A) \[{{0}^{\text{o}}}\]
B) \[{{60}^{\text{o}}}\]
C) \[{{90}^{\text{o}}}\]
D) \[{{120}^{\text{o}}}\]
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 \[\vec{B}\] \[{{90}^{\text{o}}}\].
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