3-30 할리데이 10판 솔루션 일반물리학

$$\begin{cases} \vec a=(4.0 \ut{m})\hat i + (-3.0 \ut{m})\hat j\\ \vec b= (6.0\ut{m})\hat i + (8.0\ut{m})\hat j\\ \end{cases} $$
(a)$a=?$ $$ a=\sqrt{4^2+(-3)^2}=5.0\ut{m}$$
(b)$\theta_a=?$ $$ \tan^{-1}\(\frac{-3}{4}\)\approx-0.64\ut{rad}$$
(c)$b=?$ $$ b=\sqrt{6^2+8^2}=10.0\ut{m}$$
(d)$\theta_b=?$ $$ \tan^{-1}\(\frac{8}{6}\)\approx0.93\ut{rad}$$
(e),(f) $$\vec a +\vec b = (10.0 \ut{m})\hat i + (5.0 \ut{m})\hat j$$
(e)$\abs{\vec a +\vec b}=?$ $$ \Ans=\sqrt{10^2+5^2}=5\sqrt5\ut{m}\approx11\ut{m}$$
(f)$\theta_{\vec a +\vec b}=?$ $$ \Ans=\tan^{-1}\(\frac{5}{10}\)\approx0.464\ut{rad}$$
(g),(h) $$\vec b -\vec a = (2.0 \ut{m})\hat i + (11.0 \ut{m})\hat j$$
(g)$\abs{\vec b -\vec a}=?$ $$ \Ans=\sqrt{2^2+11^2}=5\sqrt5\ut{m}\approx11.2\ut{m}$$
(h)$\theta_{\vec b -\vec a}=?$ $$ \Ans=\tan^{-1}\(\frac{11}{2}\)\approx1.4\ut{rad}$$
(i),(j) $$\vec a -\vec b = (-2.0 \ut{m})\hat i + (-11.0 \ut{m})\hat j$$
(i)$\abs{\vec a -\vec b}=?$ $$ \Ans=\sqrt{(-2)^2+(-11)^2}=5\sqrt5\ut{m}\approx11.2\ut{m}$$
(j)$\theta_{\vec a -\vec b}=?$ $$ \Ans=\tan^{-1}\(\frac{-11}{-2}\)\approx-1.8\ut{rad}$$
(k)$\phi_{(\vec b-\vec a)|(\vec a-\vec b)}$ $$ \begin{aligned} \Ans =& \phi_{(\vec b-\vec a)|(\vec a-\vec b)}\\ =&\abs{\theta_{\vec b-\vec a}-\theta_{\vec a-\vec b}}\\ =&\abs{\tan^{-1}\(\frac{11}{2}\)-\tan^{-1}\(\frac{-11}{-2}\)}\\ =&\pi\ut{rad} \end{aligned} $$