Question

given that p vector +q vector =r vector and that r vector is perpendicular to p vector. if p modulus = r modulus ,then what is the angle between p and q?

Answer 1

$$\begin{gathered} \overrightarrow{\left|r\right|}=\overrightarrow{\left|p\right|}=a\left(say\right).Then \\ \sin cemagnituderesul\tan toftwovectorsisgivenby: \\ \left|R\right|=\sqrt{A^2+B^2+2AB\hspace{0.17em}\cos \theta }whereistheanglebetweenthetwovectorsbeingaddedinourcasebetween\overrightarrow{\left|p\right|}and\left|\overrightarrow{q}\right| \\ then, \\ \overrightarrow{\left|r\right|}=\sqrt{\overrightarrow{{\left|p\right|}^2}+{\left|\overrightarrow{q}\right|}^2+2\overrightarrow{\left|p\right|}\left|\overrightarrow{q}\right|\cos \theta } \\ a=\sqrt{a^2+{\left|\overrightarrow{q}\right|}^2+2a\left|\overrightarrow{q}\right|\cos \theta } \\ squaringbothsides, \\ {a}^2=a^2+{\left|\overrightarrow{q}\right|}^2+2a\left|\overrightarrow{q}\right|\cos \theta \\ {\left|\overrightarrow{q}\right|}^2=-2a\left|\overrightarrow{q}\right|\cos \theta \\ \left|\overrightarrow{q}\right|=-2a\cos \theta \\ now, \\ IfAandBaretwovectorsatanangle\theta andtheirresul\tan tmakesanangle\alpha fromAthen \\ \tan \alpha =\frac{\left|B\right|\sin \theta }{\left|A\right|+\left|B\right|\cos \theta } \\ here \\ \alpha ={90}^o \\ \left|B\right|=\left|\overrightarrow{q}\right|=-2a\cos \theta \\ and, \\ \left|A\right|=\overrightarrow{\left|p\right|}=a \\ then \\ \tan 90=\frac{-2a\cos \theta \times \sin \theta }{a+2a\cos \theta }, \\ thisispossibleonlywhen, \\ (a+2a\cos \theta )=0(\sin ce,\tan {90}^o=\infty =\frac{1}{0}). \\ or, \\ 1-2\cos \theta =0 \\ \cos \theta =\frac{1}{2} \\ \theta ={\cos }^{-1}\left(\frac{1}{2}\right) \\ \theta ={60}^o. \end{gathered}$$