设向量a,b满足丨a丨=2,丨a-b丨=1,则a与b夹角的取值范围是?)|a-b|=1,故:|a-b|^2=(a-b)·(a-b)=|a|^2+|b|^2-2a·b=4+|b|^2-2a·b=1即:a·b=(|b|^2+3)/2,而:a·b=|a|*|b|*cos,故:cos=a·b/(2|b|)=(1/4)(3/|b|+|b|)≥sqrt(3)/2,[这句
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![设向量a,b满足丨a丨=2,丨a-b丨=1,则a与b夹角的取值范围是?)|a-b|=1,故:|a-b|^2=(a-b)·(a-b)=|a|^2+|b|^2-2a·b=4+|b|^2-2a·b=1即:a·b=(|b|^2+3)/2,而:a·b=|a|*|b|*cos,故:cos=a·b/(2|b|)=(1/4)(3/|b|+|b|)≥sqrt(3)/2,[这句](/uploads/image/z/10755437-5-7.jpg?t=%E8%AE%BE%E5%90%91%E9%87%8Fa%2Cb%E6%BB%A1%E8%B6%B3%E4%B8%A8a%E4%B8%A8%3D2%2C%E4%B8%A8a-b%E4%B8%A8%3D1%2C%E5%88%99a%E4%B8%8Eb%E5%A4%B9%E8%A7%92%E7%9A%84%E5%8F%96%E5%80%BC%E8%8C%83%E5%9B%B4%E6%98%AF%3F%EF%BC%89%7Ca-b%7C%3D1%2C%E6%95%85%EF%BC%9A%7Ca-b%7C%5E2%3D%28a-b%29%C2%B7%28a-b%29%3D%7Ca%7C%5E2%2B%7Cb%7C%5E2-2a%C2%B7b%3D4%2B%7Cb%7C%5E2-2a%C2%B7b%3D1%E5%8D%B3%EF%BC%9Aa%C2%B7b%3D%28%7Cb%7C%5E2%2B3%29%2F2%2C%E8%80%8C%EF%BC%9Aa%C2%B7b%3D%7Ca%7C%2A%7Cb%7C%2Acos%2C%E6%95%85%EF%BC%9Acos%3Da%C2%B7b%2F%282%7Cb%7C%29%3D%281%2F4%29%283%2F%7Cb%7C%2B%7Cb%7C%29%E2%89%A5sqrt%283%29%2F2%2C%5B%E8%BF%99%E5%8F%A5)
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设向量a,b满足丨a丨=2,丨a-b丨=1,则a与b夹角的取值范围是?)|a-b|=1,故:|a-b|^2=(a-b)·(a-b)=|a|^2+|b|^2-2a·b=4+|b|^2-2a·b=1即:a·b=(|b|^2+3)/2,而:a·b=|a|*|b|*cos,故:cos=a·b/(2|b|)=(1/4)(3/|b|+|b|)≥sqrt(3)/2,[这句
设向量a,b满足丨a丨=2,丨a-b丨=1,则a与b夹角的取值范围是?)
|a-b|=1,故:|a-b|^2=(a-b)·(a-b)=|a|^2+|b|^2-2a·b=4+|b|^2-2a·b=1
即:a·b=(|b|^2+3)/2,而:a·b=|a|*|b|*cos,
故:cos=a·b/(2|b|)=(1/4)(3/|b|+|b|)≥sqrt(3)/2,[这句第二个等于号后面没看懂]
故:cos∈[0,π/6]
设向量a,b满足丨a丨=2,丨a-b丨=1,则a与b夹角的取值范围是?)|a-b|=1,故:|a-b|^2=(a-b)·(a-b)=|a|^2+|b|^2-2a·b=4+|b|^2-2a·b=1即:a·b=(|b|^2+3)/2,而:a·b=|a|*|b|*cos,故:cos=a·b/(2|b|)=(1/4)(3/|b|+|b|)≥sqrt(3)/2,[这句
cos=a·b/(2|b|)=(1/4)(3/|b|+|b|)≥sqrt(3)/2 用到了必修五的基本不等式,
也就是a²+b²>=2ab这个均值不等式 3/|b|+|b>=2倍根号下3/|b|*b|=2倍根号下3
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