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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
且向量a*b=3/5 β∈(0,π)求sinβ的值
人气:106 ℃ 时间:2019-11-22 18:46:39
解答
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
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