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www.xb8833.com:2019届高中数学人教A必修4学业分层测评17 平面向量共线的坐标表示 含解析

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2019 届数学人教版精品资料 学业分层测评(十七) (建议用时:45 分钟) [学业达标] 一、选择题 1.已知向量 a=(2,3),b=(-1,2),若 ma+4b 与 a-2b 共线,则 m 的值 为( ) 1 A.2 1 C.-2 【解析】 B.2 D.-2 ma+4b=(2m-4,3m+8),a-2b=(4,-1),由 ma+4b 与 a -2b 共线,有-(2m-4)=4(3m+8),解得 m=-2,故选 D. 【答案】 D 2.已知 A,B,C 三点共线,且 A(3,-6),B(-5,2),若 C 点的横坐标为 6,则 C 点的纵坐标为( ) 【导学号:00680053】 A.-13 C.-9 【解析】 → ∥AC →, 设 C(6,y),∵AB B.9 D.13 → =(-8,8),AC → =(3,y+6), 又AB ∴-8×(y+6)-3×8=0, ∴y=-9. 【答案】 C ?1 ? 3.已知向量 a=(1-sin θ ,1),b=?2,1+sin θ?,且 a∥b,则锐角 θ 等于 ? ? ( ) A.30° C.60° 【解析】 B.45° D.75° 1 2 由 a∥b,可得(1-sin θ)(1+sin θ)-2=0,即 cos θ=± 2 ,而 θ 是锐角,故 θ=45°. 【答案】 B 4.(2016· 马鞍山期末)已知向量 a=(1,-2),b=(m,4),且 a∥b,那么 2a -b=( ) B.(0,4) D.(-4,8) A.(4,0) C.(4,-8) 【解析】 由 a∥b 知 4+2m=0,∴m=-2,2a-b=(2,-4)-(-2,4)= (4,-8).故选 C. 【答案】 C ) 5.如果向量 a=(k,1),b=(4,k)共线且方向相反,则 k 等于( A.±2 C.-2 【解析】 C. 【答案】 二、填空题 C B.2 D.0 由 a,b 共线得 k2=4,又两个向量的方向相反,故 k=-2.故选 6.已知向量 a=(-2,3),b∥a,向量 b 的起点为 A(1,2),终点 B 在坐标 轴上,则点 B 的坐标为________. → =(x-1,y 【解析】 由 b∥a,可设 b=λa=(-2λ,3λ).设 B(x,y),则AB -2)=b. ? ?-2λ=x-1, ? ?x=1-2λ, 由? ?? ? ? ?3λ=y-2, ?y=3λ+2, 又 B 点在坐标轴上,则 1-2λ=0 或 3λ+2=0, 7? ?7 ? ? 所以 B?0,2?或?3,0?. ? ? ? ? 7? ?7 ? ? 【答案】 ?0,2?或?3,0? ? ? ? ? 7.向量 a=(1,-2),向量 b 与 a 共线,且|b|=4|a|,则 b=________. 【解析】 因为 b∥a,令 b=λa=(λ,-2λ), 又|b|=4|a|, 所以(λ)2+(-2λ)2=16(1+4),故有 λ2=16,解得 λ=± 4,∴b=(4,-8)或(- 4,8). 【答案】 三、解答题 1→ → =1AB → → → 8.已知点 A(-1,2),B(2,8)及AC 3 ,DA=-3BA,求点 C,D 和CD 的坐标. → =(x +1,y -2),AB →= 【解】 设点 C(x1,y1),D(x2,y2),由题意可得AC 1 1 → =(-1-x ,2-y ),BA → =(-3,-6), (3,6),DA 2 2 → =1AB → ,DA → =-1BA → 因为AC 3 3 , 1 所以(x1+1,y1-2)=3(3,6)=(1,2), 1 (-1-x2,2-y2)=-3(-3,-6) ? ?x1+1=1, ? ?-1-x2=1, =(1,2),则有? 和? ? ?y1-2=2, ? ?2-y2=2, ? ?x1=0, ? ?x2=-2, 解得? 和? ? ?y1=4, ? ?y2=0. → =(-2,-4). 所以点 C,D 的坐标分别为(0,4)和(-2,0),所以CD 9.如图 2-3-21,在△OCB 中,A 是 CB 的中点,D 是 OB 的靠近 B 点的 → =λOA → ,求实数 λ 的值. 一个三等分点,DC 与 OA 交于点 E,若OE (4,-8)或(-4,8) 图 2-3-21 【解】 ∵C、E、D 三点共线, → =xCD →, ∴存在实数 x,有CE → -OC → =x(OD → -OC → ), ∴OE → -OC → ∴λOA ?2 → → ? =x?3OB-OC?, ? ? 又 A 是 CB 的中点, 2→ →? 1 → → → =x? ?3OB-OC?, ∴λ·2(OB +OC)-OC ? ? λ → ?λ ? → 2 → →, ∴2OB +? -1?OC=3xOB-xOC 2 ? ? 2 ?λ = 2 3x, ∴? λ ?2-1=-x, 5 4 ∴4λ-1=0,∴λ=5. [能力提升] → =i+2j,DC → =(3-x)i+(4-y)j(其中 i,j 的方 1.(2016· 温州高一检测)若AB → 与DC → 共线,则 x,y 的值可能分别 向分别与 x,y 轴正方向相同且为单位向量).AB 为( ) A.1,2 C.3,2 【解析】 B.2,2 D.2,4 → =(1,2),DC → =(3-x,4-y), 因为AB → ∥DC →, 又AB 所以 4-y-2×(3-x)=0, 即 2x-y-2=0,验知 B 合适. 【答案】 B 2.已知四边形 ABCD 是边长为 6 的正方形,E 为 AB 的中点,点 F 在 BC 上,且 BF∶FC=2∶1,AF 与 EC 相交于点 P,求四边形 APCD 的面积. 【解】 → 为 x 轴建立直角坐标系,如图所示, 以 A 为坐标原点,AB ∴A(0,0),B(6,0),C(6,6),D(0,6), F(6,4),E(3,0), → =(x,y

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