question_answer

If \[\overset{\to }{\mathop{R}}\,=\left( \overset{\to }{\mathop{A}}\,+\overset{\to }{\mathop{B}}\, \right)\], show that \[{{R}^{2}}={{A}^{2}}+{{B}^{2}}+2AB\cos \theta \] where\[\theta \] is the smaller angle between\[\overset{\to }{\mathop{A}}\,\] and \[\overset{\to }{\mathop{B}}\,\].

Answer:

                \[\text{cos}\theta =\frac{{{R}^{2}}-{{A}^{2}}-{{B}^{2}}}{2AB}=\frac{{{R}^{2}}-{{R}^{2}}}{2AB}=0\]or\[\theta ~=\text{9}0{}^\circ \].