3-43 할리데이 10판 솔루션 일반물리학

$$\begin{cases} \vec a:(3.00\ut{m},0)\\ \vec b:(4.00\ut{m},30.0^\circ)\\ \vec c:(10.0\ut{m},\frac{\pi}{2}+30.0^\circ)\\ \end{cases}$$
(a)(b) $$\vec a = 3\i,$$
(a)$a_x=?$ $$a_x = 3.00$$
(b)$a_y=?$ $$a_y = 0$$
(c)(d) $$\begin{aligned} \vec b =& 4\cos30.0^\circ\i+4\sin30.0^\circ\j\\ =&2\sqrt3\i+2\j\\ \end{aligned}$$
(c)$b_x=?$ $$b_x=2\sqrt3 \approx3.46$$
(d)$b_y=?$ $$b_y=2.00$$
(e)(f) $$\begin{aligned} \vec c =& 10\cos\(\frac{\pi}{2}+30.0^\circ\)\i+10\sin\(\frac{\pi}{2}+30.0^\circ\)\j\\ =&-5\i+5\sqrt3\j, \end{aligned}$$
(e)$c_x=?$ $$c_x=5.00$$
(f)$c_y=?$ $$c_y=5\sqrt3\approx 8.66$$
(g),(h)$\vec c = p\vec a+ q\vec b$ $$\begin{aligned} \vec c =& p\vec a+ q\vec b\\ -5\i+5\sqrt3\j=&p\(3\i\)+q\(2\sqrt3\i+2\j\)\\ =&\(3p+2\sqrt3q\)\i+\(2q\)\j \end{aligned}$$ $$\begin{cases} -5= 3p+2\sqrt3q\\ 5\sqrt3=2q \end{cases}$$
(g)$p=?$ $$p= -\frac{20}{3}\approx-6.67$$
(h)$q=?$ $$q= \frac{5 }{2}\sqrt3\approx 4.33$$