10판/3. 벡터

3-8 할리데이 10판 솔루션 일반물리학

짱세디럭스 2019. 8. 6. 18:04

{a:(0,2.5)b:(3.0,0)c:(0,4.2)\begin{cases} \vec a:(0,2.5)\\ \vec b:(-3.0,0)\\ \vec c:(0,-4.2)\\ \end{cases}
(a)

x=a+b+c=(3.0,1.7)\vec x = \vec a+\vec b+\vec c = (-3.0,-1.7)
(b)x=?\abs{\vec x}=? x=(3.0)2+(1.7)2=118910[m]3.5[m] \begin{aligned} \abs{\vec x} &= \sqrt{(-3.0)^2+(-1.7)^2}\\ &=\frac{\sqrt{1189}}{10}\ut{m}\\ &\approx 3.5\ut{m}\\ \end{aligned}
(c)θx=?\theta_x=? θx=tan1(1.73.0)2.6[rad] \begin{aligned} \theta_x&=\tan^{-1}\(\frac{-1.7}{-3.0}\)\\ &\approx -2.6\ut{rad} \end{aligned}