If x cosθ=y cosθ+2π3=zcosθ+4π3 Then xy+yz+zx2=_____

x cosθ=y cosθ+2π3=zcosθ+4π3 =kkx+ky+kz=cosθ+cosθ+2π3+cosθ+4π3 ⇒kxy+yz+zxxyz=cosθ+cosθ+2π3+cosθ-2π3 ∵cosθ+4π3=cos2π+θ-2π3 ⇒kxy+yz+zxxyz=cosθ+2cosθ cos2π3=cosθ+2-12cosθ=0 xy+yz+zx=0(or)Put θ = 0x=y−12=z−12=kx=k,y=−2k,z=−2kNow xy+yz+zx=−2k2+4k2−2k2=0

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If $x \cos θ = y \cos (θ + \frac{2 π}{3}) = z \cos (θ + \frac{4 π}{3}) T h e n \frac{x y + y z + z x}{2} =_____$

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Detailed Solution

$x \cos θ = y \cos (θ + \frac{2 π}{3}) = z \cos (θ + \frac{4 π}{3}) = k$
$\frac{k}{x} + \frac{k}{y} + \frac{k}{z} = \cos θ + \cos (θ + \frac{2 π}{3}) + \cos (θ + \frac{4 π}{3}) \Rightarrow k (\frac{x y + y z + z x}{x y z}) = \cos θ + \cos (θ + \frac{2 π}{3}) + \cos (θ - \frac{2 π}{3}) (∵ \cos (θ + \frac{4 π}{3}) = \cos (2 π + θ - \frac{2 π}{3})) \Rightarrow k (\frac{x y + y z + z x}{x y z}) = \cos θ + 2 \cos θ \cos \frac{2 π}{3} = \cos θ + 2 (- \frac{1}{2}) \cos θ = 0 x y + y z + z x = 0$
(or)
Put θ = 0