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Question

If xcosθ=y cos(θ+2π3)=z cos(θ+4π3), prove that xy+yz+zx=0

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Solution

Given xcosθ=y cos(θ+2π3)=z cos(θ+4π3)=k(say)x=kcosθy=kcos[θ+2π3]z=kcos[θ+4π3]

xy+yz+zx=k2[1cosθ cos(θ+2π3)+1cos(θ+2π3)cos(θ+4π3)+1cos(θ+4π3)cosθ]=k2[cos(θ+4π3)+cosθ+cos(θ+2π3)cosθ cos(θ+2π3)cos(θ+4π3)]=k2cosθ(12)sinθ(32)+cosθ+cosθ(12)sinθ(32)cosθ cos(θ+2π3)cos(θ+4π3)=k2cosθ+sinθ(32)+cosθsinθ(32)cosθcos(θ+2π3)cos(θ+4π3)=0


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