(f(1,2),f(r(2),2)r(2)eq f(1,2).異面直線EF與BC1所成的角是60.答案：B二、填空題6如圖，已知ABC在平面內(nèi)，BAC90，DA平面，則直線CA與DB的位置關(guān)系是_解析：eq o(CA,sup6()eq o(BD,sup6()eq o(CA,sup6()(eq o(BA,sup6()eq o(AD,sup6()eq o(CA,sup6()eq o(BA,sup6()eq o(CA,sup6()eq o(AD,sup6()000，eq o(CA,sup6()eq o(DB,sup6().答案：垂直7(2019晉中市高二期末)如圖所示，在一個(gè)直二面角AB的棱上有

8、A，B兩點(diǎn)，AC，BD分別是這個(gè)二面角的兩個(gè)面內(nèi)垂直于AB的線段，且AB4，AC6，BD8，則CD的長(zhǎng)為_(kāi)解析：eq o(CD,sup6()eq o(CA,sup6()eq o(AB,sup6()eq o(BD,sup6()eq o(AB,sup6()eq o(AC,sup6()eq o(BD,sup6()，eq o(CD,sup6()2(eq o(AB,sup6()eq o(AC,sup6()eq o(BD,sup6()2eq o(AB,sup6()2eq o(AC,sup6()22eq o(AB,sup6()eq o(AC,sup6()eq o(BD,sup6()22eq o(AB,sup

9、6()eq o(BD,sup6()2eq o(AC,sup6()eq o(BD,sup6()163664116，|eq o(CD,sup6()|2eq r(29).答案：2eq r(29)8(2019鄂東南九校聯(lián)考)在四面體OABC中，棱OA，OB，OC兩兩垂直，且OA1，OB2，OC3，G為ABC的重心，則eq o(OG,sup6()(eq o(OA,sup6()eq o(OB,sup6()eq o(OC,sup6()_.解析：eq o(OG,sup6()eq o(OA,sup6()eq o(AG,sup6()eq o(OA,sup6()eq f(1,3)(eq o(AB,sup6()eq

10、o(AC,sup6()eq o(OA,sup6()eq f(1,3)(eq o(OB,sup6()eq o(OA,sup6()(eq o(OC,sup6()eq o(OA,sup6()eq f(1,3)eq o(OB,sup6()eq f(1,3)eq o(OC,sup6()eq f(1,3)eq o(OA,sup6().eq o(OG,sup6()(eq o(OA,sup6()eq o(OB,sup6()eq o(OC,sup6()eq f(1,3)eq o(OB,sup6()eq f(1,3)eq o(OC,sup6()eq f(1,3)eq o(OA,sup6()(eq o(OA,sup

11、6()eq o(OB,sup6()eq o(OC,sup6()eq f(1,3)eq o(OB,sup6()2eq f(1,3)eq o(OC,sup6()2eq f(1,3)eq o(OA,sup6()2eq f(14,3).答案：eq f(14,3)三、解答題9(2019鄂東南九校聯(lián)考)已知在正四面體DABC中，所有棱長(zhǎng)都為1，ABC的重心為G，求DG的長(zhǎng)解：如圖，連接AG并延長(zhǎng)交BC于點(diǎn)M，連接DM，G是ABC的重心，eq o(AG,sup6()eq f(2,3)eq o(AM,sup6()，又eq o(DG,sup6()eq o(DA,sup6()eq o(AG,sup6()eq o(

12、DA,sup6()eq f(2,3)eq o(AM,sup6()eq o(DA,sup6()eq f(2,3)(eq o(DM,sup6()eq o(DA,sup6()eq o(DA,sup6()eq f(2,3)eq f(1,2)(eq o(DB,sup6()eq o(DC,sup6()eq o(DA,sup6()eq f(1,3)(eq o(DA,sup6()eq o(DB,sup6()eq o(DC,sup6()，而(eq o(DA,sup6()eq o(DB,sup6()eq o(DC,sup6()2eq o(DA,sup6()2eq o(DB,sup6()2eq o(DC,sup6(

13、)22eq o(DA,sup6()eq o(DB,sup6()2eq o(DB,sup6()eq o(DC,sup6()2eq o(DC,sup6()eq o(DA,sup6()1112(cos 60cos 60cos 60)6，|eq o(DG,sup6()|eq f(r(6),3).10如圖，PA垂直于矩形ABCD所在的平面，M，N分別為AB，PC的中點(diǎn)(1)求證：MNCD；(2)若PDA45，求證：MN平面PCD.證明：(1)設(shè)eq o(AB,sup6()a，eq o(AD,sup6()b，eq o(AP,sup6()c，則eq o(MN,sup6()eq o(MB,sup6()eq o

14、(BC,sup6()eq o(CN,sup6()eq f(1,2)eq o(AB,sup6()eq o(AD,sup6()eq f(1,2)eq o(PC,sup6()eq f(1,2)eq o(AB,sup6()eq o(AD,sup6()eq f(1,2)(eq o(PA,sup6()eq o(AB,sup6()eq o(BC,sup6()eq f(1,2)(eq o(AD,sup6()eq o(AP,sup6()eq f(1,2)(bc)，eq o(CD,sup6()eq o(AB,sup6()a，eq o(MN,sup6()eq o(CD,sup6()eq f(1,2)(bc)(a)eq f(1,2)(abac)四邊形ABCD是矩形，PA平面ABCD，ab，ac，ab0，ac0.eq o(MN,sup6()eq o(CD,sup6()0，eq o(MN,sup6()eq o(CD,sup6()，即MNCD.(2)由(1)知，MNCD，eq o(MN,sup6()eq f(1,2)(bc)，eq o(