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 \[{{90}^{o}}\]. |
You need to login to perform this action.
You will be redirected in
3 sec