已知向量a=(2cosα,2sinα) b=(sinβ,cosβ)求向量a+b的模的最小值 若向量c=（-1/2,根号3/2） 且向量a*b

p表示α,q表示β：|a+b|^2=(a+b)·(a+b)=|a|^2+|b|^2+2a·b=4+1+2(2cosp,2sinp)·(sinq,cosq)=5+4(cospsinq+sinpcosq)=5+4sin(p+q),故：|a+b|^2∈[1,9],即：|a+b|的最小值是1a·b=(2cosp,2sinp)·(sinq,cosq)=2sin(p...嗯是向量c*b=3/5我解得也是1b·c=(sinq,cosq)·(-1/2,sqrt(3)/2)=-sinq/2+sqrt(3)cosq/2=-sin(q-π/3)=3/5即：sin(q-π/3)=-3/5，而：q∈(0,π)，故：q-π/3∈(-π/3,2π/3)而：sin(q-π/3)=-3/5，故：q-π/3∈(-π/3,0]，故：cos(q-π/3)=4/5sinq=sin(q-π/3+π/3)=sin(q-π/3)cos(π/3)+cos(q-π/3)sin(π/3)=(-3/5)*(1/2)+(4/5)*(sqrt(3)/2)=(4sqrt(3)-3)/10b·c=(sinq,cosq)·(-1/2,sqrt(3)/2)=-sinq/2+sqrt(3)cosq/2=-sin(q-π/3)=3/5即：sin(q-π/3)=-3/5，而：q∈(0,π)，故：q-π/3∈(-π/3,2π/3)而：sin(q-π/3)=-3/5，故：q-π/3∈(-π/3,0]，故：cos(q-π/3)=4/5sinq=sin(q-π/3+π/3)=sin(q-π/3)cos(π/3)+cos(q-π/3)sin(π/3)=(-3/5)*(1/2)+(4/5)*(sqrt(3)/2)=(4sqrt(3)-3)/10