10판/3. 벡터

3-11 할리데이 10판 솔루션 일반물리학

짱세디럭스 2019. 8. 6. 19:05

{a=(4.0[m])i^+(3.0[m])j^b=(13.0[m])i^+(7.0[m])j^\begin{cases} \vec a =\(4.0\ut{m} \)\hat i + \(3.0\ut{m} \)\hat j\\ \vec b =\(-13.0\ut{m} \)\hat i + \(7.0\ut{m} \)\hat j\\ \end{cases}
(a)a+b=?\vec a + \vec b=? a+b=r=(4.013.0)i^+(3.0+7.0)j^=(9)i^+(10)j^ \begin{aligned} \vec a+\vec b &=\vec r = \(4.0-13.0 \)\hat i + \(3.0+7.0 \)\hat j\\ &=\(-9 \)\hat i + \(10 \)\hat j \end{aligned}
(b)r=?\abs{\vec r}=? r=(9)2+102=18113[m] \abs{\vec r}=\sqrt{(-9)^2+10^2}=\sqrt{181}\approx13\ut{m}
(c)θr=?\theta_r=? θr=tan1(109)2.3[rad]\theta_r=\tan^{-1}\(\frac{10}{-9}\)\approx 2.3\ut{rad}