If the adjacent sides of a rectangle are bar{ | vedclass

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(A) Given vectors are $\bar{a}=5\bar{m}-3\bar{n}$,$\bar{c}=-4\bar{m}-\bar{n}$,and $\bar{d}=-\bar{m}+\bar{n}$.First,calculate $\bar{x}$:$\bar{x} = \fra

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$\frac{\pi}{2}$

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(A) Given vectors are $\bar{a}=5\bar{m}-3\bar{n}$,$\bar{c}=-4\bar{m}-\bar{n}$,and $\bar{d}=-\bar{m}+\bar{n}$.First,calculate $\bar{x}$:$\bar{x} = \frac{\bar{a}+\bar{c}+\bar{d}}{3} = \frac{(5\bar{m}-3\bar{n}) + (-4\bar{m}-\bar{n}) + (-\bar{m}+\bar{n})}{3} = \frac{0\bar{m}-3\bar{n}}{3} = -\bar{n}$.Next,calculate $\bar{y}$:$\bar{y} = \frac{\bar{c}+\bar{d}}{5} = \frac{(-4\bar{m}-\bar{n}) + (-\bar{m}+\bar{n})}{5} = \frac{-5\bar{m}}{5} = -\bar{m}$.Since $\bar{m}$ and $\bar{n}$ are adjacent sides of a rectangle,they are perpendicular,so $\bar{m} \cdot \bar{n} = 0$. Assuming $|\bar{m}| = |\bar{n}| = 1$ for simplicity as they represent unit directions of the rectangle sides,the angle $	heta$ between $\bar{x} = -\bar{n}$ and $\bar{y} = -\bar{m}$ is the same as the angle between $\bar{n}$ and $\bar{m}$,which is $\frac{\pi}{2}$.

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