Question

At what angle the 2 vectors of magnitude (P+Q) and (P-Q) should act so that their resultant is root of P²+Q^{2 ?}

Answer 1

$$\begin{gathered} \mathrm{Suppose}\mathrm{the}\mathrm{two}\mathrm{forces}\mathrm{act}\mathrm{at}\mathrm{an}\mathrm{angle}\mathrm{\theta },\mathrm{then} \\ \sqrt{{\mathrm{P}}^2+{\mathrm{Q}}^2}=\sqrt{(\mathrm{P}+\mathrm{Q}{)}^2+(\mathrm{P}-\mathrm{Q}{)}^2+2(\mathrm{P}+\mathrm{Q})(\mathrm{P}-\mathrm{Q})\mathrm{cos\theta }} \\ [\mathrm{Using}\mathrm{resultant}(\mathrm{R})\mathrm{of}\mathrm{two}\mathrm{vectors}\mathrm{A}\mathrm{and}\mathrm{B},\mathrm{R}=\sqrt{{\mathrm{A}}^2+{\mathrm{B}}^2+2\mathrm{ABcos\theta }}] \\ \mathrm{On}\mathrm{squaring}, \\ {\mathrm{P}}^2+{\mathrm{Q}}^2=(\mathrm{P}+\mathrm{Q}{)}^2+(\mathrm{P}-\mathrm{Q}{)}^2+2(\mathrm{P}+\mathrm{Q})(\mathrm{P}-\mathrm{Q})\mathrm{cos\theta } \\ =2{\mathrm{P}}^2+2{\mathrm{Q}}^2+2({\mathrm{P}}^2-{\mathrm{Q}}^2)\mathrm{cos\theta } \\ =-2({\mathrm{P}}^2-{\mathrm{Q}}^2)\mathrm{cos\theta } \\ \mathrm{cos\theta }=-\frac{1}{2}\left(\frac{{\mathrm{P}}^2-{\mathrm{Q}}^2}{{\mathrm{P}}^2-{\mathrm{Q}}^2}\right)=\frac{1}{2}\left(\frac{{\mathrm{P}}^2+{\mathrm{Q}}^2}{{\mathrm{Q}}^2-{\mathrm{P}}^2}\right) \\ \mathrm{or}\mathrm{\theta }={\cos }^{-1}\left[\frac{1}{2}\right(\frac{{\mathrm{P}}^2+{\mathrm{Q}}^2}{{\mathrm{Q}}^2-{\mathrm{P}}^2}\left)\right] \\ \mathrm{Thus},\mathrm{the}\mathrm{two}\mathrm{forces}\mathrm{act}\mathrm{at}\mathrm{an}\mathrm{angle}{\cos }^{-1}\left[\frac{1}{2}\right(\frac{{\mathrm{P}}^2+{\mathrm{Q}}^2}{{\mathrm{Q}}^2-{\mathrm{P}}^2}\left)\right]. \end{gathered}$$