10판/3. 벡터

3-32 할리데이 10판 솔루션 일반물리학

짱세디럭스 2019. 8. 10. 16:46

(풀이자 주:x축이 아니라 z축을 나타내지 않았는데 오타로 추정됩니다. x축은 나타나있습니다.) {a:(4,0)b:(3,π2)c:(5,π+tan1(34))\begin{cases} a:(4,0)\\ b:(3,\frac{\pi}{2})\\ c:(5,\pi+\tan^{-1}\(\frac{3}{4}\))\\ \end{cases}
(a)a×b=?\abs{\vec a \times \vec b}=? Ans=a×b=absinϕab=431=12 \begin{aligned} \Ans=&\abs{\vec a \times \vec b}\\ =& ab\sin\phi_{ab}\\ =& 4\cdot3\cdot1\\ =&12\\ \end{aligned}
(b)Direction of a×b=?\text{Direction of }\vec a \times \vec b=? +z + z
(c)a×c=?\abs{\vec a \times \vec c}=? Ans=a×c=acsinϕac=45sin{πtan1(34)}=45sin{tan1(34)}=20(35)=12 \begin{aligned} \Ans=&\abs{\vec a \times \vec c}\\ =& ac\sin\phi_{ac}\\ =& 4\cdot5\cdot\sin\bra{\pi-\tan^{-1}\(\frac{3}{4}\)}\\ =& 4\cdot5\cdot\sin\bra{\tan^{-1}\(\frac{3}{4}\)}\\ =& 20\cdot\(\frac{3}{5}\)\\ =&12 \end{aligned}
(d)Direction of a×c=?\text{Direction of }\vec a \times \vec c=? z -z
(e)b×c=?\abs{\vec b \times \vec c}=? Ans=b×c=bcsinϕbc=35sin{π+tan1(34)π2}=35sin{π2+tan1(34)}=35cos{tan1(34)}=15(45)=12 \begin{aligned} \Ans=&\abs{\vec b \times \vec c}\\ =& bc\sin\phi_{bc}\\ =& 3\cdot5\cdot\sin\bra{\pi+\tan^{-1}\(\frac{3}{4}\)-\frac{\pi}{2}}\\ =& 3\cdot5\cdot\sin\bra{\frac{\pi}{2}+\tan^{-1}\(\frac{3}{4}\)}\\ =& 3\cdot5\cdot\cos\bra{\tan^{-1}\(\frac{3}{4}\)}\\ =& 15\cdot\(\frac{4}{5}\)\\ =&12 \end{aligned}
(f)Direction of b×c=?\text{Direction of }\vec b \times \vec c=? +z +z