The resultant of two vectors $\overrightarrow{\mathrm{P}}$ and $\overrightarrow{\mathrm{Q}}$ is…

The resultant of two vectors $\overrightarrow{\mathrm{P}}$ and $\overrightarrow{\mathrm{Q}}$ is $\overrightarrow{\mathrm{R}}$. When the direction of $\overrightarrow{\mathrm{Q}}$ is reversed, the resultant is given by $\overrightarrow{\mathrm{S}}$. Which one of the following is true for vectors $\overrightarrow{\mathrm{R}}$ and $\overrightarrow{\mathrm{S}}$ ?
  1. $\mathrm{R}^{2}-\mathrm{S}^{2}=\left(\mathrm{P}^{2}+\mathrm{Q}^{2}\right)$
  2. $\mathrm{R}^{2}-\mathrm{S}^{2}=2(\overrightarrow{\mathrm{P}} \cdot \overrightarrow{\mathrm{Q}})$
  3. $\mathrm{R}^{2}+\mathrm{S}^{2}=4(\overrightarrow{\mathrm{P}} \cdot \overrightarrow{\mathrm{Q}})$
  4. $\mathrm{R}^{2}+\mathrm{S}^{2}=2\left(\mathrm{P}^{2}+\mathrm{Q}^{2}\right)$

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 resurtants. $\begin{array}{l} R^{2}=p^{2}+Q^{2}+2 P Q \cos \theta...(1) \\ S^{2}=p^{2}+Q^{2}-2 P Q \cos \theta \quad...(2) \ \end{array}$ Adding equation (1) and (2) $R^{2}+S^{2}=2\left(P^{2}+Q^{2}\right)$ So, the corvect answer is $R^{2}+S^{2}=2\left(P^{2}+Q^{2}\right)$

Asked in: MHT CET 2020 (19 Oct Shift 1)

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