question_answer

The resultant of two vectors \[\vec{P}\]  and \[\vec{Q}\] is  perpendicular to \[\vec{P}\] and its magnitude is half that of \[\vec{Q}\]. What is the angle between \[\vec{P}\] and \[\vec{Q}\]?

Answer:

                Clearly, \[{{P}^{2}}+{{\left( \frac{Q}{2} \right)}^{2}}={{Q}^{2}}\]                 or            \[{{P}^{2}}=\frac{3}{4}{{Q}^{2}}\]             or \[P=\frac{\sqrt{3}}{2}Q\]                                   \[\therefore \]  \[\tan \theta =\frac{Q/2}{P}=\frac{Q/2}{\sqrt{3}Q/2}=\frac{1}{\sqrt{3}}\]                 or            \[\theta ={{30}^{\circ }}\]                 Angle between \[\vec{P}\]and \[\vec{Q},\]                        \[\beta ={{180}^{\circ }}-\theta ={{180}^{\circ }}-{{30}^{\circ }}=\mathbf{15}{{\mathbf{0}}^{\mathbf{o}}}\]