3-41 할리데이 11판 솔루션 일반물리학

$$\begin{cases} \vec a&=6.0\i+5.0\j\\ \vec b&=3.0\i+4.0\j\\ \end{cases}$$ $$\ab{a}$$ $$\vec a\times \vec b=(a_yb_z-b_ya_z)\i+(a_zb_x-b_za_x)\j+(a_xb_y-b_xa_y)\k,$$ $$\begin{aligned} \vec a \times \vec b &=\(6\i+5\j\)\times\(3\i+4\j\)\\ &=(6\times4-5\times3)\k\\ &=9\k \end{aligned}$$ $$\ab{b}$$ $$\vec a \cdot \vec b=a_xb_x+a_yb_y+a_zb_z,$$ $$\begin{aligned} \vec a \cdot \vec b&=\(6\i+5\j\)\cdot\(3\i+4\j\)\\ &=6\cdot3+5\cdot4\\ &=38 \end{aligned}$$ $$\ab{c}$$ $$\begin{aligned} (\vec a + \vec b)\cdot\vec b&=\(6\i+5\j+3\i+4\j\)\cdot\(3\i+4\j\)\\ &=\(9\i+9\j\)\cdot\(3\i+4\j\)\\ &=9\cdot3+9\cdot4\\ &=63 \end{aligned}$$ $$\ab{d}$$ $$\begin{aligned} \Ans &= a\cos\theta_{ab}\\ &=\frac{\vec a\cdot \vec b}{b}\\ &=\frac{38}{5}\\ &=7.6 \end{aligned}$$