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qingdaonewscom:18版高中数学第二章平面向量2.4.2平面向量数量积的坐标表示、模、夹角学业分层测评新人教A版必修4

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2.4.2 平面向量数量积的坐标表示、模、夹角 (建议用时:45 分钟) [学业达标] 一、选择题 1. 已知向量 a=(3,1), b=(x, -2), c=(0,2), 若 a⊥(b-c), 则实数 x 的值为( 4 A. 3 3 C.- 4 3 B. 4 4 D.- 3 ) 4 【解析】 b-c=(x,-4),由 a⊥(b-c)知 3x-4=0,∴x= .故选 A. 3 【答案】 A 2.已知向量 a=(1,-2),b=(x,4),且 a∥b,则|a-b|=( A.5 3 C.2 5 【解析】 ∵a∥b,∴4+2x=0, ∴x=-2,a-b=(1,-2)-(-2,4)=(3,-6), ∴|a-b|=3 5.故选 B. 【答案】 B 3.已知向量 a=(1, 3),b=(-2,2 3),则 a 与 b 的夹角是( π A. 6 π C. 3 【解析】 设 a 与 b 的夹角为 θ , 则 cos θ = π B. 4 π D. 2 ) B.3 5 D.2 2 ) a·b = |a||b| , 3 -2,2 3 2×4 1 = , 2 π 解得 θ = .故选 C. 3 【答案】 C 4.若 a=(2,3),b=(-4,7),则 a 在 b 方向上的投影为( A. C. 65 5 13 5 B. 65 D. 13 ) 【导学号:00680059】 1 【解析】 a 在 b 方向上的投影为|a|cos〈a,b〉= 2× -4 +3×7 65 = . 5 65 【答案】 A a·b 2,3 · -4,7 = = 2 2 |b | -4 +7 5.已知正方形 OABC 两边 AB,BC 的中点分别为 D 和 E,则∠DOE 的余弦值为( 1 A. 2 3 C. 5 B. 3 2 ) 4 D. 5 ? 1? 【解析】 以点 O 为原点, OA 所在直线为 x 轴建立直角坐标系, 设边长为 1, 则 D?1, ?, ? 2? ? ? E? ,1?,于是 cos∠DOE= 2 1 ? ? ?1,1?·?1,1? ? 2? ?2 ? ? ? ? ? ?1?2 2 1 +? ? · ?2? 4 = . ?1?2+12 5 ?2? ? ? 【答案】 D 二、填空题 6.已知 O A =(-2,1),O B =(0,2),且 A C ∥O B ,B C ⊥A B ,则点 C 的坐标是____ . 【解析】 设 C(x,y),则 A C =(x+2,y-1), → → → → → → → → → B C =(x,y-2),A B =(2,1). 由 A C ∥O B ,B C ⊥A B ,得 ? x+ =0, ? ? ?2x+y-2=0, ? → → → → 解得? ?x=-2, ? ?y=6, ? ∴点 C 的坐标为(-2,6). 【答案】 (-2,6) 7.若向量 a=(-2,2)与 b=(1,y)的夹角为钝角,则 y 的取值范围为________. 【解析】 若 a 与 b 夹角为 180°,则有 b=λ a(λ <0), 1=-2λ , ? ? 即?y=2λ , ? ?λ <0, 若 a 与 b 夹角 θ ∈? 1 解得 y=-1 且 λ =- ,所以 b≠λ a(λ <0)时,y≠-1;① 2 ?π ,π ?时,则只要 a·b<0 且 b≠λ a(λ <0). ? ?2 ? 当 a·b<0,有-2+2y<0,解得 y<1.② 由①②得 y<-1 或-1<y<1. 2 【答案】 (-∞,-1)∪(-1,1) 三、解答题 → → → 8.已知AB=(6,1),BC=(4,k),CD=(2,1). (1)若 A,C,D 三点共线,求 k 的值; → → (2)在(1)的条件下,求向量BC与CD的夹角的余弦值. 【导学号:70512037】 → → → 【解】 (1)因为AC=AB+BC=(10,k+1),由题意知 A,C,D 三点共线, → → 所以AC∥CD,所以 10×1-2(k+1)=0,即 k=4. → → BC·CD 12 → → → (2)因为CD=(2,1),设向量BC与CD的夹角为 θ ,则 cos θ = = = → → |BC||CD| 4 2× 5 3 10 . 10 9.已知 a=(1,1),b=(0,-2),当 k 为何值时, (1)ka-b 与 a+b 共线; (2)ka-b 与 a+b 的夹角为 120°. 【解】 ∵a=(1,1),b=(0,-2), ka-b=k(1,1)-(0,-2)=(k,k+2), a+b=(1,1)+(0,-2)=(1,-1). (1)∵ka-b 与 a+b 共线, ∴k+2-(-k)=0,∴k=-1. 即当 k=-1 时,ka-b 与 a+b 共线. (2)∵|ka-b|= k + k+ |a+b|= 1 + - 2 2 2 2 , = 2, (ka-b)·(a+b)=(k,k+2)·(1,-1) =k-k-2=-2, 而 ka-b 与 a+b 的夹角为 120°, ∴cos 120°= ka-b a+b , |ka-b||a+b| 2 1 -2 即- = 2 2 2· k + k+ 2 , 化简整理,得 k +2k-2=0,解之得 k=-1± 3. 即当 k=-1± 3时,ka-b 与 a+b 的夹角为 120°. [能力提升] 1.已知向量 a=(1,2),b=(2,-3).若向量 c 满足(c+a)∥b,c⊥(a+b),则 c 等 3 于( ) ?7 7? A.? , ? ?9 3? ?7 7? C.? , ? ?3 9? 【解析】 设 c=(x,y), 又因为 a=(1,2),b=(2,-3), 所以 c+a=(x+1,y+2). 又因为(c+a)∥b, 所以有(x+1)·(-3)-2·(y+2)=0, 即-3x-2y-7=0,① 又 a+b=(3,-1),

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