3-21 할리데이 11판 솔루션 일반물리학

$$ \begin{cases} \vec a &= \(4.0\i+2.0\j\)\ut{m}\\ \vec b &= \(0.75\i+3.3\j\)\ut{m}\\ \end{cases} $$ $$\ab{a}$$ $$ \begin{aligned} \phi_{ab}&=\cos^{-1}\frac{\vec a \cdot \vec b}{ab}\\ &=\cos^{-1}\frac{4\cdot0.75+2\cdot3.3}{\sqrt{4^2+2^2}\sqrt{0.75^2+3.3^2}}\\ &=\cos^{-1}\frac{\frac{48}{5}}{\frac{3 }{2}\sqrt{\frac{509}{5}}}\\ &=\cos^{-1}\frac{32}{\sqrt{2545}}\\ &\approx 0.8836721166534576\ut{rad}\\ &\approx 0.88\ut{rad}\\ \end{aligned} $$ $$\ab{b,c,e,f}$$ $$ \begin{aligned} \cos^{-1}\frac{\vec a \cdot \vec x}{ax}&=\frac{\pi}{2}\\ \frac{\vec a \cdot \vec x}{ax}&=\cos\frac{\pi}{2}=0\\ \end{aligned} $$ $$ \begin{aligned} 0&=\vec a \cdot \vec x\\ &=\(4\i+2\j\)\cdot\(8\cos\theta\i+8\sin\theta\j\)\\ &=4\cdot8\cos\theta+2\cdot8\sin\theta\\ \end{aligned} $$ $$ \begin{aligned} -2&=\frac{\sin\theta}{\cos\theta}=\tan\theta\\ \end{aligned} $$ $$ \therefore\begin{cases} \cos\theta=\pm\cfrac{1}{\sqrt5}\\ \sin\theta=\mp\cfrac{2}{\sqrt5} \end{cases} $$ $$ \begin{aligned} \vec x &= 8\cos\theta\i+8\sin\theta\j\\ &= \pm\frac{8}{\sqrt5}\i\mp\frac{16}{\sqrt5}\j\\ \end{aligned} $$ $$\ab{b,c}$$ $$ \begin{aligned} \vec c&=\frac{8}{\sqrt5}\i-\frac{16}{\sqrt5}\j \end{aligned} $$ $$\ab{b}$$ $$c_x=\frac{8}{\sqrt5}$$ $$\ab{c}$$ $$c_y=-\frac{16}{\sqrt5}$$ $$\ab{e,f}$$ $$ \begin{aligned} \vec d&=-\frac{8}{\sqrt5}\i+\frac{16}{\sqrt5}\j \end{aligned} $$ $$\ab{e}$$ $$d_x=-\frac{8}{\sqrt5}$$ $$\ab{c}$$ $$d_y=\frac{16}{\sqrt5}$$