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(A) Given that $\overrightarrow{a}=\overrightarrow{b}+\overrightarrow{c}$.Taking the dot product of both sides with themselves:$\overrightarrow{a} \cd

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$a^2=b^2+c^2$

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(A) Given that $\overrightarrow{a}=\overrightarrow{b}+\overrightarrow{c}$.Taking the dot product of both sides with themselves:$\overrightarrow{a} \cdot \overrightarrow{a} = (\overrightarrow{b}+\overrightarrow{c}) \cdot (\overrightarrow{b}+\overrightarrow{c})$$|\overrightarrow{a}|^2 = |\overrightarrow{b}|^2 + |\overrightarrow{c}|^2 + 2(\overrightarrow{b} \cdot \overrightarrow{c})$Since the angle between $\overrightarrow{b}$ and $\overrightarrow{c}$ is $\frac{\pi}{2}$, their dot product $\overrightarrow{b} \cdot \overrightarrow{c} = |\overrightarrow{b}| |\overrightarrow{c}| \cos(\frac{\pi}{2}) = 0$.Therefore, $a^2 = b^2 + c^2 + 0$$a^2 = b^2 + c^2$.

If $\overrightarrow{a}, \overrightarrow{b}, \overrightarrow{c}$ are three vectors such that $\overrightarrow{a}=\overrightarrow{b}+\overrightarrow{c}$ and the angle between $\overrightarrow{b}$ and $\overrightarrow{c}$ is $\frac{\pi}{2}$, then:

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