Question

A2+b2+c2=16 and x2+y2+z2=25 ,ax+by+cz=20, find (a+b+c)/(x+y+z)

Answer 1

I think ...this is the correct answer

Answer 2

The value of $\dfrac{a+b+c}{x+y+z}=\dfrac{4}{5}$.

Step-by-step explanation:

We have,

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

To find, $\dfrac{a+b+c}{x+y+z}=?$

Put a = 4, b = 0 and c = 0

Also, Put x = 5, y = 0 and z = 0

All equations satisfied.

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

$=\dfrac{4}{5}$

The value of$\dfrac{a+b+c}{x+y+z}=\dfrac{4}{5}$

Hence, the value of $\dfrac{a+b+c}{x+y+z}=\dfrac{4}{5}$.