솔루션피아

$$ \begin{cases} \Delta \vec r_{0\rarr1}&=(5.70\ut{m})\j\\ \Delta r_{1\rarr2}&=3.40\ut{m}\\ \theta_{1\rarr2}&=-\cfrac{\pi}{4}\ut{rad}\\ \Delta r_{2\rarr3}&=0.91\ut{m}\\ \theta_{2\rarr3}&=-\cfrac{3\pi}{4}\ut{rad}\\ \end{cases} $$ $$ \begin{aligned} \Delta \vec r_{1\rarr2}&=r_{1\rarr2}\cos\theta_{1\rarr2}\i+r_{1\rarr2}\sin\theta_{1\rarr2}\\ &=3.40\cos\(-\cfrac{\pi}{4}\)\i+3.40\sin\(-\cfrac{\pi}{4}..

$$ \begin{cases} \Delta \vec r_{0\rarr1}&=(-33\ut{m})\i\\ \Delta \vec r_{1\rarr2}&=(-33\ut{m})\j\\ \Delta \vec r_{2\rarr3}&=(35\ut{m})\k\\ \end{cases} $$ $$\ab{a}$$ $$ \begin{aligned} \Sigma \vec {r_a} &= (-33\ut{m})\i+(-33\ut{m})\j+(35\ut{m})\k\\ \end{aligned} $$ $$\ab{b}$$ $$ \begin{aligned} \Sigma \vec {r_b} &= (-33\ut{m})\i+(-33\ut{m})\j\\ r_b&=\sqrt{(-33)^2+(-33)^2}\\ &=33\sqrt{2}\ut{m}\\ &..

$$ \begin{cases} \vec A&=A\i\\ B&=14.0\ut{m}\\ \vec A + \vec B &= \vec C\\ \vec C &= C\j\\ C &= 3A \end{cases} $$ $$ A\i + \vec B = 3A\j $$ $$ \begin{aligned} \vec B &= (0-A)\i+(3A-0)\j+(0-0)\k\\ &=-A\i+3A\j\\ \end{aligned} $$ $$ B=\sqrt{(-A)^2+(3A)^2}=14\ut{m} $$ $$ \begin{aligned} A&=7\sqrt{\frac{2}{5}}\ut{m}\\ &\approx 4.427188724235731\ut{m}\\ &\approx 4.43\ut{m}\\ \end{aligned} $$

$$ \begin{cases} \i&:\text{East}\\ \j&:\text{North}\\ \k&:\text{Up}\\ \end{cases} $$ $$\ab{a}$$ $$ \begin{aligned} \i\cdot\k&=\(1\i+0\j+0\k\)\cdot\(0\i+0\j+1\k\)\\ &=1\cdot0+0\cdot0+0\cdot1\\ &=0 \end{aligned} $$ $$\ab{b}$$ $$ \begin{aligned} \(-\k\)\cdot\(-\j\)&=\bra{0\i+0\j+(-1)\k}\cdot\bra{0\i+(-1)\j+0\k}\\ &=0\cdot0+0\cdot(-1)+(-1)\cdot0\\ &=0 \end{aligned} $$ $$\ab{c}$$ $$ \begin{aligned} \..