Asked by shuvra
                (a2 +b2 +c2) (x2 +y2 +z2) equal (ax +by +cz) 2 prove that a:b:c equal x:y:z............here all 2 are as square 
            
            
        Answers
                    Answered by
            Reiny
            
    given: (a^2 +b^2 +c^2) (x^2 +y^2 +z^2) = (ax +by +cz)^2
a^2x^2 + a^2y^2 + a^2z^2 + b^2x^2 + b^2y^2 + b^2z^2 + c^2x^2 + c^2y^2 + c^2z^2 = a^2x^2 + b^2y^2 + c^2z^2 + 2abxy + 2acxz + 2bcyz
a^2y^2 + a^2z^2 + b^2x^2 + b^2z^2 + c^2x^2 + c^2y^2 = 2abxy + 2acxz + 2bcyz
now if a:b:c = x:y:z
then a/x = b/y = c/z = k
a = kx
b = ky
c = kz
subbing into the long equation:
LS = x^2y^2k^2 + x^2z^2k^2 + x^2y^2k^2 + y^2z^2k^2 + x^2z^2k^2 + z^2y^2k^2
= 2x^2y^2k^2 + 2x^2z^2k^2 + 2y^2z^2k^2
RS = 2(xk)(yk)(xy) + 2(xk)(zk)(xz) + 2(yk)(zk)(yz)
= 2x^2y^2k^2 + 2x^2z^2k^2 + 2y^2z^2k^2
= LS
    
a^2x^2 + a^2y^2 + a^2z^2 + b^2x^2 + b^2y^2 + b^2z^2 + c^2x^2 + c^2y^2 + c^2z^2 = a^2x^2 + b^2y^2 + c^2z^2 + 2abxy + 2acxz + 2bcyz
a^2y^2 + a^2z^2 + b^2x^2 + b^2z^2 + c^2x^2 + c^2y^2 = 2abxy + 2acxz + 2bcyz
now if a:b:c = x:y:z
then a/x = b/y = c/z = k
a = kx
b = ky
c = kz
subbing into the long equation:
LS = x^2y^2k^2 + x^2z^2k^2 + x^2y^2k^2 + y^2z^2k^2 + x^2z^2k^2 + z^2y^2k^2
= 2x^2y^2k^2 + 2x^2z^2k^2 + 2y^2z^2k^2
RS = 2(xk)(yk)(xy) + 2(xk)(zk)(xz) + 2(yk)(zk)(yz)
= 2x^2y^2k^2 + 2x^2z^2k^2 + 2y^2z^2k^2
= LS
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