3-28 할리데이 11판 솔루션 일반물리학

$$ \begin{cases} d_1&=0.30\ut{m},\theta_{1~y-}=-45\degree\\ \vec d_2&=0.250\ut{m}\i\\ d_3&=0.60\ut{m},\theta_{3~x+}=60.0\degree\\ \end{cases} $$ $$\ab{a}$$ $$ \begin{aligned} d_{1x}&=0.3\cos\(-90\degree-45\degree\)\\ &=-\frac{3}{10 \sqrt{2}}\ut{m}\\ &\approx -0.2121320343559642\ut{m}\\ &\approx -0.21\ut{m} \end{aligned} $$ $$\ab{b}$$ $$ \begin{aligned} d_{1y}&=0.3\sin\(-90\degree-45\degree\)\\ &=-\frac{3}{10 \sqrt{2}}\ut{m}\\ &\approx -0.2121320343559642\ut{m}\\ &\approx -0.21\ut{m} \end{aligned} $$ $$\ab{c}$$ $$ d_{2x}=0.250\ut{m} $$ $$\ab{d}$$ $$ d_{2y}=0 $$ $$\ab{e}$$ $$ \begin{aligned} d_{3x}&=0.6\cos\(60\degree\)\\ &=\frac{3}{10}\ut{m}\\ &= 0.3\ut{m}\\ &= 0.30\ut{m}\\ \end{aligned} $$ $$\ab{f}$$ $$ \begin{aligned} d_{3y}&=0.6\sin\(60\degree\)\\ &=\frac{3 \sqrt{3}}{10}\ut{m}\\ &\approx 0.5196152422706631\ut{m}\\ &\approx 0.520\ut{m}\\ \end{aligned} $$ $$\ab{g,h,i,j}$$ $$ \begin{aligned} \Sigma \vec d&=\vec d_1+\vec d_2+\vec d_3\\ &=\(-\frac{3}{10 \sqrt{2}}\i-\frac{3}{10 \sqrt{2}}\j\)+\(0.25\ut{m}\i\)+\(0.3\i+0.3\sqrt{3}\j\)\\ &=\frac{1}{20}\(11-3\sqrt{2}\)\i-\frac{3}{20}\(\sqrt{2}-2\sqrt{3}\)\j\\ \end{aligned} $$ $$\ab{g}$$ $$ \begin{aligned} d_x&=\frac{1}{20}\(11-3\sqrt{2}\)\ut{m}\\ &\approx 0.3378679656440358\ut{m}\\ &\approx 0.34\ut{m}\\ \end{aligned} $$ $$\ab{h}$$ $$ \begin{aligned} d_y&=-\frac{3}{20}\(\sqrt{2}-2\sqrt{3}\)\ut{m}\\ &\approx 0.3074832079146989\ut{m}\\ &\approx 0.31\ut{m}\\ \end{aligned} $$ $$\ab{i}$$ $$ \begin{aligned} d&=\sqrt{\(\frac{11-3\sqrt{2}}{20}\)^2+\(\frac{6\sqrt{3}-3\sqrt{2}}{20}\)^2}\\ &=\frac{1}{20} \sqrt{265-66 \sqrt{2}-36 \sqrt{6}}\ut{m}\\ &\approx 0.4568377013316144\ut{m}\\ &\approx 0.46\ut{m}\\ \end{aligned} $$ $$\ab{j}$$ $$ \begin{aligned} \theta_d&=\tan^{-1}\frac{-\cfrac{3}{20}\(\sqrt{2}-2\sqrt{3}\)}{\cfrac{1}{20}\(11-3\sqrt{2}\)}\\ &\approx 0.7383503902645553\ut{rad}\\ &\approx 0.74\ut{rad}\\ \end{aligned} $$ $$\ab{k,l}$$ $$ \begin{aligned} \Ans &=\vec d_2= -\vec d\\ \end{aligned} $$ $$\ab{k}$$ $$ \begin{aligned} d_2&=d\\ &=\frac{1}{20} \sqrt{265-66 \sqrt{2}-36 \sqrt{6}}\ut{m}\\ &\approx 0.4568377013316144\ut{m}\\ &\approx 0.46\ut{m}\\ \end{aligned} $$ $$\ab{l}$$ $$ \begin{aligned} \theta_{d_2}&=\theta_d+\pi\\ &=\pi+\tan^{-1}\frac{\cfrac{3}{20}\(\sqrt{2}-2\sqrt{3}\)}{\cfrac{1}{20}\(11-3\sqrt{2}\)}\\ &\approx -2.403242263325238\ut{rad}\\ &\approx -2.4\ut{rad}\\ \end{aligned} $$