3-19 할리데이 10판 솔루션 일반물리학

$$\begin{cases} \vec p_1 = \hat i + 2\hat j\\ \vec p_2 = -2\hat i + 1\hat j\\ \vec p_3 = -1\hat i + 2\hat j\\ \end{cases} $$ $$\Sigma\vec p = -2\hat i + 5\hat j$$
(a)$\abs{\Sigma\vec p}=?$ $$ \begin{aligned} \abs{\Sigma\vec p}&=\sqrt{(-2)^2+(5)^2}\\ &=\sqrt{29}\\ &\approx 5.39\ut{m}\\ \end{aligned} $$
(b)$\theta_{\Sigma\vec p}=?$ $$ \begin{aligned} \theta_{\Sigma\vec p}&=\tan^{-1}\(\frac{5}{-2}\)\\ &\approx 1.95\ut{rad} \end{aligned} $$