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(D) The magnitude of the sum of two vectors is given by $|\vec{A}+\vec{B}|^2 = |\vec{A}|^2 + |\vec{B}|^2 + 2|\vec{A}||\vec{B}| \cos 	heta$.Similarly,

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$90$

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(D) The magnitude of the sum of two vectors is given by $|\vec{A}+\vec{B}|^2 = |\vec{A}|^2 + |\vec{B}|^2 + 2|\vec{A}||\vec{B}| \cos 	heta$.Similarly,the magnitude of the difference of two vectors is given by $|\vec{A}-\vec{B}|^2 = |\vec{A}|^2 + |\vec{B}|^2 - 2|\vec{A}||\vec{B}| \cos 	heta$.Given that $|\vec{A}+\vec{B}| = |\vec{A}-\vec{B}|$,we square both sides to get $|\vec{A}+\vec{B}|^2 = |\vec{A}-\vec{B}|^2$.Substituting the expressions,we have $|\vec{A}|^2 + |\vec{B}|^2 + 2|\vec{A}||\vec{B}| \cos 	heta = |\vec{A}|^2 + |\vec{B}|^2 - 2|\vec{A}||\vec{B}| \cos 	heta$.Canceling the common terms $|\vec{A}|^2$ and $|\vec{B}|^2$ from both sides,we get $2|\vec{A}||\vec{B}| \cos 	heta = -2|\vec{A}||\vec{B}| \cos 	heta$.Rearranging the terms,we get $4|\vec{A}||\vec{B}| \cos 	heta = 0$.Since the vectors are non-zero,$|\vec{A}| 
eq 0$ and $|\vec{B}| 
eq 0$,therefore $\cos 	heta = 0$.This implies $	heta = 90^{\circ}$.

The vectors $\vec{A}$ and $\vec{B}$ are such that $|\vec{A}+\vec{B}|=|\vec{A}-\vec{B}|$. The angle between the two vectors is: (in $^{\circ}$)

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