1、A級：“四基”鞏固訓練一、選擇題1已知|a|2，|b|4，ab4，則向量a與b的夾角為()A30 B60 C150 D120答案D解析coseq f(ab,|a|b|)eq f(4,24)eq f(1,2)，0，180，120.故選D.2已知向量a，b滿足ab0，|a|1，|b|2，則|2ab|()A0 B2eq r(2) C4 D8答案B解析|2ab|24a24abb28，|2ab|2eq r(2).3若平面四邊形ABCD滿足eq o(AB,sup16()eq o(CD,sup16()0，(eq o(AB,sup16()eq o(AD,sup16()eq o(AC,sup16()0，則該四邊

2、形一定是()A直角梯形 B矩形C菱形 D正方形答案C解析由eq o(AB,sup16()eq o(CD,sup16()0，得平面四邊形ABCD是平行四邊形，由(eq o(AB,sup16()eq o(AD,sup16()eq o(AC,sup16()0，得eq o(DB,sup16()eq o(AC,sup16()0，即平行四邊形ABCD的對角線互相垂直，則該四邊形一定是菱形4若非零向量a，b滿足|a|eq f(2r(2),3)|b|，且(ab)(3a2b)，則a與b的夾角為()A.eq f(,4) B.eq f(,2) C.eq f(3,4) D答案A解析由題意，得(ab)(3a2b)3a2

3、2b2ab0，即ab3a22b2.又|a|eq f(2r(2),3)|b|，所以ab3eq blc(rc)(avs4alco1(f(2r(2),3)|b|)22b2eq f(2,3)b2，所以cosa，beq f(ab,|a|b|)eq f(f(2,3)b2,f(2r(2),3)b2)eq f(r(2),2)，所以a，beq f(,4).5已知a，b是平面內兩個互相垂直的單位向量，若非零向量c滿足(ac)(bc)0，則|c|的最大值是()A1 B2 C.eq r(2) D.eq f(r(2),2)答案C解析因為|a|b|1，ab0，(ac)(bc)c(ab)|c|2|c|ab|cos|c|20

4、，其中為c與ab的夾角，所以|c|ab|coseq r(2)coseq r(2)，所以|c|的最大值是eq r(2)，故選C.二、填空題6已知向量a，b的夾角為45，且|a|4，eq blc(rc)(avs4alco1(f(1,2)ab)(2a3b)12，則|b|_；b在a上的投影向量的模等于_答案eq r(2)1解析ab|a|b|cosa，b4|b|cos452eq r(2)|b|，又eq blc(rc)(avs4alco1(f(1,2)ab)(2a3b)|a|2eq f(1,2)ab3|b|216eq r(2)|b|3|b|212，解得|b|eq r(2)或|b|eq f(2r(2),3)

5、(舍去)b在a上的投影向量的模為|b|cosa，b|eq r(2)cos451.7已知兩個單位向量a，b的夾角為60，cta(1t)b.若bc0，則t_.答案2解析由題意，將bcta(1t)bb整理，得tab(1t)0，又abeq f(1,2)，所以t2.8在平行四邊形ABCD中，AD1，BAD60，E為CD的中點若eq o(AC,sup16()eq o(BE,sup16()1，則AB的長為_答案eq f(1,2)解析因為eq o(BE,sup16()eq o(BA,sup16()eq o(AD,sup16()eq o(DE,sup16()eq o(AB,sup16()eq o(AD,sup1

6、6()eq f(1,2)eq o(AB,sup16()eq o(AD,sup16()eq f(1,2)eq o(AB,sup16()，所以eq o(AC,sup16()eq o(BE,sup16()(eq o(AB,sup16()eq o(AD,sup16()eq blc(rc)(avs4alco1(o(AD,sup16()f(1,2)o(AB,sup16()eq o(AD,sup16()2eq f(1,2)eq o(AD,sup16()eq o(AB,sup16()eq f(1,2)eq o(AB,sup16()21eq f(1,2)1|eq o(AB,sup16()|cos60eq f(1

7、,2)|eq o(AB,sup16()|21，所以eq f(1,4)|eq o(AB,sup16()|eq f(1,2)|eq o(AB,sup16()|20，解得|eq o(AB,sup16()|eq f(1,2).三、解答題9已知a，b是兩個非零向量，當atb(tR)的模取得最小值時，(1)求t的值(用a，b表示)；(2)求證：b與atb垂直解(1)|atb|2a2t2b22tabb2eq blc(rc)(avs4alco1(tf(ab,b2)2a2eq f(ab2,b2).當teq f(ab,b2)時，|atb|取得最小值(2)證明：因為(atb)babtb2abeq f(ab,b2)b

8、20，所以atb與b垂直B級：“四能”提升訓練1設向量a，b，c滿足abc0，(ab)c，ab，若|a|1，則|a|2|b|2|c|2的值是_答案4解析由abc0，得(abc)20，得a2b2c22(abbcca)0.又(ab)c，ab，(ab)c0，ab0.acbc.a2b2c24bc，b2c214bc.由abc0，得bca，故(bc)21，即b2c22bc1.由得bc1，故a2b2c24，即|a|2|b|2|c|24.2在四邊形ABCD中，已知AB9，BC6，eq o(CP,sup16()2eq o(PD,sup16().(1)若四邊形ABCD是矩形，求eq o(AP,sup16()eq

9、o(BP,sup16()的值；(2)若四邊形ABCD是平行四邊形，且eq o(AP,sup16()eq o(BP,sup16()6，求eq o(AB,sup16()與eq o(AD,sup16()夾角的余弦值解(1)因為四邊形ABCD是矩形，所以eq o(AD,sup16()eq o(DC,sup16()0，由eq o(CP,sup16()2eq o(PD,sup16()，得eq o(DP,sup16()eq f(1,3)eq o(DC,sup16()，eq o(CP,sup16()eq f(2,3)eq o(CD,sup16()eq f(2,3)eq o(DC,sup16().所以eq o(

10、AP,sup16()eq o(BP,sup16()(eq o(AD,sup16()eq o(DP,sup16()(eq o(BC,sup16()eq o(CP,sup16()eq blc(rc)(avs4alco1(o(AD,sup16()f(1,3)o(DC,sup16()eq blc(rc)(avs4alco1(o(AD,sup16()f(2,3)o(DC,sup16()eq o(AD,sup16()2eq f(1,3)eq o(AD,sup16()eq o(DC,sup16()eq f(2,9)eq o(DC,sup16()236eq f(2,9)8118.(2)由題意，eq o(AP,

11、sup16()eq o(AD,sup16()eq o(DP,sup16()eq o(AD,sup16()eq f(1,3)eq o(DC,sup16()eq o(AD,sup16()eq f(1,3)eq o(AB,sup16()，eq o(BP,sup16()eq o(BC,sup16()eq o(CP,sup16()eq o(BC,sup16()eq f(2,3)eq o(CD,sup16()eq o(AD,sup16()eq f(2,3)eq o(AB,sup16()，所以eq o(AP,sup16()eq o(BP,sup16()eq blc(rc)(avs4alco1(o(AD,su

12、p16()f(1,3)o(AB,sup16()eq blc(rc)(avs4alco1(o(AD,sup16()f(2,3)o(AB,sup16()eq o(AD,sup16()2eq f(1,3)eq o(AB,sup16()eq o(AD,sup16()eq f(2,9)eq o(AB,sup16()236eq f(1,3)eq o(AB,sup16()eq o(AD,sup16()1818eq f(1,3)eq o(AB,sup16()eq o(AD,sup16().又eq o(AP,sup16()eq o(BP,sup16()6，所以18eq f(1,3)eq o(AB,sup16()eq o(AD,sup16(