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

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When $\vec{A}$ and $\vec{B}$ are perpendicular.

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(B) The magnitude of the sum of two vectors is given by $|\vec{A} + \vec{B}| = \sqrt{A^2 + B^2 + 2AB \cos 	heta}$.The magnitude of the difference of two vectors is given by $|\vec{A} - \vec{B}| = \sqrt{A^2 + B^2 - 2AB \cos 	heta}$.Setting these equal: $\sqrt{A^2 + B^2 + 2AB \cos 	heta} = \sqrt{A^2 + B^2 - 2AB \cos 	heta}$.Squaring both sides: $A^2 + B^2 + 2AB \cos 	heta = A^2 + B^2 - 2AB \cos 	heta$.This simplifies to $4AB \cos 	heta = 0$.Since $A$ and $B$ are non-zero vectors,we must have $\cos 	heta = 0$,which implies $	heta = 90^\circ$.Thus,the magnitudes are equal when $\vec{A}$ and $\vec{B}$ are perpendicular to each other.

When are the magnitudes of $\vec{A} + \vec{B}$ and $\vec{A} - \vec{B}$ equal?

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