3-32 할리데이 11판 솔루션 일반물리학

$$\begin{cases} \vec a &= 3.0\i+4.0\j-3.0\k\\ \vec b &= -2.0\i+2.0\j+3.0\k\\ \vec c &= 4.0\i+3.0\j+4.0\k\\ \end{cases}$$ $$\ab{a}$$ $$\begin{aligned} \vec r &= \vec a-\vec b+\vec c\\ &=9.0\i+5.0\j-2.0\k\\ \end{aligned}$$ $$\ab{b}$$ $$\begin{aligned} \theta_{r~z+}&=\cos^{-1}\frac{r_z}{r}\\ &=\cos^{-1}\frac{-2}{\sqrt{9^2+5^2+(-2)^2}}\\ &=\cos ^{-1}\left(-\sqrt{\frac{2}{55}}\right)\\ &\approx 1.762663888280588\ut{rad}\\ &\approx 1.8\ut{rad}\\ \end{aligned}$$ $$\ab{c}$$ $$\begin{aligned} \Ans &= a\cos(\theta_{a~b})\\ &=\frac{\vec a\cdot\vec b}{b}\\ &=\frac{3\cdot(-2)+4\cdot2+(-3)\cdot3}{\sqrt{(-2)^2+2^2+3^2}}\\ &=-\frac{7}{\sqrt{17}}\\ &\approx -1.697749375254331\\ &\approx -1.7\\ \end{aligned}$$ $$\ab{d}$$ $$\vec a\times \vec b=(a_yb_z-b_ya_z)\i+(a_zb_x-b_za_x)\j+(a_xb_y-b_xa_y)\k,$$ $$\begin{aligned} \Ans &= a\sin(\theta_{a~b})\\ &=\frac{\abs{\vec a\times\vec b}}{b}\\ &=\frac{\abs{(3\i+4\j-3\k)\times(-2\i+2\j+3\k)}}{\sqrt{(-2)^2+2^2+3^2}}\\ &=\frac{23}{\sqrt{17}}\\ &\approx 5.578319375835658\\ &\approx 5.6\\ \end{aligned}$$