| The vectors \[\vec{A}\] and \[\vec{B}\] are such that a: |
| \[\left| \vec{A}+\vec{B} \right|=\left| \vec{A}-\vec{B} \right|\] |
| The angle between the two vectors is: [AIPMT (S) 2006] |
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 |
| \[{{A}^{2}}+{{B}^{2}}+2AB\cos \theta ={{A}^{2}}+{{B}^{2}}-2AB\cos \theta \] |
| or \[4AB\cos \theta =0\] |
| As \[AB\ne 0\] |
| \[\therefore \] \[\cos \theta =0=\cos {{90}^{o}}\] |
| \[\therefore \] \[\theta ={{90}^{0}}\] |
| Hence, angle between \[\vec{A}\] and \[\vec{B}\] is \[{{90}^{o}}\]. |
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