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

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(C) Let the two vectors be $\vec{A}$ and $\vec{B}$.Given that the magnitude of their sum is equal to the magnitude of their difference: $|\vec{A} + \v

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$90$

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(C) Let the two vectors be $\vec{A}$ and $\vec{B}$.Given that the magnitude of their sum is equal to the magnitude of their difference: $|\vec{A} + \vec{B}| = |\vec{A} - \vec{B}|$.Squaring both sides,we get: $|\vec{A} + \vec{B}|^2 = |\vec{A} - \vec{B}|^2$.Using the property $|\vec{a} \pm \vec{b}|^2 = A^2 + B^2 \pm 2AB \cos 	heta$,we have:$A^2 + B^2 + 2AB \cos 	heta = A^2 + B^2 - 2AB \cos 	heta$.Subtracting $A^2 + B^2$ from both sides,we get: $2AB \cos 	heta = -2AB \cos 	heta$.This simplifies to: $4AB \cos 	heta = 0$.Since $A$ and $B$ are non-zero vectors,$\cos 	heta = 0$.Therefore,$	heta = 90^{\circ}$.

If the magnitude of the sum of two vectors is equal to the magnitude of the difference of two vectors,then the angle between them is: (in $^{\circ}$)

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