The resultant of vec{A} + vec{B} is vec{R}_1. | vedclass

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(C) Given,$\vec{R}_1 = \vec{A} + \vec{B}$ and $\vec{R}_2 = \vec{A} - \vec{B}$.Using the parallelogram law of vector addition,the magnitude of the resu

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

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(C) Given,$\vec{R}_1 = \vec{A} + \vec{B}$ and $\vec{R}_2 = \vec{A} - \vec{B}$.Using the parallelogram law of vector addition,the magnitude of the resultant is given by $R^2 = A^2 + B^2 + 2AB \cos 	heta$.For $\vec{R}_1$,the magnitude squared is $R_1^2 = A^2 + B^2 + 2AB \cos 	heta$.For $\vec{R}_2$,the angle between $\vec{A}$ and $-\vec{B}$ is $(180^\circ - 	heta)$,so $R_2^2 = A^2 + B^2 + 2AB \cos(180^\circ - 	heta) = A^2 + B^2 - 2AB \cos 	heta$.Adding these 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 = 2(A^2 + B^2)$.

The resultant of $\vec{A} + \vec{B}$ is $\vec{R}_1$. On reversing the vector $\vec{B}$,the resultant becomes $\vec{R}_2$. What is the value of $R_1^2 + R_2^2$?

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