A) \[{{0}^{o}}\]
B) \[{{60}^{o}}\]
C) \[{{90}^{o}}\]
D) \[{{120}^{o}}\]
Correct Answer: C
Solution :
We have given |
\[|\vec{A}+\vec{B}|=\,|\vec{A}-\vec{B}|\] |
Squaring both the sides, we obtain |
\[{{\left| \vec{A}+\vec{B} \right|}^{2}}=\,{{\left| \vec{A}-\vec{B} \right|}^{2}}\] |
or \[(\vec{A}+\vec{B})\,.\,(\vec{A}+\vec{B})\,=(\vec{A}-\vec{B})\,.(\vec{A}-\vec{B})\] |
or \[\vec{A}.\vec{A}+\vec{A}.\vec{B}+\vec{B}.\vec{A}+\vec{B}.\vec{B}=\vec{A}.\vec{A}-\vec{A}.\vec{B}\]\[-\vec{B}\,.\,\vec{A}+\vec{B}\,.\,\vec{B}\] |
or \[\vec{A}.\vec{B}+\vec{A}.\vec{B}=-\vec{A}.\vec{B}-\vec{A}.\vec{B}\]\[(\because \,\,\vec{B}.\vec{A}=\vec{A}.\vec{B})\] |
or \[4\vec{A}.\vec{B}=0\] |
or \[\vec{A}\,.\,\vec{B}=0\] |
Since dot product of \[\vec{A}\] and \[\vec{B}\] is zero hence, \[\vec{A}\]and \[\vec{B}\] are mutually perpendicular i.e., angle between and \[\vec{A}\] is \[\vec{B}\]\[90{}^\circ \] |
You need to login to perform this action.
You will be redirected in
3 sec