11판/3. 벡터

3-22 할리데이 11판 솔루션 일반물리학

짱세디럭스 2024. 1. 23. 21:15
{a=8i^b=6j^c=8i^6j^ \begin{cases} \vec a&=8\i\\ \vec b&=6\j\\ \vec c&=-8\i-6\j\\ \end{cases} (a,b)\ab{a,b} a×b=(8i^)×(6j^)=48(i^×j^)=48k^ \begin{aligned} \vec a\times\vec b&=\(8\i\)\times\(6\j\)\\ &=48\(\i\times\j\)\\ &=48\k\\ \end{aligned} (a)\ab{a} a×b=48\abs{\vec a\times\vec b}=48 (b)\ab{b} +z\text{+z} (c,d)\ab{c,d} a×c=(8i^)×(8i^6j^)=8×(8)(i^×i^)+8×(6)(i^×j^)=048k^=48k^ \begin{aligned} \vec a\times\vec c&=\(8\i\)\times\(-8\i-6\j\)\\ &=8\times(-8)\(\i\times\i\)+8\times(-6)\(\i\times\j\)\\ &=0-48\k\\ &=-48\k \end{aligned} (c)\ab{c} a×c=48\abs{\vec a\times\vec c}=48 (d)\ab{d} -z\text{-z} (e,f)\ab{e,f} b×c=(6j^)×(8i^6j^)=6×(8)(j^×i^)+6×(6)(j^×j^)=48(1)(i^×j^)+0=48k^ \begin{aligned} \vec b\times\vec c&=\(6\j\)\times\(-8\i-6\j\)\\ &=6\times(-8)\(\j\times\i\)+6\times(-6)\(\j\times\j\)\\ &=-48(-1)\(\i\times\j\)+0\\ &=48\k\\ \end{aligned} (e)\ab{e} b×c=48\abs{\vec b\times\vec c}=48 (f)\ab{f} +z\text{+z}