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

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Answer 1 · 2016-12-21T15:56:33

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 · 2016-12-21T16:21:10

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#