3-2 할리데이 11판 솔루션 일반물리학

$$ \begin{cases} \vec A&=A\i\\ B&=14.0\ut{m}\\ \vec A + \vec B &= \vec C\\ \vec C &= C\j\\ C &= 3A \end{cases} $$ $$ A\i + \vec B = 3A\j $$ $$ \begin{aligned} \vec B &= (0-A)\i+(3A-0)\j+(0-0)\k\\ &=-A\i+3A\j\\ \end{aligned} $$ $$ B=\sqrt{(-A)^2+(3A)^2}=14\ut{m} $$ $$ \begin{aligned} A&=7\sqrt{\frac{2}{5}}\ut{m}\\ &\approx 4.427188724235731\ut{m}\\ &\approx 4.43\ut{m}\\ \end{aligned} $$