1、高考资源网（） 您身边的高考专家第二章测评B(高考体验卷)(时间：90分钟满分：100分)一、选择题(本大题共10小题，每小题5分，共50分在每小题给出的四个选项中，只有一项是符合题目要求的)1(2013辽宁高考)已知点A(1,3)，B(4，1)，则与向量eq o(AB,sup6()同方向的单位向量为()A.eq blc(rc)(avs4alco1(f(3,5)，f(4,5) B.eq blc(rc)(avs4alco1(f(4,5)，f(3,5) C.eq blc(rc)(avs4alco1(f(3,5)，f(4,5) D.eq blc(rc)(avs4alco1(f(4,5)，f(3,5)
2、2(2013福建高考)在四边形ABCD中， EMBED Equation.DSMT4 (1,2)， EMBED Equation.DSMT4 (4,2)，则该四边形的面积为()A.eq r(5) B2eq r(5) C5 D103(2013安徽高考)在平面直角坐标系中，O是坐标原点，两定点A，B满足|Oeq o(A,sup6()|Oeq o(B,sup6()|Oeq o(A,sup6()Oeq o(B,sup6()2，则点集P|Oeq o(P,sup6()eq o(OA,sup6()eq o(OB,sup6()，|1，R所表示的区域的面积是()A2 EMBED Equation.DSMT4 B
3、2 EMBED Equation.DSMT4 C4 EMBED Equation.DSMT4 D4 EMBED Equation.DSMT4 INCLUDEPICTURER36.EPS4(2013陕西高考)已知向量a(1，m)，b(m,2)，若ab，则实数m等于()A EMBED Equation.DSMT4 B. EMBED Equation.DSMT4 C EMBED Equation.DSMT4 或 EMBED Equation.DSMT4 D05(2013大纲全国高考)已知向量m(1,1)，n(2,2)，若(mn)(mn)，则()A4 B3 C2 D16(2013湖北高考)已知点A(1
4、,1)，B(1,2)，C(2，1)，D(3,4)，则向量 EMBED Equation.DSMT4 在 EMBED Equation.DSMT4 方向上的投影为()A.eq f(3r(2),2) B.eq f(3r(15),2)Ceq f(3r(2),2) Deq f(3r(15),2)7(2013课标全国高考改编)已知正方形ABCD的边长为2，E为CD的中点，则 EMBED Equation.DSMT4 EMBED Equation.DSMT4 ()A1 B2 C1 D2INCLUDEPICTURER39.EPS8(2013课标全国高考改编)已知两个单位向量a，b的夹角为60，cta(1t)
5、b.若bc0，则t()A1 B2 C1 D29(2012广东高考)若向量 EMBED Equation.DSMT4 (2,3)， EMBED Equation.DSMT4 (4,7)，则 EMBED Equation.DSMT4 ()A(2，4) B(2,4)C(6,10) D(6，10)10(2012重庆高考)设x，yR，向量a(x,1)，b(1，y)，c(2，4)，且ac，bc，则|ab|()A. EMBED Equation.DSMT4 B. EMBED Equation.DSMT4 C2 EMBED Equation.DSMT4 D10二、填空题(本大题共5小题，每小题5分，共25分把
6、答案填在题中的横线上)11(2013江苏高考)设D，E分别是ABC的边AB，BC上的点，AD EMBED Equation.DSMT4 AB，BE EMBED Equation.DSMT4 BC.若 EMBED Equation.DSMT4 1 EMBED Equation.DSMT4 2 EMBED Equation.DSMT4 (1，2为实数)，则12的值为_INCLUDEPICTURER40.EPS12(2013四川高考)如图，在平行四边形ABCD中，对角线AC与BD交于点O， EMBED Equation.DSMT4 EMBED Equation.DSMT4 EMBED Equatio
7、n.DSMT4 .则_.INCLUDEPICTURER41.EPS13(2013江西高考)设e1，e2为单位向量，且e1，e2的夹角为 EMBED Equation.DSMT4 ，若ae13e2，b2e1，则向量a在b方向上的投影为_14(2013天津高考)在平行四边形ABCD中，AD1，BAD60，E为CD的中点若 EMBED Equation.DSMT4 EMBED Equation.DSMT4 1，则AB的长为 15(2013北京高考)向量a，b，c在正方形网格中的位置如图所示，若cab(，R)，则 EMBED Equation.DSMT4 _.INCLUDEPICTURER43.EPS
8、三、解答题(本大题共4小题，共25分解答时应写出文字说明、证明过程或演算步骤)16(本小题6分)(2013山东高考改编)已知向量 EMBED Equation.DSMT4 与 EMBED Equation.DSMT4 的夹角为120，且| EMBED Equation.DSMT4 |3，| EMBED Equation.DSMT4 |2，若 EMBED Equation.DSMT4 EMBED Equation.DSMT4 EMBED Equation.DSMT4 ，且 EMBED Equation.DSMT4 EMBED Equation.DSMT4 ，求实数的值17(本小题6分)(2013
9、浙江高考改编)设ABC，P0是边AB上一定点，满足P0B EMBED Equation.DSMT4 AB，且对于边AB上任一点P，恒有 EMBED Equation.DSMT4 EMBED Equation.DSMT4 EMBED Equation.DSMT4 EMBED Equation.DSMT4 ，求证ACBC.18(本小题6分)(2012江苏高考改编)如图，在矩形ABCD中，AB EMBED Equation.DSMT4 ，BC2，点E为BC的中点，点F在边CD上，若 EMBED Equation.DSMT4 EMBED Equation.DSMT4 EMBED Equation.DS
10、MT4 ，求 EMBED Equation.DSMT4 EMBED Equation.DSMT4 的值INCLUDEPICTURER45.EPS19(本小题7分)(2012上海高考改编)在平行四边形ABCD中，A EMBED Equation.DSMT4 ，边AB，AD的长分别为2,1.若M，N分别是边BC，CD上的点，且满足 EMBED Equation.DSMT4 EMBED Equation.DSMT4 ，求 EMBED Equation.DSMT4 EMBED Equation.DSMT4 的取值范围参考答案1. 解析：与eq o(AB,sup6()同方向的单位向量为eq f(o(AB
11、,sup6(),|o(AB,sup6()|)eq f(3，4),r(32(4)2)eq blc(rc)(avs4alco1(f(3,5)，f(4,5)，故选A.答案：A2. 解析：因为 EMBED Equation.DSMT4 EMBED Equation.DSMT4 1(4)220，所以 EMBED Equation.DSMT4 EMBED Equation.DSMT4 .又| EMBED Equation.DSMT4 |eq r(122)eq r(5)，| EMBED Equation.DSMT4 |eq r(4)222)eq r(164)2eq r(5)，S四边形ABCD EMBED E
12、quation.DSMT4 | EMBED Equation.DSMT4 | EMBED Equation.DSMT4 |5.答案：C3. 解析：以Oeq o(A,sup6()，Oeq o(B,sup6()为邻边作一个平行四边形，将其放置在如图平面直角坐标系中，使A，B两点关于x轴对称，由已知|Oeq o(A,sup6()|Oeq o(B,sup6()|Oeq o(A,sup6()Oeq o(B,sup6()2，可得出AOB60，点A( EMBED Equation.DSMT4 ，1)，点B( EMBED Equation.DSMT4 ，1)，点D(2 EMBED Equation.DSMT4
13、 ，0)现设P(x，y)，则由Oeq o(P,sup6()eq o(OA,sup6()eq o(OB,sup6()得(x，y)( EMBED Equation.DSMT4 ，1)( EMBED Equation.DSMT4 ，1)，即eq blcrc (avs4alco1(r(3)()x，,y.)由于|1，R，可得eq blcrc (avs4alco1(r(3)xr(3)，,1y1，)画出动点P(x，y)满足的区域为如图阴影部分，故所求区域的面积为2 EMBED Equation.DSMT4 24 EMBED Equation.DSMT4 .INCLUDEPICTURER37.EPS答案：D4
14、. 解析：由ab知12m20，即m EMBED Equation.DSMT4 或 EMBED Equation.DSMT4 .答案：C5. 解析：由(mn)(mn)|m|2|n|20(1)21(2)2403.故选B.答案：B6. 解析：由题意可知 EMBED Equation.DSMT4 (2,1)， EMBED Equation.DSMT4 (5,5)，故 EMBED Equation.DSMT4 在 EMBED Equation.DSMT4 方向上的投影为eq f(o(AB,sup6()o(CD,sup6(),|o(CD,sup6()|)eq f(15,r(50)eq f(3r(2),2)
15、.答案：A7. 解析：以AB所在直线为x轴，AD所在直线为y轴建立平面直角坐标系，如图所示，则点A的坐标为(0,0)，点B的坐标为(2,0)，点D的坐标为(0,2)，点E的坐标为(1,2)，则 EMBED Equation.DSMT4 (1,2)， EMBED Equation.DSMT4 (2,2)，所以 EMBED Equation.DSMT4 EMBED Equation.DSMT4 2.答案：B8. 解析：因为cta(1t)b，所以bctab(1t)|b|2.又因为|a|b|1，且a与b夹角为60，bc，所以0t|a|b|cos 60(1t)，0 EMBED Equation.DSMT
16、4 t1t.所以t2.答案：D9. 解析：因为 EMBED Equation.DSMT4 (2,3)， EMBED Equation.DSMT4 (4,7)，所以 EMBED Equation.DSMT4 EMBED Equation.DSMT4 EMBED Equation.DSMT4 EMBED Equation.DSMT4 EMBED Equation.DSMT4 (2,3)(4,7)(24,37)(2，4)答案：A10. 解析：由ac，得ac2x40，解得x2.由bc得 EMBED Equation.DSMT4 EMBED Equation.DSMT4 ，解得y2，所以a(2,1)，b
17、(1，2)，ab(3，1)，|ab| EMBED Equation.DSMT4 ，故选B.答案：B11. 解析：由题意作图如图因为在ABC中， EMBED Equation.DSMT4 EMBED Equation.DSMT4 EMBED Equation.DSMT4 EMBED Equation.DSMT4 EMBED Equation.DSMT4 EMBED Equation.DSMT4 EMBED Equation.DSMT4 EMBED Equation.DSMT4 EMBED Equation.DSMT4 EMBED Equation.DSMT4 ( EMBED Equation.D
18、SMT4 EMBED Equation.DSMT4 ) EMBED Equation.DSMT4 EMBED Equation.DSMT4 EMBED Equation.DSMT4 EMBED Equation.DSMT4 1 EMBED Equation.DSMT4 2 EMBED Equation.DSMT4 ，所以1 EMBED Equation.DSMT4 ，2 EMBED Equation.DSMT4 .故12 EMBED Equation.DSMT4 .答案： EMBED Equation.DSMT4 12. 解析：由平行四边形法则知 EMBED Equation.DSMT4 EM
19、BED Equation.DSMT4 EMBED Equation.DSMT4 2 EMBED Equation.DSMT4 ，所以2.答案：213. 解析：因为ab(e13e2)2e12e EMBED Equation.DSMT4 6e1e22612cos EMBED Equation.DSMT4 5，所以a在b上的射影为 EMBED Equation.DSMT4 EMBED Equation.DSMT4 .答案： EMBED Equation.DSMT4 14解析：如图所示，在平行四边形ABCD中， EMBED Equation.DSMT4 EMBED Equation.DSMT4 EMB
20、ED Equation.DSMT4 ， EMBED Equation.DSMT4 EMBED Equation.DSMT4 Ceq o(E,sup6() EMBED Equation.DSMT4 EMBED Equation.DSMT4 EMBED Equation.DSMT4 .INCLUDEPICTURER42.EPS所以 EMBED Equation.DSMT4 EMBED Equation.DSMT4 ( EMBED Equation.DSMT4 EMBED Equation.DSMT4 ) EMBED Equation.DSMT4 EMBED Equation.DSMT4 | EMB
21、ED Equation.DSMT4 |2| EMBED Equation.DSMT4 |2 EMBED Equation.DSMT4 EMBED Equation.DSMT4 EMBED Equation.DSMT4 EMBED Equation.DSMT4 | EMBED Equation.DSMT4 |2 EMBED Equation.DSMT4 | EMBED Equation.DSMT4 |11，解方程得| EMBED Equation.DSMT4 | EMBED Equation.DSMT4 (舍去| EMBED Equation.DSMT4 |0)，所以线段AB的长为 EMBED
22、Equation.DSMT4 .答案： EMBED Equation.DSMT4 15. 解析：可设aij，i，j为单位向量且ij，则b6i2j，ci3j.由cab(6)i(2)j，所以 EMBED Equation.DSMT4 解得 EMBED Equation.DSMT4 所以 EMBED Equation.DSMT4 4.答案：416. 解：因为 EMBED Equation.DSMT4 EMBED Equation.DSMT4 EMBED Equation.DSMT4 ， EMBED Equation.DSMT4 EMBED Equation.DSMT4 ，又 EMBED Equati
23、on.DSMT4 EMBED Equation.DSMT4 EMBED Equation.DSMT4 ，所以( EMBED Equation.DSMT4 EMBED Equation.DSMT4 )( EMBED Equation.DSMT4 EMBED Equation.DSMT4 )0.所以 EMBED Equation.DSMT4 2 EMBED Equation.DSMT4 EMBED Equation.DSMT4 EMBED Equation.DSMT4 EMBED Equation.DSMT4 EMBED Equation.DSMT4 20，即4(1)32 EMBED Equati
24、on.DSMT4 90，即7120，所以 EMBED Equation.DSMT4 .17. 解：设 EMBED Equation.DSMT4 t EMBED Equation.DSMT4 (0t1)，INCLUDEPICTURER44.EPS所以Peq o(C,sup6() EMBED Equation.DSMT4 EMBED Equation.DSMT4 t EMBED Equation.DSMT4 EMBED Equation.DSMT4 ，所以 EMBED Equation.DSMT4 EMBED Equation.DSMT4 (t EMBED Equation.DSMT4 )(t E
25、MBED Equation.DSMT4 EMBED Equation.DSMT4 )t2 EMBED Equation.DSMT4 2t EMBED Equation.DSMT4 EMBED Equation.DSMT4 .由题意 EMBED Equation.DSMT4 EMBED Equation.DSMT4 EMBED Equation.DSMT4 EMBED Equation.DSMT4 ，即t2 EMBED Equation.DSMT4 2t EMBED Equation.DSMT4 EMBED Equation.DSMT4 EMBED Equation.DSMT4 EMBED Eq
26、uation.DSMT4 EMBED Equation.DSMT4 eq blc(rc)(avs4alco1(f(1,4)2 EMBED Equation.DSMT4 2eq f(1,4) EMBED Equation.DSMT4 EMBED Equation.DSMT4 ，即当t EMBED Equation.DSMT4 时 EMBED Equation.DSMT4 EMBED Equation.DSMT4 取得最小值由二次函数的性质可知： EMBED Equation.DSMT4 EMBED Equation.DSMT4 ，即： EMBED Equation.DSMT4 EMBED Equ
27、ation.DSMT4 EMBED Equation.DSMT4 EMBED Equation.DSMT4 2，所以 EMBED Equation.DSMT4 EMBED Equation.DSMT4 0.取AB中点M，则 EMBED Equation.DSMT4 EMBED Equation.DSMT4 EMBED Equation.DSMT4 EMBED Equation.DSMT4 EMBED Equation.DSMT4 EMBED Equation.DSMT4 ，所以 EMBED Equation.DSMT4 EMBED Equation.DSMT4 0，即ABMC.所以ACBC.1
28、8. 解：由 EMBED Equation.DSMT4 EMBED Equation.DSMT4 EMBED Equation.DSMT4 得， EMBED Equation.DSMT4 ( EMBED Equation.DSMT4 EMBED Equation.DSMT4 ) EMBED Equation.DSMT4 ，即 EMBED Equation.DSMT4 EMBED Equation.DSMT4 EMBED Equation.DSMT4 EMBED Equation.DSMT4 EMBED Equation.DSMT4 ，又因为 EMBED Equation.DSMT4 EMBED
29、 Equation.DSMT4 ，所以 EMBED Equation.DSMT4 EMBED Equation.DSMT4 0，所以 EMBED Equation.DSMT4 EMBED Equation.DSMT4 EMBED Equation.DSMT4 ，故 EMBED Equation.DSMT4 EMBED Equation.DSMT4 ( EMBED Equation.DSMT4 EMBED Equation.DSMT4 )( EMBED Equation.DSMT4 EMBED Equation.DSMT4 ) EMBED Equation.DSMT4 EMBED Equatio
30、n.DSMT4 EMBED Equation.DSMT4 EMBED Equation.DSMT4 EMBED Equation.DSMT4 EMBED Equation.DSMT4 EMBED Equation.DSMT4 EMBED Equation.DSMT4 0 EMBED Equation.DSMT4 ( EMBED Equation.DSMT4 EMBED Equation.DSMT4 ) EMBED Equation.DSMT4 | EMBED Equation.DSMT4 |20 EMBED Equation.DSMT4 EMBED Equation.DSMT4 EMBED E
31、quation.DSMT4 EMBED Equation.DSMT4 2 EMBED Equation.DSMT4 | EMBED Equation.DSMT4 |22 EMBED Equation.DSMT4 22 EMBED Equation.DSMT4 .19. 解：如图，设 EMBED Equation.DSMT4 EMBED Equation.DSMT4 ，INCLUDEPICTURER46.EPS则0,1， EMBED Equation.DSMT4 EMBED Equation.DSMT4 ( EMBED Equation.DSMT4 EMBED Equation.DSMT4 )(
32、 EMBED Equation.DSMT4 EMBED Equation.DSMT4 )( EMBED Equation.DSMT4 EMBED Equation.DSMT4 )( EMBED Equation.DSMT4 (1) EMBED Equation.DSMT4 ) EMBED Equation.DSMT4 EMBED Equation.DSMT4 (1) EMBED Equation.DSMT4 EMBED Equation.DSMT4 EMBED Equation.DSMT4 EMBED Equation.DSMT4 (1) EMBED Equation.DSMT4 EMBED Equation.DSMT4 12 EMBED Equation.DSMT4 (1)(4)1(1)(1)1442(1)26.因为0,1，所以 EMBED Equation.DSMT4 EMBED Equation.DSMT4 2,5高考资源网版权所有，侵权必究！
Copyright@ 2020-2024 m.ketangku.com网站版权所有