52. Find the angle between the vectors: a) $\vec{a}=(4 ; 0)$ and $\vec{b}=(2 ;-2)$; b) $\vec{a}=(5 ;-3)$ and $\vec{b}=(3 ; 5)$.
Solution. a) Using formula (5), we find
$\cos \varphi=\frac{4 \cdot 2-0 \cdot(-2)}{\sqrt{16+0} \cdot \sqrt{4+4}}=\frac{2}{2 \sqrt{2}}=\frac{\sqrt{2}}{2} ; \varphi=\arccos \frac{\sqrt{2}}{2}=\frac{\pi}{4}$.

b) We have

$$
\cos \varphi=\frac{5 \cdot 3+(-3) \cdot 5}{\sqrt{25+9} \cdot \sqrt{9+25}}=\frac{0}{34}=0 ; \varphi=\arccos 0=\frac{\pi}{2}
$$

i.e., the given vectors are perpendicular.