3-13 할리데이 10판 솔루션 일반물리학

$$\begin{cases} \vec a:(1.00,0)\\ \vec b:(1.0,\theta_b)\\ \vec a+\vec b=\vec r:(1.0,\theta_r)\\ \end{cases} $$ $$ \begin{aligned} \vec a&=(1.00\cos0 )\hat i + (1.00\sin0 )\hat j\\ &=\hat i \end{aligned} $$ $$ \begin{aligned} \vec b&=(1.0\cos\theta_b )\hat i + (1.0\sin\theta_b )\hat j\\ &=(\cos\theta_b )\hat i + (\sin\theta_b )\hat j \end{aligned} $$ $$ \vec r = (1+\cos\theta_b )\hat i + (\sin\theta_b )\hat j $$
(a)$\theta_b=?$ $$ \begin{aligned} \abs{\vec r}&=\sqrt{(1+\cos\theta_b )^2+(\sin\theta_b )^2}\\ &=\sqrt{2} \sqrt{\cos\theta_b+1}=1.0\\ \end{aligned} $$ $$ \cos\theta_b=-\frac{1}{2}$$ $$ \theta_b = \pm \frac{2}{3}\pi$$
(b)$\abs{\vec a - \vec b}=?$ $$ \begin{aligned} \vec b&=(\cos\theta_b )\hat i + (\sin\theta_b )\hat j\\ &=\(-\frac{1}{2}\)\hat i + \bra{\sin\(\pm \frac{2}{3}\pi\)}\hat j\\ &=\(-\frac{1}{2}\)\hat i + \(\pm\frac{\sqrt{3}}{2}\)\hat j\\ \end{aligned} $$ $$ \begin{aligned} \Ans &= \abs{\vec a - \vec b}\\ &=\abs{\(1+\frac{1}{2}\)\hat i + \(\mp\frac{\sqrt{3}}{2}\)\hat j}\\ &=\abs{\(\frac{3}{2}\)\hat i + \(\mp\frac{\sqrt{3}}{2}\)\hat j}\\ &=\sqrt{\(\frac{3}{2}\)^2+\(\mp\frac{\sqrt{3}}{2}\)^2}\\ &=\sqrt{3}\ut{m}\\ &\approx 1.7\ut{m}\\ \end{aligned} $$