10판/3. 벡터

3-9 할리데이 10판 솔루션 일반물리학

짱세디럭스 2019. 8. 6. 18:22

{a=(5.0)i^(4.0)j^+(2.0)k^b=(2.0m)i^+(2.0m)j^+(5.0m)k^\begin{cases} \vec a = (5.0)\hat i - (4.0)\hat j + (2 . 0)\hat k\\ \vec b = (-2.0m)\hat i +(2 .0m)\hat j + (5.0m)\hat k\\ \end{cases}
(a)a+b\vec a + \vec b a+b=(5.02.0m)i^+(4.0+2.0m)j^+(2.0+5.0m)k^\vec a + \vec b = \(5.0-2.0m \)\hat i + \(-4.0+2.0m \)\hat j + \(2.0+5.0m \)\hat k
(b)ab\vec a- \vec b ab=(5.0+2.0m)i^+(4.02.0m)j^+(2.05.0m)k^\vec a - \vec b = \(5.0+2.0m \)\hat i + \(-4.0-2.0m \)\hat j + \(2.0-5.0m \)\hat k
(c)ab+c=0,c=?\vec a - \vec b + \vec c = 0, \vec c=? c=ba=(5.002.00m)i^+(4.002.00m)j^+(2.00+5.00m)k^ \begin{aligned} \vec c&=\vec b-\vec a\\ &=\(-5.00-2.00m \)\hat i + \(4.00-2.00m \)\hat j + \(-2.00+5.00m \)\hat k \end{aligned}