3-17 할리데이 10판 솔루션 일반물리학

$$\begin{cases} \vec a:(50,30^\circ)\\ \vec b:(50,195^\circ)\\ \vec c:(50,315^\circ)\\ \end{cases} $$ $$ \begin{aligned} \vec a &= (50\cos30^\circ)\hat i + (50\sin30^\circ)\hat j\\ &=\(25\sqrt3\)\hat i + 25\hat j \end{aligned} $$ $$ \begin{aligned} \vec b &= (50\cos195^\circ)\hat i + (50\sin195^\circ)\hat j\\ &=\(-25 \sqrt{2+\sqrt{3}}\)\hat i + \(25\sqrt{2-\sqrt{3}}\)\hat j \end{aligned} $$ $$ \begin{aligned} \vec c &= (50\cos315^\circ)\hat i + (50\sin315^\circ)\hat j\\ &=\(25\sqrt2\)\hat i + \(-25\sqrt2\)\hat j \end{aligned} $$
(a) $\abs{\vec a+\vec b+\vec c} =?$ $$ \begin{aligned} \vec r &= \vec a+\vec b+\vec c \\ &= \(25 \sqrt{2}+25 \sqrt{3}-25\sqrt{2+\sqrt{3}}\)\hat i + \(25-25 \sqrt{2}+25\sqrt{2-\sqrt{3}}\)\hat j\\ \end{aligned} $$ $$ \begin{aligned} \Ans&=\sqrt{\(25 \sqrt{2}+25 \sqrt{3}-25\sqrt{2+\sqrt{3}}\)^2+\(25-25 \sqrt{2}+25\sqrt{2-\sqrt{3}}\)^2}\\ &=25 \sqrt{2 \left(2-\sqrt{2}\right)\left(3-\sqrt{3}\right)}\\ &\approx 30\ut{m} \end{aligned} $$
(b) $\theta_{\vec a+\vec b+\vec c}=?$ $$ \begin{aligned} \theta_{\vec a+\vec b+\vec c}&=\tan^{-1}\(\frac{25-25 \sqrt{2}+25\sqrt{2-\sqrt{3}}}{25 \sqrt{2}+25 \sqrt{3}-25\sqrt{2+\sqrt{3}}}\)\\ &=\cot ^{-1}\left(3+2 \sqrt{2}+2\sqrt{3}+\sqrt{6}\right)\\ &\approx 0.085\ut{rad} \end{aligned} $$
(c) $\abs{\vec a-\vec b+\vec c} =?$ $$ \begin{aligned} \vec r &= \vec a-\vec b+\vec c \\ &= \(25 \sqrt{2}+25 \sqrt{3}+25\sqrt{2+\sqrt{3}}\)\hat i + \(25-25 \sqrt{2}-25\sqrt{2-\sqrt{3}}\)\hat j\\ \end{aligned} $$ $$ \begin{aligned} \Ans&=\sqrt{\left(-25+25 \sqrt{2}+25\sqrt{2-\sqrt{3}}\right)^2+\left(25\sqrt{2}+25 \sqrt{3}+25\sqrt{2+\sqrt{3}}\right)^2}\\ &=50 \sqrt{3+\sqrt{3}+\sqrt{2+\sqrt{3}}}\\ &\approx 130\ut{m} \end{aligned} $$
(d) $\theta_{\vec a-\vec b+\vec c}=?$ $$ \begin{aligned} \theta_{\vec a-\vec b+\vec c}&=\tan ^{-1}\left(\frac{25-25 \sqrt{2}-25\sqrt{2-\sqrt{3}}}{25 \sqrt{2}+25\sqrt{3}+25\sqrt{2+\sqrt{3}}}\right)\\ &=-\cot^{-1}\left(3+\sqrt{6}\right)\\ &\approx -0.18\ut{rad} \end{aligned} $$
(e) $(\vec a + \vec b)-(\vec c + \vec d)=0, \vec d=?$ $$ \vec d = \vec a + \vec b-\vec c$$ $$ \begin{aligned} \vec d &= \vec a-\vec b+\vec c \\ &= \(25 \sqrt{2}+25 \sqrt{3}+25\sqrt{2+\sqrt{3}}\)\hat i + \(25-25 \sqrt{2}-25\sqrt{2-\sqrt{3}}\)\hat j\\ &=\bra{\frac{25\left(-3-\sqrt{3}+\sqrt{6}\right)}{\sqrt{2}}}\hat i + \bra{25\left(1+\sqrt{2+\sqrt{3}}\right)}\hat j\\ \end{aligned} $$ $$ \begin{aligned} \Ans&=\sqrt{\frac{625}{2}\left(3+\sqrt{3}-\sqrt{6}\right)^2+625\left(1+\sqrt{2+\sqrt{3}}\right)^2}\\ &=50 \sqrt{3+\sqrt{3}-\sqrt{2+\sqrt{3}}}\\ &\approx 84\ut{m} \end{aligned} $$
(f) $\theta_{\vec d}=?$ $$ \begin{aligned} \theta_{\vec d}&=\tan ^{-1}\left(\frac{25+25 \sqrt{2}+25\sqrt{2-\sqrt{3}}}{-25 \sqrt{2}+25\sqrt{3}-25\sqrt{2+\sqrt{3}}}\right)\\ &=\pi -\tan^{-1}\left(1+\sqrt{\frac{2}{3}}\right)\\ &\approx -1.1\ut{rad} \end{aligned} $$