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wwwsb8366com:高中数学人教A必修4学业分层测评18 平面向量数量积的物理背景及其含义 含解析

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学业分层测评(十八) (建议用时:45 分钟) [学业达标] 一、选择题 2 1.已知|b|=3,a 在 b 方向上的投影是3,则 a· b 为( 1 A.3 C.3 【解析】 由数量积的几何意义知 4 B.3 D.2 ) 2 a· b=3×3=2,故选 D. 【答案】 D 2. 设 e1 和 e2 是互相垂直的单位向量, 且 a=3e1+2e2, b=-3e1+4e2, 则 a· b 等于( ) B.-1 D.2 因为|e1|=|e2|=1,e1·e2=0,所以 a· b=(3e1+2e2)· (-3e1+4e2)= A.-2 C.1 【解析】 -9|e1|2+8|e2|2+6e1·e2=-9×12+8×12+6×0=-1.故选 B. 【答案】 B 3.若向量 a 与 b 的夹角为 60°,|b|=4,且(a+2b)· (a-3b)=-72,则 a 的 模为( A.2 C.6 【解析】 ) B.4 D.12 ∵(a+2b)· (a-3b)=a2-a· b-6b2 =|a|2-|a|· |b|cos 60°-6|b|2 =|a|2-2|a|-96=-72, ∴|a|2-2|a|-24=0, ∴|a|=6. 【答案】 C 4.(2016· 宁波期末)已知向量 a,b 满足|a|=2,|b|=1,a·b=1,则向量 a 与 a-b 的夹角为( ) 【导学号:00680056】 π A. 6 5π C. 6 【解析】 |a-b|= (a-b)2= π B. 3 2π D. 3 a2+b2-2a· b= 3, 设向量 a 与 a-b 的夹角为 θ,则 cos θ= a· (a-b) |a||a-b| = 3 = 2 ,又 θ∈[0,π], 2× 3 22-1 π 所以 θ= 6 .故选 A. 【答案】 A ) 5.已知点 A,B,C 满足||=3,||=4,||=5,则·+·+·的值是( A.-25 C.-24 【解析】 B.25 D.24 因为||2+||2=9+16=25=||2, 所以∠ABC=90°, 所以原式=·+(+)=0+·=-2=-25. 【答案】 二、填空题 6.已知 a⊥b,|a|=2,|b|=1,且 3a+2b 与 λa-b 垂直,则 λ 等于________. 【解析】 ∵(3a+2b)⊥(λa-b), A ∴(λa-b)· (3a+2b)=0, ∴3λa2+(2λ-3)a· b-2b2=0. 又∵|a|=2,|b|=1,a⊥b, 1 ∴12λ-2=0,∴λ=6. 【答案】 1 6 7. 已知|a|=|b|=|c|=1, 且满足 3a+mb+7c=0, 其中 a 与 b 的夹角为 60°, 则实数 m=________. 【解析】 ∵3a+mb+7c=0,∴3a+mb=-7c, ∴(3a+mb)2=(-7c)2,化简得 9+m2+6ma· b=49. 1 又 a· b=|a||b|cos 60°=2,∴m2+3m-40=0, 解得 m=5 或 m=-8. 【答案】 三、解答题 8.已知向量 a、b 的长度|a|=4,|b|=2. (1)若 a、b 的夹角为 120° ,求|3a-4b|; (2)若|a+b|=2 3,求 a 与 b 的夹角 θ. 【解】 ? 1? (1)a· b=|a||b|cos 120°=4×2×?-2?=-4. ? ? 5 或-8 又|3a-4b|2=(3a-4b)2=9a2-24a· b+16b2 =9×42-24×(-4)+16×22=304, ∴|3a-4b|=4 19. (2)∵|a+b|2=(a+b)2=a2+2a· b+b2 =42+2a· b+22=(2 3)2, a· b -4 1 ∴a· b=-4,∴cos θ=|a||b|= =-2. 4×2 2π 又 θ∈[0,π],∴θ= 3 . 9.在△ABC 中,=a,=b,=c,且 a· b=b· c=c· a,试判断△ABC 的形状. 【解】 如图,a+b+c=0. 则 a+b=-c, 即(a+b)2=(-c)2, 故 a2+2a· b+b2=c2.① 同理,a2+2a· c+c2=b2, ② b2+2b· c+c2=a2. 由①-②,得 b2-c2=c2-b2,即 2b2=2c2, 故|b|=|c|. 同理,由①-③,得|a|=|c|.故|a|=|b|=|c|, 故△ABC 为等边三角形. [能力提升] 1.(2016· 玉溪高一检测)已知|a|=2|b|≠0,且关于 x 的方程 x2+|a|x+a· b=0 有实根,则 a 与 b 的夹角的取值范围是( π? ? A.?0, ? 6? ? ?π 2 ? C.? , π ? ?3 3 ? 【解析】 ) ?π ? B.? ,π ? ?3 ? ?π ? D.? ,π ? ?6 ? ③ 因为 Δ=a2-4|a|· |b|cos θ(θ 为向量 a 与 b 夹角). 若方程有实根,则有 Δ≥0 即 a2-4|a|· |b|cos θ≥0, 又|a|=2|b|,∴4|b|2-8|b|2cos θ≥0, 1 ∴cos θ≤2,又 0≤θ≤π, π ∴ 3 ≤θ≤π. 【答案】 B 2.已知单位向量 e1,e2 的夹角为 60°,求向量 a=e1+e2,b=e2-2e1 的夹 角. 【解】 ∵e1,e2 为单位向量且夹角为 60°, 1 ∴e1·e2=1×1×cos 60°=2. 1 3 ∵a· b=(e1+e2)· (e2-2e1)=-2-e1·e2+1=-2-2+1=-2, |a|= a2= = (e1+e2)2 1 1+2×2+1= 3, (e2-2e1)2 |b|= b2= = 1 1+4-4×2= 3, 1 1 =-2. 3× 3 a· b 3 ∴cos θ=|a||b|=-2× 又∵θ∈[0°,180°],∴θ=120°, ∴a 与 b 的夹角为 120°.

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