Just a reverse
a x + b y + c z = 2 x a + y b + z c = 0 a 2 x 2 + b 2 y 2 + c 2 z 2 = ?
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2 solutions
Yay! same way but I chose α , β , γ
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I'm not sure that this is valid but isn't the equation in the bottom row just the square of the equation in the top row. So that the answer to the equation in the third row will be the square of the answer to the equation in the top row . And sqrt 2 ^2 =2
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This is what chew-seong sir did. u are a donkey
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@Department 8 – Ok, thanks, Lakshya!
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@Sean Silverman – Dude please don't go on donkey part my friends hacked my account
@Sean Silverman – Hello sir, How are you? Regarding some mischief word in my account, I assume that my account was hacked by some of my colleagues. Request you to kindly ignore the same and let us continue our conversation as earlier.
Yep the very same method
I had it most of the way. I forget a lot of the formulae so I couldn't get that last step you showed. Good Job!
a 2 x 2 + b 2 y 2 + c 2 z 2 = ( a x + b y + c z ) 2 − 2 ( a b x y + b c y z + c a z x ) = ( 2 ) 2 − 2 ( a b c c x y + a y z + b z x ) = 2 − a b c 2 x y z ( x y z c x y + a y z + a z x ) = 2 − a b c 2 x y z ( z c + x a + y b ) = 2 − 2 a b c x y z ( 0 ) = 2
This is like I used :)
Let A = a x , B = b y and C = c z . Then we have A + B + C = 2 and A 1 + B 1 + C 1 = 0 . If we sum the fractions, we have A B C A B + A C + B C = 0 , or simply A B + A C + B C = 0 .
Finally, we want to find A 2 + B 2 + C 2 , which is ( A + B + C ) 2 − 2 ( A B + A C + B C ) . Substituting the values that we know, we get ( 2 ) 2 − 2 ( 0 ) = 2 .