3-24 할리데이 10판 솔루션 일반물리학

$$\begin{cases} \vec A = A_x\hat i\\ B=6.0\\ \vec A + \vec B = \vec r\\ \vec r = R_y\hat j\\ r=2A\\ \end{cases} $$ $$ \vec B = (6\cos\theta_B)\hat i + (6\sin\theta_B)\hat j$$ $$ \Ans = A = \abs{A_x}$$ $$ \begin{aligned} \vec r =& \vec A + \vec B\\ =& (A_x+6\cos\theta_B)\hat i + (6\sin\theta_B)\hat j \end{aligned} $$ $$\therefore \begin{cases} A_x+6\cos\theta_B=0\\ 6\sin\theta_B=2A_x \end{cases} $$ $$ \therefore \abs{A_x} = \abs{\pm\frac{6}{\sqrt5}}=\frac{6}{\sqrt5}$$