Question

If a²+ b²+ c²= 1, and x²+ y²+ z²= 1, show that ax + by + cz = 1



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Answer 1

$\huge\red{Answer♡}$

This is a problem of Cyclic expression, in which it's considered

a/x = b/y = c/z = k ……….(1) ( which can be proved)

=> a = kx => a² = k² x²

b = ky => b² = k²y²

c = kz => c² = k²z²

=> a²+b²+c²= k²x² + k²y²+ k²z²

=> a²+b²+ c² = k²( x²+ y²+z²)

=> (a²+b²+ c²)/(x²+ y²+ z²) = k²

Hence, 16/25 = k²

=> k = 4/5, -4/5. But since ratios are always positive. So k= -4/5 is ruled out.

So, a/x + b/y + c/z = 4/5 …. by eq (1)

=> 5a = 4x

5b = 4y

5c = 4z

=> 5a + 5b + 5c = 4x + 4y + 4z

=> 5( a+b+ c) = 4( x+ y+ z)

=> (a+b+ c) / ( x+ y+ z) = 4/5

$\small\pink{hope it helps♡}$

Answer 2

answer refers to the pic