Question 72

If $$a^2 + b^2 + c^2 = 16, x^2 + y^2 + z^2 = 25 and ax + by + cz = 20$$, then the value of $$\frac{a + b + c}{x + y + z}$$

Solution

$$a^2 + b^2 + c^2 = 16, x^2 + y^2 + z^2 = 25 and ax + by + cz = 20$$

let a = 0, b= 0 ,x=0,y=0

we get

$$0^2 + 0^2 + c^2 = 16 , c^2 = 16 , c= 4 $$

$$ 0^2 + 0^2 + z^2 = 25 , z^2 = 25 ,z=5 $$

putting value of c and z

$$ 0x + 0y + cz = 20 $$

satisfy the above equation

putting the values

$$\frac{a + b + c}{x + y + z}$$ = $$\frac{0 + 0 + 4}{0 + 0 + 5}$$

$$\frac{4}{5}$$

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