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If a + b2 + c2 = 16, x2 + y2 + z2 = 25 and ax + by + cz = 20, then the value of (a+b+c)/(x+y+z)

Q: Answer-If a + b2 + c2 = 16, x2 + y2 + z2 = 25 and ax + by + cz = 20, then the value of (a+b+c)/(x+y+z)

a2 b2 c2 16, x2 y2 z2 25 and ax by cz 20 let a 0, b 0 ,x0,y0 we get 02 02 c2 16, c2 16, c 4 02 02 z2 25 , z2 25, z 5 putting value of c and z 0x 0y cz 20 satisfy the above equation putting the values abc xyz 004 005 45

A 3/5

B 5/3

C 4/5

D 5/4

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If a + b2 + c2 = 16, x2 + y2 + z2 = 25 and ax + by + cz = 20, then the value of (a+b+c)/(x+y+z)

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If a + b2 + c2 = 16, x2 + y2 + z2 = 25 and ax + by + cz = 20, then the value of (a+b+c)/(x+y+z)

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