Question
If a + b2 + c2 = 16, x2
+ y2 + z2 = 25 and ax + by + cz = 20, then the value of (a+b+c)/(x+y+z)Solution
a2 + b2 + c2 = 16, x2 + y2 + z2 = 25 and ax + by + cz = 20 let a = 0, b= 0 ,x=0,y=0 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 (a+b+c) / (x+y+z) = (0+0+4) / (0+0+5) 4/5
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