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

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(D) Given that $|\vec{A} + \vec{B}| = |\vec{A} - \vec{B}|$.Squaring both sides,we get:$|\vec{A} + \vec{B}|^2 = |\vec{A} - \vec{B}|^2$Using the propert

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$90$

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(D) Given that $|\vec{A} + \vec{B}| = |\vec{A} - \vec{B}|$.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 have:$(\vec{A} + \vec{B}) \cdot (\vec{A} + \vec{B}) = (\vec{A} - \vec{B}) \cdot (\vec{A} - \vec{B})$$A^2 + B^2 + 2\vec{A} \cdot \vec{B} = A^2 + B^2 - 2\vec{A} \cdot \vec{B}$$4\vec{A} \cdot \vec{B} = 0$Since $\vec{A} \cdot \vec{B} = AB \cos 	heta$,we have:$4AB \cos 	heta = 0$Assuming $A 
eq 0$ and $B 
eq 0$,we get $\cos 	heta = 0$.Therefore,$	heta = 90^\circ$.

If $|\vec{A} + \vec{B}| = |\vec{A} - \vec{B}|$,then the angle between $\vec{A}$ and $\vec{B}$ is ........ $^\circ$.

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