11판/3. 벡터

3-40 할리데이 11판 솔루션 일반물리학

짱세디럭스 2024. 1. 25. 16:17
{East:+i^North:+j^up:+k^ \begin{cases} \text{East}:+\i\\ \text{North}:+\j\\ \text{up}:+\k\\ \end{cases} (a)\ab{a} Ans=(j^)×(i^)=(1)(1)(j^×i^)=(j^×i^)=(1)(i^×j^)=(1)(k^)=k^Down \begin{aligned} \Ans&=\(-\j\)\times\(-\i\)\\ &=(-1)(-1)\(\j\times\i\)\\ &=\(\j\times\i\)\\ &=(-1)\(\i\times\j\)\\ &=(-1)\(\k\)\\ &=-\k\\ &\text{Down} \end{aligned} (b)\ab{b} Ans=(k^)×(j^)=(1)(1)(k^×j^)=(k^×j^)=(1)(j^×k^)=(1)(i^)=i^West \begin{aligned} \Ans&=\(-\k\)\times\(-\j\)\\ &=(-1)(-1)\(\k\times\j\)\\ &=\(\k\times\j\)\\ &=(-1)\(\j\times\k\)\\ &=(-1)\(\i\)\\ &=-\i\\ &\text{West} \end{aligned} (c)\ab{c} Ans=i^×k^=(1)(k^×i^)=(1)(j^)=j^South \begin{aligned} \Ans&=\i\times\k\\ &=(-1)\(\k\times\i\)\\ &=(-1)\(\j\)\\ &=-\j\\ &\text{South} \end{aligned} (d)\ab{d} Ans=i^i^=11cos0=1 \begin{aligned} \Ans&=\i\cdot\i\\ &=1\cdot1\cdot\cos0\\ &=1 \end{aligned} (e)\ab{e} Ans=(j^)×(j^)=11sin0=0 \begin{aligned} \Ans&=\(\j\)\times\(\j\)\\ &=1\cdot1\cdot\sin0\\ &=0 \end{aligned}