(完整版)高中数学平面向量知识点总结.pdf

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1、高中数学必修4 之平面向量知识点归纳一.向量的基本概念与基本运算1、向量的概念：向量：既有大小又有方向的量向量不能比较大小，但向量的模可以比较大小零向量：长度为0 的向量，记为0，其方向是任意的，0与任意向量平行单位向量：模为1 个单位长度的向量平行向量（共线向量）：方向相同或相反的非零向量相等向量：长度相等且方向相同的向量2、向量加法：设,ABa BCbuu u ru uu rrr，则a+br=ABBCuuu ruu u r=ACuuu r（1）aaa00；（2）向量加法满足交换律与结合律；ABBCCDPQQRARuuu ruuu ru uu ruuu ruuu ruuu rL，但这时必须“

2、首尾相连”3、向量的减法：相反向量：与a长度相等、方向相反的向量，叫做a的相反向量向量减法：向量a加上b的相反向量叫做a与b的差，作图法：ba可以表示为从b的终点指向a的终点的向量（a、b有共同起点）4、实数与向量的积：实数与向量a的积是一个向量，记作a，它的长度与方向规定如下：（）aa；（）当0时，a的方向与a的方向相同；当0时，a的方向与a的方向相反；当0时，0a，方向是任意的5、两个向量共线定理：向量b与非零向量a共线有且只有一个实数，使得b=a6、平面向量的基本定理：如果21,ee是一个平面内的两个不共线向量，那么对这一平面内的任一向量a，有且只有一对实数21,使：2211eea，其中

3、不共线的向量21,ee叫做表示这一平面内所有向量的一组基底二.平面向量的坐标表示1平面向量的坐标表示：平面内的任一向量ar可表示成axiyjrrr，记作ar=(x,y)。2平面向量的坐标运算：(1)若1122,ax ybxyrr，则1212,abxxyyrr(2)若2211,yxByxA，则2121,ABxxyyuuu r(3)若ar=(x,y)，则ar=(x,y)(4)若1122,ax ybxyrr，则1221/0abx yx yrr(5)若1122,ax ybxyrr，则1212a bxxyyrr若abrr，则02121yyxx三平面向量的数量积1两个向量的数量积：已知两个非零向量ar与b

4、r，它们的夹角为，则arbr=ar brcos叫做ar与br的数量积（或内积）规定00arr2向量的投影：brcos=|a barrrR，称为向量br在ar方向上的投影投影的绝对值称为射影3数量积的几何意义：arbr等于ar的长度与br在ar方向上的投影的乘积4向量的模与平方的关系：22|a aaarrrr5乘法公式成立：2222ababababrrrrrrrr；2222abaa bbrrrrrr222aa bbrrrr6平面向量数量积的运算律：交换律成立：a bb arrrr对实数的结合律成立：aba babRrrrrrr分配律成立：abca cb crrrrrrrcabrrr特别注意：（1

5、）结合律不成立：ab ca bcrrrrrr；（2）消去律不成立a ba crrr r不能得到bcrr（3）a brr=0不能得到ar=0r或br=0r7两个向量的数量积的坐标运算：已知两个向量1122(,),(,)ax ybxyrr，则arbr=1212x xy y8向量的夹角：已知两个非零向量ar与br，作OAuu u r=ar,OBuuu r=br,则 AOB=（001800）叫做向量ar与br的夹角cos=cos,aba bab?rrrrrr=222221212121yxyxyyxx当且仅当两个非零向量ar与br同方向时，=00，当且仅当ar与br反方向时=1800，同时0r与其它任何

6、非零向量之间不谈夹角这一问题9垂直：如果ar与br的夹角为 900则称ar与br垂直，记作arbr10两个非零向量垂直的充要条件：abab O02121yyxx平面向量数量积的性质一、选择题1在 ABC中，ABAC，D，E分别是 AB，AC的中点，则()A AB 与 AC 共线 B DE 与 CB 共线C AD 与 AE 相等D AD 与 BD 相等2下列命题正确的是()A向量 AB 与 BA 是两平行向量B若 a，b 都是单位向量，则abC若 AB DC，则 A，B，C，D 四点构成平行四边形D两向量相等的充要条件是它们的始点、终点相同3平面直角坐标系中，O 为坐标原点，已知两点A(3，1)

7、，B(1，3)，若点 C满足 OC OA OB，其中，R，且 1，则点 C的轨迹方程为()A3x2y110 B(x1)2(y1)2 5 C2xy0 D x2y50 4已知 a、b 是非零向量且满足(a2b)a，(b2a)b，则 a 与 b 的夹角是A6B3C23D565已知四边形ABCD是菱形，点 P在对角线AC上(不包括端点A，C)，则 AP A(AB AD)，(0，1)B (AB BC)，(0，22)C(AB AD)，(0，1)D(AB BC)，(0，22)6 ABC中，D，E，F分别是 AB，BC，AC的中点，则DF ()(第 1 题)文档编码:CK5L8Y3U6V5 HG3B8X5M1

8、F5 ZY9S4W1A10O6文档编码:CK5L8Y3U6V5 HG3B8X5M1F5 ZY9S4W1A10O6文档编码:CK5L8Y3U6V5 HG3B8X5M1F5 ZY9S4W1A10O6文档编码:CK5L8Y3U6V5 HG3B8X5M1F5 ZY9S4W1A10O6文档编码:CK5L8Y3U6V5 HG3B8X5M1F5 ZY9S4W1A10O6文档编码:CK5L8Y3U6V5 HG3B8X5M1F5 ZY9S4W1A10O6文档编码:CK5L8Y3U6V5 HG3B8X5M1F5 ZY9S4W1A10O6文档编码:CK5L8Y3U6V5 HG3B8X5M1F5 ZY9S4W1A10

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14、X5M1F5 ZY9S4W1A10O6文档编码:CK5L8Y3U6V5 HG3B8X5M1F5 ZY9S4W1A10O6文档编码:CK5L8Y3U6V5 HG3B8X5M1F5 ZY9S4W1A10O6文档编码:CK5L8Y3U6V5 HG3B8X5M1F5 ZY9S4W1A10O6A EF EDB EF DEC EF ADD EF AF7若平面向量a 与 b 的夹角为60，|b|4，(a2b)(a3b)72，则向量 a 的模为()A2 B4 C 6 D12 8 点 O 是三角形 ABC所在平面内的一点，满足 OA OB OB OC OC OA，则点 O 是 ABC的()A三个内角的角平分线的

15、交点B三条边的垂直平分线的交点C三条中线的交点D三条高的交点9在四边形ABCD中，AB a2b，BC 4ab，DC 5a3b，其中 a，b 不共线，则四边形ABCD为()A平行四边形B矩形C梯形D菱形10 如图，梯形 ABCD中，|AD|BC|，EF AB CD 则相等向量是()A AD 与 BCB OA 与 OBC AC 与 BDD EO 与 OF二、填空题11已知向量OA(k，12)，OB(4，5)，OC(k，10)，且 A，B，C 三点共线，则k12已知向量a(x3，x23x4)与 MN 相等，其中M(1，3)，N(1，3)，则 x13 已知平面上三点A，B，C满足|AB|3，|BC|4

16、，|CA|5，则 AB BC BC CA CA AB 的值等于14给定两个向量a(3，4)，b(2，1)，且(amb)(ab)，则实数m等于15已知 A，B，C三点不共线，O 是 ABC内的一点，若OA OB OC 0，则 O 是 ABC的16设平面内有四边形ABCD和点 O，OA a，OB b，OC c,OD d，若acbd，则四边形ABCD的形状是三、解答题17已知点A(2，3)，B(5，4)，C(7，10)，若点 P 满足 AP AB AC(R)，试求 为何值时，点P在第三象限内？(第 10 题)文档编码:CK5L8Y3U6V5 HG3B8X5M1F5 ZY9S4W1A10O6文档编码:

17、CK5L8Y3U6V5 HG3B8X5M1F5 ZY9S4W1A10O6文档编码:CK5L8Y3U6V5 HG3B8X5M1F5 ZY9S4W1A10O6文档编码:CK5L8Y3U6V5 HG3B8X5M1F5 ZY9S4W1A10O6文档编码:CK5L8Y3U6V5 HG3B8X5M1F5 ZY9S4W1A10O6文档编码:CK5L8Y3U6V5 HG3B8X5M1F5 ZY9S4W1A10O6文档编码:CK5L8Y3U6V5 HG3B8X5M1F5 ZY9S4W1A10O6文档编码:CK5L8Y3U6V5 HG3B8X5M1F5 ZY9S4W1A10O6文档编码:CK5L8Y3U6V5 H

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23、档编码:CK5L8Y3U6V5 HG3B8X5M1F5 ZY9S4W1A10O6文档编码:CK5L8Y3U6V5 HG3B8X5M1F5 ZY9S4W1A10O6文档编码:CK5L8Y3U6V5 HG3B8X5M1F5 ZY9S4W1A10O618如图，已知ABC，A(7，8)，B(3，5)，C(4，3)，M，N，D 分别是AB，AC，BC的中点，且MN 与 AD 交于 F，求 DF 19如图，在正方形ABCD中，E，F分别为 AB，BC的中点，求证：AFDE(利用向量证明)20已知向量a(cos ，sin )，向量 b(3，1)，则|2ab|的最大值(第 18 题)(第 19 题)文档编码:

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27、码:CK5L8Y3U6V5 HG3B8X5M1F5 ZY9S4W1A10O6文档编码:CK5L8Y3U6V5 HG3B8X5M1F5 ZY9S4W1A10O6文档编码:CK5L8Y3U6V5 HG3B8X5M1F5 ZY9S4W1A10O6文档编码:CK5L8Y3U6V5 HG3B8X5M1F5 ZY9S4W1A10O6文档编码:CK5L8Y3U6V5 HG3B8X5M1F5 ZY9S4W1A10O6文档编码:CK5L8Y3U6V5 HG3B8X5M1F5 ZY9S4W1A10O6文档编码:CK5L8Y3U6V5 HG3B8X5M1F5 ZY9S4W1A10O6文档编码:CK5L8Y3U6V5

28、 HG3B8X5M1F5 ZY9S4W1A10O6文档编码:CK5L8Y3U6V5 HG3B8X5M1F5 ZY9S4W1A10O6文档编码:CK5L8Y3U6V5 HG3B8X5M1F5 ZY9S4W1A10O6文档编码:CK5L8Y3U6V5 HG3B8X5M1F5 ZY9S4W1A10O6文档编码:CK5L8Y3U6V5 HG3B8X5M1F5 ZY9S4W1A10O6文档编码:CK5L8Y3U6V5 HG3B8X5M1F5 ZY9S4W1A10O6文档编码:CK5L8Y3U6V5 HG3B8X5M1F5 ZY9S4W1A10O6文档编码:CK5L8Y3U6V5 HG3B8X5M1F5

29、ZY9S4W1A10O6文档编码:CK5L8Y3U6V5 HG3B8X5M1F5 ZY9S4W1A10O6文档编码:CK5L8Y3U6V5 HG3B8X5M1F5 ZY9S4W1A10O6文档编码:CK5L8Y3U6V5 HG3B8X5M1F5 ZY9S4W1A10O6文档编码:CK5L8Y3U6V5 HG3B8X5M1F5 ZY9S4W1A10O6文档编码:CK5L8Y3U6V5 HG3B8X5M1F5 ZY9S4W1A10O6文档编码:CK5L8Y3U6V5 HG3B8X5M1F5 ZY9S4W1A10O6文档编码:CK5L8Y3U6V5 HG3B8X5M1F5 ZY9S4W1A10O6文

30、档编码:CK5L8Y3U6V5 HG3B8X5M1F5 ZY9S4W1A10O6文档编码:CK5L8Y3U6V5 HG3B8X5M1F5 ZY9S4W1A10O6文档编码:CK5L8Y3U6V5 HG3B8X5M1F5 ZY9S4W1A10O6文档编码:CK5L8Y3U6V5 HG3B8X5M1F5 ZY9S4W1A10O6一、选择题1B 解析：如图，AB 与 AC，AD 与 AE 不平行，AD 与 BD 共线反向2A 解析：两个单位向量可能方向不同，故B 不对若 AB DC，可能 A，B，C，D 四点共线，故C 不对两向量相等的充要条件是大小相等，方向相同，故D 也不对3D 解析：提示：设O

31、C(x，y)，OA(3，1)，OB(1，3)，OA(3，)，OB(，3)，又OAOB(3 ，3)，(x，y)(3 ，3)，33yx，又 1，由此得到答案为D4B 解析：(a2b)a，(b2a)b，(a2b)aa22ab0，(b2a)bb22ab0，a2b2，即|a|b|a|22|a|b|cos 2|a|2cos 解得 cos 21 a 与 b 的夹角是35A 解析：由平行四边形法则，AB AD AC，又 AB BC AC，由 的范围和向量数乘的长度，(0，1)6D 解析：如图，AF DE，DF DE EF EF AF 7C 解析：由(a2b)(a3b)72，得 a2ab6b2 72而|b|4，

32、ab|a|b|cos 60 2|a|，|a|22|a|96 72，解得|a|68D 解析：由OA OB OB OC OC OA，得 OA OB OC OA,即 OA(OC OB)0，故 BC OA 0，BC OA，同理可证AC OB，O 是 ABC的三条高的交点9C 解析：AD AB BC DC 8a2b2 BC，AD BC 且|AD|BC|四边形 ABCD为梯形10D 解析：AD 与 BC，AC 与 BD，OA 与 OB 方向都不相同，不是相等向量(第 1 题)文档编码:CK5L8Y3U6V5 HG3B8X5M1F5 ZY9S4W1A10O6文档编码:CK5L8Y3U6V5 HG3B8X5M

33、1F5 ZY9S4W1A10O6文档编码:CK5L8Y3U6V5 HG3B8X5M1F5 ZY9S4W1A10O6文档编码:CK5L8Y3U6V5 HG3B8X5M1F5 ZY9S4W1A10O6文档编码:CK5L8Y3U6V5 HG3B8X5M1F5 ZY9S4W1A10O6文档编码:CK5L8Y3U6V5 HG3B8X5M1F5 ZY9S4W1A10O6文档编码:CK5L8Y3U6V5 HG3B8X5M1F5 ZY9S4W1A10O6文档编码:CK5L8Y3U6V5 HG3B8X5M1F5 ZY9S4W1A10O6文档编码:CK5L8Y3U6V5 HG3B8X5M1F5 ZY9S4W1A1

34、0O6文档编码:CK5L8Y3U6V5 HG3B8X5M1F5 ZY9S4W1A10O6文档编码:CK5L8Y3U6V5 HG3B8X5M1F5 ZY9S4W1A10O6文档编码:CK5L8Y3U6V5 HG3B8X5M1F5 ZY9S4W1A10O6文档编码:CK5L8Y3U6V5 HG3B8X5M1F5 ZY9S4W1A10O6文档编码:CK5L8Y3U6V5 HG3B8X5M1F5 ZY9S4W1A10O6文档编码:CK5L8Y3U6V5 HG3B8X5M1F5 ZY9S4W1A10O6文档编码:CK5L8Y3U6V5 HG3B8X5M1F5 ZY9S4W1A10O6文档编码:CK5L8

35、Y3U6V5 HG3B8X5M1F5 ZY9S4W1A10O6文档编码:CK5L8Y3U6V5 HG3B8X5M1F5 ZY9S4W1A10O6文档编码:CK5L8Y3U6V5 HG3B8X5M1F5 ZY9S4W1A10O6文档编码:CK5L8Y3U6V5 HG3B8X5M1F5 ZY9S4W1A10O6文档编码:CK5L8Y3U6V5 HG3B8X5M1F5 ZY9S4W1A10O6文档编码:CK5L8Y3U6V5 HG3B8X5M1F5 ZY9S4W1A10O6文档编码:CK5L8Y3U6V5 HG3B8X5M1F5 ZY9S4W1A10O6文档编码:CK5L8Y3U6V5 HG3B8X

36、5M1F5 ZY9S4W1A10O6文档编码:CK5L8Y3U6V5 HG3B8X5M1F5 ZY9S4W1A10O6文档编码:CK5L8Y3U6V5 HG3B8X5M1F5 ZY9S4W1A10O6文档编码:CK5L8Y3U6V5 HG3B8X5M1F5 ZY9S4W1A10O6文档编码:CK5L8Y3U6V5 HG3B8X5M1F5 ZY9S4W1A10O6文档编码:CK5L8Y3U6V5 HG3B8X5M1F5 ZY9S4W1A10O6文档编码:CK5L8Y3U6V5 HG3B8X5M1F5 ZY9S4W1A10O6文档编码:CK5L8Y3U6V5 HG3B8X5M1F5 ZY9S4W1

37、A10O6文档编码:CK5L8Y3U6V5 HG3B8X5M1F5 ZY9S4W1A10O6文档编码:CK5L8Y3U6V5 HG3B8X5M1F5 ZY9S4W1A10O6文档编码:CK5L8Y3U6V5 HG3B8X5M1F5 ZY9S4W1A10O6文档编码:CK5L8Y3U6V5 HG3B8X5M1F5 ZY9S4W1A10O6文档编码:CK5L8Y3U6V5 HG3B8X5M1F5 ZY9S4W1A10O6文档编码:CK5L8Y3U6V5 HG3B8X5M1F5 ZY9S4W1A10O6文档编码:CK5L8Y3U6V5 HG3B8X5M1F5 ZY9S4W1A10O6文档编码:CK5

38、L8Y3U6V5 HG3B8X5M1F5 ZY9S4W1A10O6文档编码:CK5L8Y3U6V5 HG3B8X5M1F5 ZY9S4W1A10O6文档编码:CK5L8Y3U6V5 HG3B8X5M1F5 ZY9S4W1A10O6文档编码:CK5L8Y3U6V5 HG3B8X5M1F5 ZY9S4W1A10O6文档编码:CK5L8Y3U6V5 HG3B8X5M1F5 ZY9S4W1A10O6文档编码:CK5L8Y3U6V5 HG3B8X5M1F5 ZY9S4W1A10O6文档编码:CK5L8Y3U6V5 HG3B8X5M1F5 ZY9S4W1A10O6文档编码:CK5L8Y3U6V5 HG3B

39、8X5M1F5 ZY9S4W1A10O6文档编码:CK5L8Y3U6V5 HG3B8X5M1F5 ZY9S4W1A10O6文档编码:CK5L8Y3U6V5 HG3B8X5M1F5 ZY9S4W1A10O6二、填空题1132解析：A，B，C三点共线等价于AB，BC 共线，AB OB OA(4，5)(k，12)(4k，7)，BC OC OB(k，10)(4，5)(k 4，5)，又 A，B，C 三点共线，5(4k)7(k4)，k3212 1解析：M(1，3)，N(1，3)，MN(2，0)，又 aMN，043232xxx解得411xxx或 x 113 25解析：思路1：AB 3，BC 4，CA 5，A

40、BC为直角三角形且ABC90，即 AB BC，AB BC 0，AB BC BC CA CA AB BC CA CA AB CA(BC AB)(CA)2 2CA 25思路 2：AB 3，BC 4，CA 5，ABC 90，cosCABCAAB53，cosBCACABC54根据数积定义，结合图(右图)知 AB BC 0，BC CA BC CA cosACE 45(54)16，CA AB CA AB cosBAD35(53)9AB BC BC CA CA AB 0169 2514323解析：amb(32m，4m)，ab(1，5)(amb)(a b)，(amb)(a b)(32m)1(4m)50m323

41、15答案：重心解析：如图，以OA，OC 为邻边作AOCF交 AC于点 E，则 OF OA OC，又OA OC OB，(第 15 题)D(第 13 题)文档编码:CK5L8Y3U6V5 HG3B8X5M1F5 ZY9S4W1A10O6文档编码:CK5L8Y3U6V5 HG3B8X5M1F5 ZY9S4W1A10O6文档编码:CK5L8Y3U6V5 HG3B8X5M1F5 ZY9S4W1A10O6文档编码:CK5L8Y3U6V5 HG3B8X5M1F5 ZY9S4W1A10O6文档编码:CK5L8Y3U6V5 HG3B8X5M1F5 ZY9S4W1A10O6文档编码:CK5L8Y3U6V5 HG3

42、B8X5M1F5 ZY9S4W1A10O6文档编码:CK5L8Y3U6V5 HG3B8X5M1F5 ZY9S4W1A10O6文档编码:CK5L8Y3U6V5 HG3B8X5M1F5 ZY9S4W1A10O6文档编码:CK5L8Y3U6V5 HG3B8X5M1F5 ZY9S4W1A10O6文档编码:CK5L8Y3U6V5 HG3B8X5M1F5 ZY9S4W1A10O6文档编码:CK5L8Y3U6V5 HG3B8X5M1F5 ZY9S4W1A10O6文档编码:CK5L8Y3U6V5 HG3B8X5M1F5 ZY9S4W1A10O6文档编码:CK5L8Y3U6V5 HG3B8X5M1F5 ZY9S

43、4W1A10O6文档编码:CK5L8Y3U6V5 HG3B8X5M1F5 ZY9S4W1A10O6文档编码:CK5L8Y3U6V5 HG3B8X5M1F5 ZY9S4W1A10O6文档编码:CK5L8Y3U6V5 HG3B8X5M1F5 ZY9S4W1A10O6文档编码:CK5L8Y3U6V5 HG3B8X5M1F5 ZY9S4W1A10O6文档编码:CK5L8Y3U6V5 HG3B8X5M1F5 ZY9S4W1A10O6文档编码:CK5L8Y3U6V5 HG3B8X5M1F5 ZY9S4W1A10O6文档编码:CK5L8Y3U6V5 HG3B8X5M1F5 ZY9S4W1A10O6文档编码:

44、CK5L8Y3U6V5 HG3B8X5M1F5 ZY9S4W1A10O6文档编码:CK5L8Y3U6V5 HG3B8X5M1F5 ZY9S4W1A10O6文档编码:CK5L8Y3U6V5 HG3B8X5M1F5 ZY9S4W1A10O6文档编码:CK5L8Y3U6V5 HG3B8X5M1F5 ZY9S4W1A10O6文档编码:CK5L8Y3U6V5 HG3B8X5M1F5 ZY9S4W1A10O6文档编码:CK5L8Y3U6V5 HG3B8X5M1F5 ZY9S4W1A10O6文档编码:CK5L8Y3U6V5 HG3B8X5M1F5 ZY9S4W1A10O6文档编码:CK5L8Y3U6V5 H

45、G3B8X5M1F5 ZY9S4W1A10O6文档编码:CK5L8Y3U6V5 HG3B8X5M1F5 ZY9S4W1A10O6文档编码:CK5L8Y3U6V5 HG3B8X5M1F5 ZY9S4W1A10O6文档编码:CK5L8Y3U6V5 HG3B8X5M1F5 ZY9S4W1A10O6文档编码:CK5L8Y3U6V5 HG3B8X5M1F5 ZY9S4W1A10O6文档编码:CK5L8Y3U6V5 HG3B8X5M1F5 ZY9S4W1A10O6文档编码:CK5L8Y3U6V5 HG3B8X5M1F5 ZY9S4W1A10O6文档编码:CK5L8Y3U6V5 HG3B8X5M1F5 ZY

46、9S4W1A10O6文档编码:CK5L8Y3U6V5 HG3B8X5M1F5 ZY9S4W1A10O6文档编码:CK5L8Y3U6V5 HG3B8X5M1F5 ZY9S4W1A10O6文档编码:CK5L8Y3U6V5 HG3B8X5M1F5 ZY9S4W1A10O6文档编码:CK5L8Y3U6V5 HG3B8X5M1F5 ZY9S4W1A10O6文档编码:CK5L8Y3U6V5 HG3B8X5M1F5 ZY9S4W1A10O6文档编码:CK5L8Y3U6V5 HG3B8X5M1F5 ZY9S4W1A10O6文档编码:CK5L8Y3U6V5 HG3B8X5M1F5 ZY9S4W1A10O6文档编

47、码:CK5L8Y3U6V5 HG3B8X5M1F5 ZY9S4W1A10O6文档编码:CK5L8Y3U6V5 HG3B8X5M1F5 ZY9S4W1A10O6文档编码:CK5L8Y3U6V5 HG3B8X5M1F5 ZY9S4W1A10O6文档编码:CK5L8Y3U6V5 HG3B8X5M1F5 ZY9S4W1A10O6文档编码:CK5L8Y3U6V5 HG3B8X5M1F5 ZY9S4W1A10O6文档编码:CK5L8Y3U6V5 HG3B8X5M1F5 ZY9S4W1A10O6OF 2 OE OB O 是 ABC的重心16答案：平行四边形解析：acbd，abdc，BA CD 四边形 ABC

48、D为平行四边形三、解答题17 1解析：设点P的坐标为(x，y)，则 AP(x，y)(2，3)(x2，y3)AB AC(5，4)(2，3)(7，10)(2，3)(3，)(5，7)(35，7)AP AB AC，(x2，y3)(35，17)713532yx即7455yx要使点 P在第三象限内，只需074055解得 118 DF(47，2)解析：A(7，8)，B(3，5)，C(4，3)，AB(4，3)，AC(3，5)又 D 是 BC的中点，AD 21(AB AC)21(43，35)21(7，8)(27，4)又 M，N 分别是 AB，AC的中点，F是 AD 的中点，DF FD 21AD 21(27，4)

49、(47，2)19证明：设AB a，AD b，则 AF a21b，EDb21aAF ED(a21b)(b21a)21b221a243ab又AB AD，且 AB AD，a2 b2，ab0AF ED 0，AF ED 本题也可以建平面直角坐标系后进行证明20分析：思路1：2ab(2cos 3，2sin 1)，|2a b|2(2cos 3)2(2sin 1)284sin 43cos 又 4sin 43cos 8(sin cos3cos sin3)8sin(3)，最大值为8，|2a b|2的最大值为16，|2ab|的最大值为4思路 2：将向量2a，b 平移，使它们的起点与原点重合，则|2a b|表示 2a

50、，b终点间的距离|2a|2，所以 2a 的终点是以原点为圆心，2 为半径的圆上的动点P，b 的终点是该圆上的一个定点Q，由圆的知识可知，|PQ|的最大值为直径的长为4(第 18 题)(第 19 题)文档编码:CK5L8Y3U6V5 HG3B8X5M1F5 ZY9S4W1A10O6文档编码:CK5L8Y3U6V5 HG3B8X5M1F5 ZY9S4W1A10O6文档编码:CK5L8Y3U6V5 HG3B8X5M1F5 ZY9S4W1A10O6文档编码:CK5L8Y3U6V5 HG3B8X5M1F5 ZY9S4W1A10O6文档编码:CK5L8Y3U6V5 HG3B8X5M1F5 ZY9S4W1A