If |vec{A}+vec{B}|=|vec{A}-vec{B}|,then the 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 the pro

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$90$

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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 terms:$A^2 + B^2 + 2AB \cos 	heta = A^2 + B^2 - 2AB \cos 	heta$.Subtracting $A^2 + B^2$ from both sides:$2AB \cos 	heta = -2AB \cos 	heta$.$4AB \cos 	heta = 0$.Since $A$ and $B$ are non-zero vectors,$\cos 	heta = 0$.Therefore,$	heta = 90^{\circ}$.

If $|\vec{A}+\vec{B}|=|\vec{A}-\vec{B}|$,then the angle between vector $\vec{A}$ and $\vec{B}$ is: (in $^{\circ}$)

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