If xcosθ=y cos(θ+2π3)=z cos(θ+4π3), prove that xy+yz+zx=0
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)]=k2⎡⎢⎣cosθ(−12)−sinθ(√32)+cosθ+cosθ(−12)−sinθ(√32)cosθ cos(θ+2π3)cos(θ+4π3)⎤⎥⎦=k2⎡⎢⎣−cosθ+sinθ(√32)+cosθ−sinθ(√32)cosθcos(θ+2π3)cos(θ+4π3)⎤⎥⎦=0