Question : If $a^2+b^2+c^2=16$, $x^2+y^2+z^2=25$ and $ax+by+cz=20$, then the value of $\frac{a+b+c}{x+y+z}$ is:
Option 1: $\frac{3}{5}$
Option 2: $\frac{5}{3}$
Option 3: $\frac{4}{5}$
Option 4: $\frac{5}{4}$
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Correct Answer: $\frac{4}{5}$
Solution :
Given: $a^2+b^2+c^2=16$----(equation 1)
$x^2+y^2+z^2=25$-----(equation 2)
$ax+by+cz=20$-----(equation 3)
Substitute the values of $b=0$ and $c=0$ in the given equation 1, we get,
$a^2+0^2+0^2=16$
$⇒a^2=16$
$\therefore a=4$
Substitute the values of $y=0$ and $z=0$ in the given equation 2, we get,
$x^2+0^2+0^2=25$
$⇒x^2=25$
$\therefore x=5$
Substitute the values in the given equation 3, we get,
$4\times5+0\times0+0\times0=20$
$⇒20=20$
So, the above equation is satisfied.
Substitute the values in the given below equation, we get,
$\therefore\frac{a+b+c}{x+y+z}=\frac{4+0+0}{5+0+0}=\frac{4}{5}$
Hence, the correct answer is $\frac{4}{5}$.
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