10판/3. 벡터

3-13 할리데이 10판 솔루션 일반물리학

짱세디럭스 2019. 8. 6. 20:36

{a:(1.00,0)b:(1.0,θb)a+b=r:(1.0,θr)\begin{cases} \vec a:(1.00,0)\\ \vec b:(1.0,\theta_b)\\ \vec a+\vec b=\vec r:(1.0,\theta_r)\\ \end{cases} a=(1.00cos0)i^+(1.00sin0)j^=i^ \begin{aligned} \vec a&=(1.00\cos0 )\hat i + (1.00\sin0 )\hat j\\ &=\hat i \end{aligned} b=(1.0cosθb)i^+(1.0sinθb)j^=(cosθb)i^+(sinθb)j^ \begin{aligned} \vec b&=(1.0\cos\theta_b )\hat i + (1.0\sin\theta_b )\hat j\\ &=(\cos\theta_b )\hat i + (\sin\theta_b )\hat j \end{aligned} r=(1+cosθb)i^+(sinθb)j^ \vec r = (1+\cos\theta_b )\hat i + (\sin\theta_b )\hat j
(a)θb=?\theta_b=? r=(1+cosθb)2+(sinθb)2=2cosθb+1=1.0 \begin{aligned} \abs{\vec r}&=\sqrt{(1+\cos\theta_b )^2+(\sin\theta_b )^2}\\ &=\sqrt{2} \sqrt{\cos\theta_b+1}=1.0\\ \end{aligned} cosθb=12 \cos\theta_b=-\frac{1}{2} θb=±23π \theta_b = \pm \frac{2}{3}\pi
(b)ab=?\abs{\vec a - \vec b}=? b=(cosθb)i^+(sinθb)j^=(12)i^+{sin(±23π)}j^=(12)i^+(±32)j^ \begin{aligned} \vec b&=(\cos\theta_b )\hat i + (\sin\theta_b )\hat j\\ &=\(-\frac{1}{2}\)\hat i + \bra{\sin\(\pm \frac{2}{3}\pi\)}\hat j\\ &=\(-\frac{1}{2}\)\hat i + \(\pm\frac{\sqrt{3}}{2}\)\hat j\\ \end{aligned} Ans=ab=(1+12)i^+(32)j^=(32)i^+(32)j^=(32)2+(32)2=3[m]1.7[m] \begin{aligned} \Ans &= \abs{\vec a - \vec b}\\ &=\abs{\(1+\frac{1}{2}\)\hat i + \(\mp\frac{\sqrt{3}}{2}\)\hat j}\\ &=\abs{\(\frac{3}{2}\)\hat i + \(\mp\frac{\sqrt{3}}{2}\)\hat j}\\ &=\sqrt{\(\frac{3}{2}\)^2+\(\mp\frac{\sqrt{3}}{2}\)^2}\\ &=\sqrt{3}\ut{m}\\ &\approx 1.7\ut{m}\\ \end{aligned}