If the magnitude of the sum of two vectors is | vedclass

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(A) Let the two vectors be $\vec{A}$ and $\vec{B}$.The magnitude of the sum of $\vec{A}$ and $\vec{B}$ is given by: $|\vec{A} + \vec{B}| = \sqrt{A^2 +

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$90$

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(A) Let the two vectors be $\vec{A}$ and $\vec{B}$.The magnitude of the sum of $\vec{A}$ and $\vec{B}$ is given by: $|\vec{A} + \vec{B}| = \sqrt{A^2 + B^2 + 2AB \cos 	heta}$.The magnitude of the difference of $\vec{A}$ and $\vec{B}$ is given by: $|\vec{A} - \vec{B}| = \sqrt{A^2 + B^2 - 2AB \cos 	heta}$.Given that $|\vec{A} + \vec{B}| = |\vec{A} - \vec{B}|$,we square both sides:$A^2 + B^2 + 2AB \cos 	heta = A^2 + B^2 - 2AB \cos 	heta$.Subtracting $A^2 + B^2$ from both sides,we get:$2AB \cos 	heta = -2AB \cos 	heta$.$4AB \cos 	heta = 0$.Since $A$ and $B$ are non-zero vectors,$4AB 
eq 0$,therefore $\cos 	heta = 0$.This implies $	heta = 90^o$.

If the magnitude of the sum of two vectors is equal to the magnitude of the difference of the two vectors,the angle between these vectors is ........ $^o$.

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