3-4 할리데이 11판 솔루션 일반물리학

$$\begin{cases} \Delta \vec r_{0\rarr1}&=(5.70\ut{m})\j\\ \Delta r_{1\rarr2}&=3.40\ut{m}\\ \theta_{1\rarr2}&=-\cfrac{\pi}{4}\ut{rad}\\ \Delta r_{2\rarr3}&=0.91\ut{m}\\ \theta_{2\rarr3}&=-\cfrac{3\pi}{4}\ut{rad}\\ \end{cases}$$ $$\begin{aligned} \Delta \vec r_{1\rarr2}&=r_{1\rarr2}\cos\theta_{1\rarr2}\i+r_{1\rarr2}\sin\theta_{1\rarr2}\\ &=3.40\cos\(-\cfrac{\pi}{4}\)\i+3.40\sin\(-\cfrac{\pi}{4}\)\j\\ &=\frac{17}{5 \sqrt{2}}\i-\frac{17}{5 \sqrt{2}}\j\\ \end{aligned}$$ $$\begin{aligned} \Delta \vec r_{2\rarr3}&=r_{2\rarr3}\cos\theta_{2\rarr3}\i+r_{2\rarr3}\sin\theta_{2\rarr3}\\ &=0.91\cos\(-\cfrac{3\pi}{4}\)\i+0.91\sin\(-\cfrac{3\pi}{4}\)\j\\ &=-\frac{91}{100 \sqrt{2}}\i-\frac{91}{100 \sqrt{2}}\j\\ \end{aligned}$$ $$\begin{aligned} \Sigma \Delta \vec r&=\Delta \vec r_{0\rarr1}+\Delta \vec r_{1\rarr2}+\Delta \vec r_{2\rarr3}\\ &=\(5.7\j\)+\(\frac{17}{5 \sqrt{2}}\i-\frac{17}{5 \sqrt{2}}\j\)+\(-\frac{91}{100 \sqrt{2}}\i-\frac{91}{100 \sqrt{2}}\j\)\\ &=\frac{249}{100 \sqrt{2}}\i+\frac{1}{200} \left(1140-431 \sqrt{2}\right)\j\\ \end{aligned}$$ $$\begin{cases} \Sigma \Delta r&=\cfrac{1}{100} \sqrt{448781-245670 \sqrt{2}}\ut{m}\\ \theta&=\tan ^{-1}\bra{\cfrac{1}{249}\left(570 \sqrt{2}-431\right)} \end{cases}$$ $$\ab{a}$$ $$\begin{aligned} \Sigma \Delta r&=\cfrac{1}{100} \sqrt{448781-245670 \sqrt{2}}\ut{m}\\ &\approx 3.183569602377208\ut{m}\\ &\approx 3.18\ut{m}\\ \end{aligned}$$ $$\ab{b}$$ $$\begin{aligned} \theta&=\tan ^{-1}\bra{\cfrac{1}{249}\left(570 \sqrt{2}-431\right)}\\ &\approx 0.9847671395607246\ut{rad}\\ &\approx 0.985\ut{rad} \end{aligned}$$