專題二 第2講 向量共線定理的應(yīng)用

1、4/4第2講向量共線定理的應(yīng)用向量共線定理可以解決一些向量共線，點(diǎn)共線問題，也可由共線求參數(shù)；對于線段的定比分點(diǎn)問題，用向量共線定理求解則更加簡潔例1(1)若點(diǎn)M是ABC所在平面內(nèi)一點(diǎn)，且滿足|3eq o(AM,sup6()eq o(A B,sup6()eq o(AC,sup 6()|0，則ABM與ABC的面積之比等于()A.eq f(3,4) B.eq f(1,4) C.eq f(1,3) D.eq f(1,2)答案C解析|3eq o(AM,sup6()eq o(AB,sup6()eq o(AC,sup 6()|0，3eq o(AM,sup6()eq o(AB,sup6()eq o(AC,s

2、up 6()0，eq o(AB,sup6()eq o(AC,sup 6()3eq o(AM,sup6().設(shè)BC的中點(diǎn)為G，則eq o(AB,sup6()eq o(AC,sup 6()2eq o(AG,sup6()，3eq o(AM,sup6()2eq o(AG,sup6()，即eq o(AM,sup6()eq f(2,3)eq o(AG,sup6()，點(diǎn)M在線段AG上，且eq f(o(|A M,sup6()|,o(|A G,sup6()|)eq f(2,3).eq f(SABM,SABG)eq f(|o(AM,sup6()|,|o(AG,sup6()|)eq f(2,3)，易得eq f(SA

3、BG,SABC)eq f(|o(BG,sup6()|,|o(BC,sup6()|)eq f(1,2)，eq f(SABM,SABC)eq f(SABM,SABG)eq f(SABG,SABC)eq f(2,3)eq f(1,2)eq f(1,3)，即ABM與ABC的面積之比等于eq f(1,3).(2)在ABC中，eq o(AN,sup6()eq f(1,2)eq o(AC,sup 6()，P是BN上的一點(diǎn)，若eq o(AP,sup6()meq o(AB,sup6()eq f(3,8)eq o(AC,sup6()，則實(shí)數(shù)m的值為_答案eq f(1,4)解析方法一B，P，N三點(diǎn)共線，eq o(B

4、P,sup6()eq o(PN,sup6()，存在實(shí)數(shù)，使得eq o(BP,sup6()eq o(PN,sup6()(0)，eq o(AP,sup6()eq o(AB,sup6()(eq o(AN,sup6()eq o(AP,sup6()，0，eq o(AP,sup6()eq f(1,1) eq o(AB,sup6()eq f(,1)eq o(AN,sup6().eq o(AN,sup6()eq f(1,2)eq o(AC,sup 6()，eq o(AP,sup6()meq o(AB,sup6()eq f(3,8)eq o(AC,sup 6()，eq o(AP,sup6()meq o(AB,s

5、up6()eq f(3,4)eq o(AN,sup6()，eq blcrc (avs4alco1(f(1,1)m，,f(,1)f(3,4)，)解得eq blcrc (avs4alco1(3，,mf(1,4).)方法二eq o(AN,sup6()eq f(1,2)eq o(AC,sup 6()，eq o(AP,sup6()meq o(AB,sup6()eq f(3,8)eq o(AC,sup 6()，eq o(AP,sup6()meq o(AB,sup6()eq f(3,4)eq o(AN,sup6().B，P，N三點(diǎn)共線，meq f(3,4)1，meq f(1,4).例2(1)(2020河北省

6、石家莊一中質(zhì)檢)在ABC中，D 為線段AC的中點(diǎn)，點(diǎn)E在邊BC上，且BEeq f(1,2)EC，AE與BD交于點(diǎn)O，則eq o(AO,sup6()等于()A.eq f(1,2)eq o(AB,sup6()eq f(1,4)eq o(AC,sup 6() B.eq f(1,4)eq o(AB,sup6()eq f(1,4)eq o(AC,sup 6()C.eq f(1,4)eq o(AB,sup6()eq f(1,2)eq o(AC,sup 6() D.eq f(1,2)eq o(AB,sup6()eq f(1,2)eq o(AC,sup 6()答案A解析如圖，設(shè)eq o(AO,sup6()eq

7、 o(AE,sup6()(0)，又eq o(AE,sup6()eq o(AB,sup6()eq f(1,3)eq o(BC,sup6()eq f(2,3)eq o(AB,sup6()eq f(1,3)eq o(AC,sup6()，eq o(AO,sup6()eq f(2,3)eq o(AB,sup6()eq f(1,3)eq o(AC,sup6()eq f(2,3)eq o(AB,sup6()eq f(2,3)eq o(AD,sup6().又B，O，D三點(diǎn)共線，eq f(2,3)eq f(2,3)1，eq f(3,4)，eq o(AO,sup6()eq f(1,2)eq o(AB,sup6()

8、eq f(1,4)eq o(AC,sup6().(2)在ABC中，過中線AD的中點(diǎn)E任作一直線分別交AB，AC于M，N兩點(diǎn)，設(shè)eq o(AM,sup6()xeq o(AB,sup6()， eq o(AN,sup6()yeq o(AC,sup6()(xy0)，則4xy的最小值是_答案eq f(9,4)解析由D為BC的中點(diǎn)知，eq o(AD,sup6()eq f(1,2)eq o(AB,sup6()eq f(1,2)eq o(AC,sup6()，又eq o(AM,sup6()xeq o(AB,sup6()，eq o(AN,sup6()yeq o(AC,sup6()(xy0)，E為AD的中點(diǎn)，故eq

9、 o(AE,sup6()eq f(1,2)eq o(AD,sup6()eq f(1,4x)eq o(AM,sup6()eq f(1,4y)eq o(AN,sup6()，M，E，N三點(diǎn)共線，eq f(1,4x)eq f(1,4y)1，4xy(4xy)eq blc(rc)(avs4alco1(f(1,4x)f(1,4y)eq f(y,4x)eq f(x,y)eq f(5,4)2eq r(f(y,4x)f(x,y)eq f(5,4)eq f(9,4)，當(dāng)且僅當(dāng)eq f(y,4x)eq f(x,y)，即xeq f(3,8)，yeq f(3,4)時取等號4xy的最小值為eq f(9,4).(1)若eq

10、o(OA,sup6()eq o(OB,sup6()eq o(OC,sup6()(，為常數(shù))，則A，B，C三點(diǎn)共線的充要條件是1.(2)使用條件“兩條線段的交點(diǎn)”時，可轉(zhuǎn)化成兩次向量共線，進(jìn)而確定交點(diǎn)位置1.如圖，ABC中，ADDB，AEEC，CD與BE交于點(diǎn)F，設(shè)eq o(AB,sup6()a，eq o(AC,sup6()b，eq o(AF,sup6()xayb，則(x，y)等于()A.eq blc(rc)(avs4alco1(f(1,2)，f(1,2) B.eq blc(rc)(avs4alco1(f(2,3)，f(2,3)C.eq blc(rc)(avs4alco1(f(1,3)，f(1,

11、3) D.eq blc(rc)(avs4alco1(f(2,3)，f(1,2)答案C解析由題意得，eq o(AF,sup6()xaybxeq o(AB,sup6()2yeq o(AE,sup6()，B，F(xiàn)，E三點(diǎn)共線，x2y1，同理，eq o(AF,sup6()2xeq o(AD,sup6()yeq o(AC,sup6()，D，F(xiàn)，C三點(diǎn)共線，2xy1，由得xyeq f(1,3)，(x，y)eq blc(rc)(avs4alco1(f(1,3)，f(1,3).2(2020河北省石家莊二中調(diào)研)已知在ABC中，ABAC3，D為邊BC上一點(diǎn)，eq o(AB,sup6()eq o(AD,sup6()

12、6，eq o(AC,sup6()eq o(AD,sup6()eq f(15,2)，則eq o(AB,sup6()eq o(AC,sup6()的值為_答案eq f(9,2)解析D為邊BC上一點(diǎn)，可設(shè)eq o(BD,sup6()eq o(BC,sup6()，eq o(A D,sup6()eq o(AB,sup6()Beq o(D,sup6()(1)eq o(AB,sup6()eq o(AC,sup6().eq blcrc (avs4alco1(o(AB,sup6()o(AD,sup6()19o(AB,sup6()o(AC,sup6()6，,o(AC,sup6()o(AD,sup6()1o(AB,s

13、up6()o(AC,sup6()9f(15,2) ，)得，9eq o(AB,sup6()eq o(AC,sup6()eq f(27,2)，eq o(AB,sup6()eq o(AC,sup6()eq f(9,2).3如圖，在直角梯形ABCD中，ABCD，DAB90，ADAB4，CD1，動點(diǎn)P在邊BC上，且滿足eq o(AP,sup6()meq o(AB,sup6()neq o(AD,sup6()(m，n均為正實(shí)數(shù))，則eq f(1,m)eq f(1,n)的最小值為_答案eq f(74r(3),4)解析設(shè)eq o(AB,sup6()a，eq o(AD,sup6()b，則eq o(BC,sup6()eq o(BA,sup6()eq o(AD,sup6()eq o(DC,sup6()abeq f(1,4)beq f(3,4)ab.設(shè)eq o(BP,sup6()eq o(BC,sup6()，則eq o(AP,sup6()eq o(AB,sup6()eq o(BP,sup6()eq blc(rc)(avs4alco1(1f(3,4)ab.因?yàn)閑q o(AP,sup6()manb，所以1eq f(3,4)m，n，消去得meq f(3,4)n1，eq f(1,m)eq f