The resultant of two vectors overrightarrow{P | vedclass

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(B) Let $	heta$ be the angle between vectors $\overrightarrow{P}$ and $\overrightarrow{Q}$.According to the law of vector addition,the magnitude of t

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$2(P^{2} + Q^{2})$

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(B) Let $	heta$ be the angle between vectors $\overrightarrow{P}$ and $\overrightarrow{Q}$.According to the law of vector addition,the magnitude of the resultant $R_{1}$ is given by:$R_{1}^{2} = P^{2} + Q^{2} + 2PQ \cos 	heta$  --- $(1)$When the direction of $\overrightarrow{Q}$ is reversed,the new vector is $-\overrightarrow{Q}$. The angle between $\overrightarrow{P}$ and $-\overrightarrow{Q}$ becomes $(\pi - 	heta)$.The magnitude of the new resultant $R_{2}$ is given by:$R_{2}^{2} = P^{2} + Q^{2} + 2PQ \cos(\pi - 	heta)$Since $\cos(\pi - 	heta) = -\cos 	heta$,we have:$R_{2}^{2} = P^{2} + Q^{2} - 2PQ \cos 	heta$  --- $(2)$Adding equation $(1)$ and equation $(2)$:$R_{1}^{2} + R_{2}^{2} = (P^{2} + Q^{2} + 2PQ \cos 	heta) + (P^{2} + Q^{2} - 2PQ \cos 	heta)$$R_{1}^{2} + R_{2}^{2} = 2(P^{2} + Q^{2})$

The resultant of two vectors $\overrightarrow{P}$ and $\overrightarrow{Q}$ has a magnitude $R_{1}$. If the direction of $\overrightarrow{Q}$ is reversed,the resultant has a magnitude $R_{2}$. The value of $(R_{1}^{2} + R_{2}^{2})$ is:

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