If vectors $P,Q$ and $R$ have magnitude $5,12$ and $13\text{units}$ and $\overrightarrow{P}+\overrightarrow{Q}=\overrightarrow{R}$, the angle between $Q$ and $R$ is ${\cos }^{-1}\frac{x}{y}$. Find $(y-x)$.

Given,magnitude of vector, $P=5\text{units}$
$Q=12\text{units}$
$R=13\text{units}$
and $\overrightarrow{P}+\overrightarrow{Q}=\overrightarrow{R}$

Vector diagram of these vectors can be drawn as,
From figure, angle between $\overrightarrow{Q}$ and $\overrightarrow{R}$
$\cos \theta =\frac{Q}{R}=\frac{12}{13}$

$\Rightarrow \theta ={\cos }^{-1}\left(\frac{12}{13}\right)$

$\Rightarrow x=12,y=13$

so,$y-x=13-12=1$

Final answer:$1$