Resultant of two vectors $\overrightarrow{\mathrm{P}}$ and $\overrightarrow{\mathrm{Q}}$ is of magnitude…
Resultant of two vectors $\overrightarrow{\mathrm{P}}$ and $\overrightarrow{\mathrm{Q}}$ is of magnitude $\mathrm{R}_{1}$. If direction of $\overrightarrow{\mathrm{Q}}$ is reversed, the resultant is of magnitude $\mathrm{R}_{2}$. The value of $\left(\mathrm{R}_{1}^{2}+\mathrm{R}_{2}^{2}\right)$ is $[\cos (\pi-\theta)=-\cos \theta]$
$\left(\mathrm{P}^{2}+\mathrm{Q}^{2}\right)$
$2\left(\mathrm{P}^{2}+\mathrm{Q}^{2}\right)$
$2\left(\mathrm{P}^{2}-\mathrm{Q}^{2}\right)$
$\left(\mathrm{P}^{2}-\mathrm{Q}^{2}\right)$
Solution
Given,
$P+Q=R$
After reversing direction of $R$, we gel
$\begin{array}{l}
-R=-P-Q \\
S=-P-Q
\end{array}$
So let angle between $P$ and $Q$ be $Q$
So resultants.
$\begin{array}{l}
R^{2}=P^{2}+Q^{2}+2 P Q \cos \theta \\
S^{2}=P^{2}+Q^{2}-2 P Q \cos \theta \quad \frac{-(1)}{(2)}
\end{array}$
Adthing equation (1) and (2)
$R^{2}+S^{2}=2\left(P^{2}+Q^{2}\right)$
So, the correct answer is $R^{2}+S^{2}=2\left(P^{2}+Q^{2}\right)$
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