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Question

Suppose the three vectors a,b,c on a plane satisfy the condition that |a|=|b|=|c|=|a+b|=1;c is perpendicular to a and b.c>0,, then
(1)Find the angle formed by 2a+b and b
(2)If the vector c is expressed as a linear combination λa+μb then find the ordered pair (λ,μ)

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Solution

|a|=|b|=|c|=|a+b|=1a.c=0|a+b|2=12|a|2+|b|+2|a|.|b|=12+2cosθ=1cosθ=12θ=120Now(2a+b).b=|2a+b|.|b|cosα4+1+4×12cosα=1+1cosα=0so,angle,α=90c=λa+μbca=λa2+μba0=λ+μ×(12)μ=2λc=λa+2λb|c|=λ2(a2+4b2+4(12))1=λ2(1+42)λ=13μ=23(λ,μ)=(13,23)

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