momo22166:高中数学人教a必修4评19 平面向量数量积的坐标表示、模、夹角 含解析

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学业分层测评(十九) (建议用时:45 分钟) [学业达标] 一、选择题 1.(2016· 开封质检)已知向量 a=(3,1),b=(x,-2),c=(0,2),若 a⊥(b -c),则实数 x 的值为( A. 4 3 3 4 ) B. 3 4 4 3 C.- D.- 【解析】 b-c=(x,-4),由 a⊥(b-c)知 3x-4=0, 4 ∴x= .故选 A. 3 【答案】 A 2.(2016· 马鞍山质检)已知向量 a=(1,-2),b=(x,4),且 a∥b,则|a- b|=( ) 3 5 ∵a∥b,∴4+2x=0, B .3 D.2 5 2 A.5 C.2 【解析】 ∴x=-2,a-b=(1,-2)-(-2,4)=(3,-6), ∴|a-b|=3 【答案】 B 3),b=(-2,2 3),则 a 与 b 的夹角是( ) 5.故选 B. 3.已知向量 a=(1, A. π 6 π 3 设 a 与 b 的夹角为 θ, B. π 4 π 2 C. D. 【解析】 a·b (1, 则 cos θ= = |a||b| π 解得 θ= .故选 C. 3 【答案】 C 3)· (-2,2 2×4 3) 1 = , 2 4.若 a=(2,3),b=(-4,7),则 a 在 b 方向上的投影为( 【00680059】 A. 65 5 13 5 B. 65 ) C. D. 13 a· b |b| = (2,3)· (-4,7) (-4)2+72 【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 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( A. 1 2 ) B. 3 2 C. 3 5 D. 4 5 【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? ? ??1 ? 1 , , 1 ? ? ? ? 2??2 4 ? ? = . ?1?2 ?1?2 5 ? ? ? ? 2 2 1 +? ? · ?2? +1 ? 2? ? ? ? ?1 ? 1? ? ? ? ? 1,则 D?1, ?,E? ,1?,于是 cos∠DOE= 2? ? ?2 ? 【答案】 二、填空题 D → =(-2,1),OB → =(0,2),且AC → ∥OB → ,BC → ⊥AB → ,则点 C 的坐标 6.已知OA 是________. 【解析】 → =(x+2,y-1), 设 C(x,y),则AC → =(x,y-2),AB → =(2,1). BC → ∥OB → ,BC → ⊥AB → ,得 由AC ? ? ?2(x+2)=0, ?x=-2, ? 解得? ? ? ?2x+y-2=0, ?y=6, ∴点 C 的坐标为(-2,6). 【答案】 (-2,6) 7.(2016· 德州高一检测)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=λa(λ<0) 1=-2λ ? ? 1 即?y=2λ ,解得 y=-1 且 λ=- ,所以 b≠λa(λ<0)时 y≠-1;① 2 ? ?λ<0 ?π ? ? ? 若 a 与 b 夹角 θ∈? ,π?时,则只要 a· b<0 且 b≠λa(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 【答案】 三、解答题 → =(6,1),BC → =(4,k),CD → =(2,1). 8.已知AB (1)若 A,C,D 三点共线,求 k 的值; → 与CD → 的夹角的余弦值. (2)在(1)的条件下,求向量BC → =AB → +BC → =(10,k+1),由题意知 A,C,D 三点共线, 【解】 (1)因为AC → ∥CD → ,所以 10×1-2(k+1)=0,即 k=4. 所以AC → ·CD → BC → → → (2)因为CD=(2,1),设向量BC与CD的夹角为 θ,则 cos θ= = → ||CD →| |BC 12 4 2× 5 3 10 . 10 (-∞,-1)∪(-1,1) = 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|= |a+b|= k2+(k+2)2, 2, 12+(-1)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°, (ka-b)· (a+b) ∴cos 120°= , |ka-b||a+b| 1 即- = 2 , 2· k2+(k+2)2 -2 化简整理,得 k2+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 等于( ) ? 7 7? ? ? B.?- ,- ? 9? ? 3 ? 7 7? ? ? D.?- ,- ? 3? ? 9 设 c=(x,y), ?7 7? ? ? A.? , ? ?9 3? ?7 7? ? ? C.? , ? ?3 9? 【解析】 又因为 a=(1,2),b=(2

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