3-19 할리데이 11판 솔루션 일반물리학

$$ \begin{cases} \vec d_1&=-d_1\j\\ \vec d_2&=d_2\i\\ \end{cases} $$ $$\ab{a}$$ $$\frac{\vec d_2}{5}= \frac{d_2}{5}\i$$ $$\therefore +x$$ $$\ab{b}$$ $$\frac{\vec d_1}{-5}= \frac{-d_1}{-5}\j=\frac{d_1}{5}\j$$ $$\therefore +y$$ $$\ab{c}$$ $$ \begin{aligned} \vec d_1\cdot\vec d_2&=\(-d_1\j\)\cdot\(d_2\i\)\\ &=\(0\i-d_1\j\)\cdot\(d_2\i+0\j\)\\ &=0\cdot d_2+(-d_1)\cdot0\\ &=0 \end{aligned} $$ $$\ab{d}$$ $$ \begin{aligned} \vec d_1\cdot\(\frac{\vec d_2}{5}\)&=\(-d_1\j\)\cdot\(\frac{d_2}{5}\i\)\\ &=\(0\i-d_1\j\)\cdot\(\frac{d_2}{5}\i+0\j\)\\ &=0\cdot \frac{d_2}{5}+(-d_1)\cdot0\\ &=0 \end{aligned} $$ $$\ab{e}$$ $$ \begin{aligned} \vec d_1\times\vec d_2&=\(-d_1\j\)\times\(d_2\i\)\\ &=(-d_1d_2)\(\j\times\i\)\\ &=(-d_1d_2)(-1)\(\i\times\j\)\\ &=(d_1d_2)\(\i\times\j\)\\ &=d_1d_2\k \end{aligned} $$ $$\therefore +z$$ $$\ab{f}$$ $$ \begin{aligned} \vec d_2\times\vec d_1&=\(d_2\i\)\times\(-d_1\j\)\\ &=(-d_2d_1)\(\i\times\j\)\\ &=-d_1d_2\k \end{aligned} $$ $$\therefore -z$$ $$\ab{g}$$ $$ \begin{aligned} \vec d_1\times\vec d_2&=d_1d_2\k,\\ \abs{\vec d_1\times\vec d_2}&=d_1d_2 \end{aligned} $$ $$\ab{h}$$ $$ \begin{aligned} \vec d_2\times\vec d_1&=-d_1d_2\k,\\ \abs{\vec d_2\times\vec d_1}&=d_1d_2 \end{aligned} $$ $$\ab{i}$$ $$ \begin{aligned} \vec d_1\times\frac{\vec d_2}{5}&=\(-d_1\j\)\times\(\frac{d_2}{5}\i\)\\ &=\(-d_1\frac{d_2}{5}\)\(\j\times\i\)\\ &=\(-d_1\frac{d_2}{5}\)(-1)\(\i\times\j\)\\ &=\frac{d_1d_2}{5}\(\i\times\j\)\\ &=\frac{d_1d_2}{5}\k\\ \end{aligned} $$ $$\therefore \abs{\vec d_1\times\frac{\vec d_2}{5}}=\frac{d_1d_2}{5} $$ $$\ab{j}$$ $$\therefore +z$$