3-38 할리데이 11판 솔루션 일반물리학

$$\begin{cases} \vec d_1 &= \(-2.0\i+3.0\j+4.0\k\)\ut{m}\\ \vec d_2 &= \(-2.0\i-4.0\j+3.0\k\)\ut{m}\\ \vec d_3 &= \(2.0\i+3.0\j+2.5\k\)\ut{m}\\ \end{cases}$$ $$\ab{a}$$ $$\vec a \cdot \vec b=a_xb_x+a_yb_y+a_zb_z,$$ $$\begin{aligned} \Ans&=\vec d_1\cdot\(\vec d_2+\vec d_3\)\\ &=\vec d_1\cdot\bra{\(-2\i-4\j+3\k\)+\(2\i+3\j+2.5\k\)}\\ &=\(-2\i+3\j+4\k\)\cdot\(-\j+5.5\k\)\\ &=(-2)\cdot0+3\cdot(-1)+4\cdot5.5\\ &=19\ut{m} \end{aligned}$$ $$\ab{b}$$ $$\vec a\times \vec b=(a_yb_z-b_ya_z)\i+(a_zb_x-b_za_x)\j+(a_xb_y-b_xa_y)\k,$$ $$\begin{aligned} \Ans&=\vec d_1\cdot\(\vec d_2\times\vec d_3\)\\ &=\vec d_1\cdot\bra{\(-2\i-4\j+3\k\)\times\(2\i+3\j+2.5\k\)}\\ &=\(-2\i+3\j+4\k\)\cdot\(-19\i+11\j+2\k\)\\ &=(-2)\cdot(-19)+3\cdot11+4\cdot2\\ &=79\ut{m} \end{aligned}$$ $$\ab{c}$$ $$\begin{aligned} \Ans&=\vec d_1\times\(\vec d_2+\vec d_3\)\\ &=\vec d_1\times\bra{\(-2\i-4\j+3\k\)+\(2\i+3\j+2.5\k\)}\\ &=\(-2\i+3\j+4\k\)\times\(-\j+5.5\k\)\\ &=\(20.5\i+11\j+2\k\)\ut{m}\\ &\approx\(21\i+11\j+2.0\k\)\ut{m}\\ \end{aligned}$$