A) \[90{}^\circ \]
B) \[40{}^\circ \]
C) \[45{}^\circ \]
D) \[60{}^\circ \]
Correct Answer: A
Solution :
Given \[|\vec{A}+\vec{B}|=|\vec{A}-\vec{B}|\] squaring, \[|\vec{A}+\vec{B}{{|}^{2}}=|\vec{A}-\vec{B}{{|}^{2}}\] or\[(\vec{A}+\vec{B}).=(\vec{A}+\vec{B})=(\vec{A}-\vec{B}).(\vec{A}-\vec{B})\] \[\Rightarrow \]\[\vec{A}.\vec{A}+\vec{A}.\vec{B}+\vec{B}.\vec{A}+\vec{B}.\vec{B}\] \[=\vec{A}.\vec{A}-\vec{A}.\vec{B}-\vec{B}.\vec{A}+\vec{B}.\vec{B}\] As \[\vec{A}.\vec{B}=\vec{B}.\vec{A}\] and\[\vec{A}.\vec{A}={{A}^{2}}etc.\] \[\therefore \]\[{{A}^{2}}+2\vec{A}.\,\vec{B}+\vec{B}={{A}^{2}}-2\vec{A}.\vec{B}+{{B}^{2}}\] or \[4\vec{A}.\,\vec{B}=0\] \[\Rightarrow \] \[\vec{A}.\,\vec{B}=0\] This implies that \[\vec{A}\] and \[\vec{B}\] are mutually perpendicular\[(or\text{ = }90{}^\circ )\]
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