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(A) Given that $(\overrightarrow{A} + \overrightarrow{B})$ is perpendicular to $(\overrightarrow{A} - \overrightarrow{B})$.Since the dot product of tw

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(A) Given that $(\overrightarrow{A} + \overrightarrow{B})$ is perpendicular to $(\overrightarrow{A} - \overrightarrow{B})$.Since the dot product of two perpendicular vectors is zero,we have:$(\overrightarrow{A} + \overrightarrow{B}) \cdot (\overrightarrow{A} - \overrightarrow{B}) = 0$Expanding the dot product:$\overrightarrow{A} \cdot \overrightarrow{A} - \overrightarrow{A} \cdot \overrightarrow{B} + \overrightarrow{B} \cdot \overrightarrow{A} - \overrightarrow{B} \cdot \overrightarrow{B} = 0$Using the commutative property of the dot product,$\overrightarrow{A} \cdot \overrightarrow{B} = \overrightarrow{B} \cdot \overrightarrow{A}$,the middle terms cancel out:$|\overrightarrow{A}|^2 - |\overrightarrow{B}|^2 = 0$$|\overrightarrow{A}|^2 = |\overrightarrow{B}|^2$$|\overrightarrow{A}| = |\overrightarrow{B}|$Therefore,the ratio of their magnitudes is $\frac{|\overrightarrow{A}|}{|\overrightarrow{B}|} = 1$.

If for two vectors $\overrightarrow{A}$ and $\overrightarrow{B}$,the sum $(\overrightarrow{A} + \overrightarrow{B})$ is perpendicular to the difference $(\overrightarrow{A} - \overrightarrow{B})$,then the ratio of their magnitudes is:

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