3-18 할리데이 11판 솔루션 일반물리학

$$\begin{cases} A&=11.0,\theta_A=100\degree\\ \vec B &= -6.50\i-8.40\j\\ \end{cases}$$ $$\vec A = 11\cos100\degree\i+11\sin100\degree\j,$$ $$\ab{a}$$ $$\begin{aligned} 3\vec A\cdot\vec B =&~ 3\cdot11\cos100\degree\cdot(-6.50)\\ &+3\cdot11\sin100\degree\cdot(-8.40)\\ \approx&-235.7411750254275\\ \approx&-236\\ \end{aligned}$$ $$\ab{b}$$ $$\vec a\times \vec b=(a_yb_z-b_ya_z)\i+(a_zb_x-b_za_x)\j+(a_xb_y-b_xa_y)\k,$$ $$\begin{aligned} \vec C &=4\vec A \times 3\vec B\\ &=4\(11\cos100\degree\i+11\sin100\degree\j\)\times3\(-6.5\i-8.4\j\)\\ &=\(\frac{5544}{5} \sin 10\degree+858 \cos 10\degree\)\k\\ &\approx 1037.506151481567\k\\ &\approx 1.04\times10^3\k\\ \end{aligned}$$ $$\ab{c}$$ $$\begin{aligned} C&=\frac{5544}{5} \sin 10\degree+858 \cos 10\degree\\ &\approx 1037.506151481567\\ &\approx 1.04\times10^3\\ \end{aligned}$$ $$\begin{aligned} \phi_C&=0 \end{aligned}$$ $$\theta_C : \text{Indeterminate}$$ $$\ab{d}$$ $$\vec a \cdot \vec b=ab\cos\phi_{ab},$$ $$\begin{aligned} \phi_{AC}&=\cos\frac{\vec A\cdot\vec C}{AC}\\ &=\cos\frac{0}{AC}\\ &=\frac{\pi}{2}\ut{rad}\\ \end{aligned}$$ $$\ab{e}$$ $$\vec A = 11\cos100\degree\i+11\sin100\degree\j,$$ $$\begin{aligned} \vec D &= \vec A +3.00\k \\ &= 11\cos100\degree\i+11\sin100\degree\j+3\k\\ &\approx -1.910129954336234\i+10.83288528313429\j+3\k\\ &\approx -1.91\i+10.8\j+3\k\\ \end{aligned}$$ $$\begin{aligned} D&= \sqrt{121 \sin ^210\degree+121 \cos ^210\degree+9}\\ &\approx 11.40175425099138\\ &\approx 11.4\\ \end{aligned}$$ $$\begin{aligned} \phi_D&=\tan ^{-1}\frac{\sqrt{121 \sin ^210\degree+121 \cos ^210\degree}}{3}\\ &\approx 1.304544277643971\ut{rad}\\ &\approx 1.30\ut{rad}\\ \end{aligned}$$ $$\begin{aligned} \theta_D&=\theta_A=100\degree \end{aligned}$$