3-23 할리데이 11판 솔루션 일반물리학

$$ \begin{cases} \theta_{a~y+}&=50.0\degree\\ a&=1.70\\ a_x&=0\\ a_z&>0\\ \end{cases} $$ $$ \begin{cases} \phi_{b~x+}&=70.0\degree\\ b&=2.90\\ b_y&=0\\ b_z&>0\\ \end{cases} $$ $$ \begin{aligned} \vec a&=a\cos\theta\j+a\sin\theta\k\\ &=1.7\cos50\degree\j+1.7\sin50\degree\k\\ \end{aligned} $$ $$ \begin{aligned} \vec b&=b\cos\theta\i+b\sin\theta\k\\ &=2.9\cos70\degree\i+2.9\sin70\degree\k\\ \end{aligned} $$ $$\ab{a}$$ $$ \begin{aligned} \vec a\cdot\vec b=&\(1.7\cos50\degree\j+1.7\sin50\degree\k\)\cdot\(2.9\cos70\degree\i+2.9\sin70\degree\k\)\\ =&0\cdot2.9\cos70\degree+1.7\cos50\degree\cdot0+1.7\sin50\degree\cdot2.9\sin70\degree\\ =&4.93\sin50\degree\sin70\degree\\ =&\frac{493}{800} \csc 10\degree\\ \approx& 3.548842310237264\\ \approx& 3.55 \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 a\times\vec b=&\(1.7\cos50\degree\j+1.7\sin50\degree\k\)\times\(2.9\cos70\degree\i+2.9\sin70\degree\k\)\\ =&\frac{493}{100} \sin40\degree \cos 20\degree\i+\frac{493}{100} \sin 20\degree \cos40\degree\j\\ &-\frac{493}{100} \sin 20\degree \sin40\degree\k\\ \approx&2.977832273626414\i+1.291672967030868\j\\ &-1.083842310237264\k\\ \approx&2.98\i+1.29\j-1.08\k\\ \end{aligned} $$ $$\ab{c}$$ $$ \begin{aligned} \phi_{ab}&=\cos^{-1}\frac{\vec a\cdot \vec b}{ab}\\ &=\cos^{-1}\frac{\frac{493}{800} \csc 10\degree\\}{1.7\cdot2.9}\\ &=\cos^{-1}\(\frac{1}{8}\csc\degree\)\\ &=0.7672154454689077\ut{rad}\\ &=0.767\ut{rad}\\ \end{aligned} $$