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(x/a) + (y/b) + (z/c)=√2(a/x) + (b/y) + (c/z)=0Find (x²/a²) + (y²/b²) + (z²/c²) |
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Answer»
Explanation: Suppose you have three points in R 3
p 1 = ( a , b , c )
p 2 = ( x , y , )
p 0 = ( 0 , 0 , 0 )
The RESPECTIVE LENGTHS of sides p 1 − p 0 and p 2 − p 0
are ∥ p 1 − p 0 ∥ = √ a 2 + b 2 + c 2
∥ p 2 − p 0 ∥ = √ x 2 + y 2 + z 2
and their scalar product ( p 1 − p 0 ) . ( p 2 − p 0 ) = a x + b y + c z
but ( p 1 − p 0 ) . ( p 2 − p 0 ) = ∥ p 1 − p 0 ∥ ∥ p 2 − p 0 ∥ cos ( ˆ p 1 p 0 p 2 ) so cos ( ˆ p 1 p 0 p 2 ) = a x + b y + c z √ a 2 + b 2 + c 2 √ x 2 + y 2 + z 2 = 30 5 × 6 = 1 Then the sides p 1 − p 0 and p 2 − p 0 are ALIGNED so ( p 1 − p 0 ) = λ ( p 2 − p 0 ) and of course λ = 5 6
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