The resultant of two vectors vec{A} and vec{B | vedclass

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(C) Given that $\vec{R_1} = \vec{A} + \vec{B}$.Taking the magnitude squared: $R_1^2 = |\vec{A} + \vec{B}|^2 = A^2 + B^2 + 2AB \cos 	heta$,where $	he

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$2(A^2 + B^2)$

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(C) Given that $\vec{R_1} = \vec{A} + \vec{B}$.Taking the magnitude squared: $R_1^2 = |\vec{A} + \vec{B}|^2 = A^2 + B^2 + 2AB \cos 	heta$,where $	heta$ is the angle between $\vec{A}$ and $\vec{B}$.When vector $\vec{B}$ is reversed,the new resultant is $\vec{R_2} = \vec{A} - \vec{B}$.Taking the magnitude squared: $R_2^2 = |\vec{A} - \vec{B}|^2 = A^2 + B^2 - 2AB \cos 	heta$.Adding the two equations:$R_1^2 + R_2^2 = (A^2 + B^2 + 2AB \cos 	heta) + (A^2 + B^2 - 2AB \cos 	heta)$.$R_1^2 + R_2^2 = 2A^2 + 2B^2 = 2(A^2 + B^2)$.

The resultant of two vectors $\vec{A}$ and $\vec{B}$ is $\vec{R_1}$. If vector $\vec{B}$ is reversed,the resultant becomes $\vec{R_2}$. What is the value of $R_1^2 + R_2^2$?

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