A) \[{{90}^{o}}\]
B) \[{{60}^{o}}\]
C) \[{{75}^{o}}\]
D) \[{{45}^{o}}\]
Correct Answer: A
Solution :
As we have given, \[\left| \vec{A}+\vec{B} \right|=\,\left| \vec{A}-\vec{B} \right|\] \[or\sqrt{{{A}^{2}}+{{B}^{2}}+2AB\cos \theta }\] \[=\sqrt{{{A}^{2}}+{{B}^{2}}+2AB\cos \theta }\] where \[\theta \] is the angle between \[\vec{A}\,and\,\vec{B}\] Squaring both sides, we have \[{{A}^{2}}+{{B}^{2}}+2AB\cos \theta ={{A}^{2}}+{{B}^{2}}-2AB\cos \theta \] \[or4AB\cos \theta =0\] \[AsAB\ne 0\] \[\therefore \cos \theta =0=\cos {{90}^{o}}\] \[\therefore \theta ={{90}^{0}}\] Hence, angle between \[\vec{A}\] and \[\vec{B}\] is \[{{90}^{o}}\].
You need to login to perform this action.
You will be redirected in
3 sec
