10판/3. 벡터

3-15 할리데이 10판 솔루션 일반물리학

짱세디럭스 2019. 8. 6. 21:27

$$\begin{cases} \vec a:(10.0, 30^\circ)\\ \vec b:(10.0, 30+105^\circ)\\ \end{cases} $$ $$ \begin{aligned} \vec a &= (10\cos30^\circ)\hat i + (10\sin30^\circ)\hat j\\ &=5\sqrt{3}\hat i + 5\hat j\\ \end{aligned} $$ $$ \begin{aligned} \vec b &= (10\cos135^\circ)\hat i + (10\sin135^\circ)\hat j\\ &=-5\sqrt2\hat i + 5\sqrt2\hat j\\ \end{aligned} $$ $$\vec r = \(5\sqrt3-5\sqrt2\)\hat i + \(5+5\sqrt2\)\hat j$$
(a)$\abs{\vec r}=?$ $$ \begin{aligned} r&=\sqrt{\(5\sqrt3-5\sqrt2\)^2+\(5+5\sqrt2\)^2}\\ &=5 \sqrt{2\left(4+\sqrt{2}-\sqrt{6}\right)}\ut{m}\\ &\approx 12.2\ut{m} \end{aligned} $$
(b)$\theta_r=?$ $$\theta_r=\tan^{-1}\(\frac{5+5\sqrt2}{5\sqrt3-5\sqrt2}\)=\frac{11}{24}\pi\approx1.44\ut{rad} $$