3-23 할리데이 10판 솔루션 일반물리학

$$b=\sqrt{3^2+4^2}=5$$ $$\theta_a=\tan^{-1}\(\frac{-1}{1}\)=-\frac{\pi}{4}$$ $$\begin{aligned} \therefore \vec r&:\(5,-\frac{\pi}{4}\)\\ &=\bra{5\cos(-\frac{\pi}{4})}\hat i+\bra{5\sin(-\frac{\pi}{4})}\hat j\\ &=\(\frac{5}{\sqrt2}\)\hat i + \(-\frac{5}{\sqrt2}\)\hat j\\ \end{aligned}$$