10판/3. 벡터

3-7 할리데이 10판 솔루션 일반물리학

짱세디럭스 2019. 8. 6. 17:34

$$\begin{cases} \vec a:(3\ut{m},\theta_a)\\ \vec b:(4\ut{m},\theta_b)\\ \theta_a-\theta_b=\phi \end{cases} $$ $$\begin{cases} \vec a = 3\cos\theta_a\hat i+3\sin\theta_a\hat j\\ \vec b = 4\cos\theta_b\hat i+4\sin\theta_b\hat j\\ \end{cases} $$ $$\vec a+\vec b = \(3\cos\theta_a+4\cos\theta_b\)\hat i+\(3\sin\theta_a+4\sin\theta_b\)\hat j$$ $$ \begin{aligned} \abs{\vec a+\vec b} &= \sqrt{\(3\cos\theta_a+4\cos\theta_b\)^2+\(3\sin\theta_a+4\sin\theta_b\)^2}\\ &=\sqrt{25+24\cos\(\theta_a-\theta_b\)}\\ &=\sqrt{25+24\cos\phi}\\ \end{aligned} $$
(a)7m $$ 7 =\sqrt{25+24\cos\phi_a} $$ $$ \phi_a = 0$$
(b)1m $$ 1 =\sqrt{25+24\cos\phi_b} $$ $$ \phi_b = \pm \pi $$
(c)5m $$ 5 =\sqrt{25+24\cos\phi_c} $$ $$ \phi_c = \pm \frac{\pi}{2} $$