3-26 할리데이 11판 솔루션 일반물리학

$$\begin{cases} \vec d_1&=3.0\i+4.0\j-7.0\k\\ \vec d_2&=-1.0\i-4.0\j+3.0\k\\ \vec d_3&=-5.0\i+3.0\j+2.0\k\\ \end{cases}$$ $$\ab{a}$$ $$\begin{aligned} \vec r&=\vec d_1-\vec d_2+\vec d_3\\ &=-i+11\j-8\k\\ &=-1.0\i+11\j-8.0\k\\ \end{aligned}$$ $$\ab{b}$$ $$\begin{aligned} \theta_{a~z+}&=\cos^{-1}\frac{r_z}{r}\\ &=\cos^{-1}\frac{-8}{\sqrt{(-1)^2+11^2+(-8)^2}}\\ &=\cos ^{-1}\left(-4 \sqrt{\frac{2}{93}}\right)\\ &\approx 2.197636334573072\ut{rad}\\ &\approx 2.2\ut{rad}\\ \end{aligned}$$ $$\ab{c}$$ $$\begin{aligned} \Ans &= d_1\cos(\theta_{d1~d2})\\ &=\frac{\vec d_1\cdot\vec d_2}{d_2}\\ &=\frac{3\cdot(-1)+4\cdot(-4)+(-7)\cdot3}{\sqrt{(-1)^2+(-4)^2+3^2}}\\ &=-20 \sqrt{\frac{2}{13}}\\ &\approx -7.844645405527362\\ &\approx -7.8\\ \end{aligned}$$ $$\ab{d}$$ $$\vec a\times \vec b=(a_yb_z-b_ya_z)\i+(a_zb_x-b_za_x)\j+(a_xb_y-b_xa_y)\k,$$ $$\begin{aligned} \Ans &= d_1\sin(\theta_{d1~d2})\\ &=\frac{\abs{\vec d_1\times\vec d_2}}{d_2}\\ &=\frac{\abs{(3\i+4\j-7\k)\times(-\i-4\j+3\k)}}{\sqrt{(-1)^2+(-4)^2+3^2}}\\ &=9 \sqrt{\frac{2}{13}}\\ &\approx 3.530090432487313\\ &\approx 3.5\\ \end{aligned}$$