For two vectors vec{a} and vec{b},if |vec{a} | vedclass

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(C) Given the condition $|\vec{a} + \vec{b}| = |\vec{a} - \vec{b}|$.Squaring both sides,we get $|\vec{a} + \vec{b}|^2 = |\vec{a} - \vec{b}|^2$.Using t

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$\frac{\pi}{2}$

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(C) Given the condition $|\vec{a} + \vec{b}| = |\vec{a} - \vec{b}|$.Squaring both sides,we get $|\vec{a} + \vec{b}|^2 = |\vec{a} - \vec{b}|^2$.Using the property $|\vec{v}|^2 = \vec{v} \cdot \vec{v}$,we expand the expression:$(\vec{a} + \vec{b}) \cdot (\vec{a} + \vec{b}) = (\vec{a} - \vec{b}) \cdot (\vec{a} - \vec{b})$.$|\vec{a}|^2 + |\vec{b}|^2 + 2(\vec{a} \cdot \vec{b}) = |\vec{a}|^2 + |\vec{b}|^2 - 2(\vec{a} \cdot \vec{b})$.Canceling $|\vec{a}|^2$ and $|\vec{b}|^2$ from both sides,we get $2(\vec{a} \cdot \vec{b}) = -2(\vec{a} \cdot \vec{b})$.This implies $4(\vec{a} \cdot \vec{b}) = 0$,so $\vec{a} \cdot \vec{b} = 0$.Since the dot product of two non-zero vectors is zero,the vectors must be perpendicular to each other.Therefore,the angle between $\vec{a}$ and $\vec{b}$ is $\frac{\pi}{2}$.

For two vectors $\vec{a}$ and $\vec{b}$,if $|\vec{a} + \vec{b}| = |\vec{a} - \vec{b}|$,then the angle between $\vec{a}$ and $\vec{b}$ is:

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