Question

If `xcostheta=ycos(theta+(2pi)/3)=zcos(theta+(4pi)/3)=1/k` calculate `xy+yz+zx=` ?

Answer 1

Calling `lambda = x cos(theta)` we have

`(x+y+z)^2=(lambda/cos(theta)+lambda/cos(theta+(2pi)/3)+lambda/cos(theta+(4pi)/3))^2 = (9 lambda^2)/cos(3theta)^2`

Now making `x^2+y^2+z^2` we arrive at the same result. Considering

`(x+y+z)^2=x^2+y^2+z^2+2(xy+xz+yz)`

we conclude

`xy+xz+yz=0`

Answer 2

Let
`xcostheta=ycos(theta+(2pi)/3)=zcos(theta+(4pi)/3)=1/k`

So

`1/x=kcostheta`

`1/y=kcos(theta+(2pi)/3)`

`1/z=kcos(theta+(4pi)/3)`

Now

Given
x,y,z are non-zero real numbers

`1/x+1/y+1/z`

`=k(costheta+cos(theta+(2pi)/3)+cos(theta+(4pi)/3))`

`=k(costheta+2cos(pi+theta)cos(pi/3))`

`=k(costheta-2*costhetaxx1/2)`

`=k(costheta-costheta)=kxx0=0`

`=>1/x+1/y+1/z=0`

`=>(xy+yz+zx)/(xyz)=0`

`=>xy+yz+zx=0`