A) \[\overset{\to }{\mathop{A}}\,\text{ }is\text{ }parallel\text{ }to\text{ }\overset{\to }{\mathop{B}}\,\]
B) \[\overset{\to }{\mathop{A}}\,\text{ }is\text{ }anti-parallel\text{ }to\text{ }\overset{\to }{\mathop{B}}\,\]
C) \[\overset{\to }{\mathop{A}}\,\text{ }and\,\,\overset{\to }{\mathop{B}}\,\,are\,\,equal\,\,in\,\,magnitude\]
D) \[\overset{\to }{\mathop{A}}\,\text{ }and\overset{\to }{\mathop{B}}\,are\,\,mutually\,\,perpendicular\]
Correct Answer: D
Solution :
Key Idea: Scalar, product of a vector by itself is equal to the square of the magnitude of that vector. Given, \[|\vec{A}+\vec{B}|=|\vec{A}-\vec{B}|\] Squaring both sides, we get \[|\vec{A}+\vec{B}{{|}^{2}}=|\vec{A}-\vec{B}{{|}^{2}}\] \[(\vec{A}+\vec{B}).(\vec{A}+\vec{B})=(\vec{A}-\vec{B}).(\vec{A}-\vec{B})\] \[\vec{A}.\vec{A}+\vec{A}.\vec{B}+\vec{B}.\vec{A}+\vec{B}.\vec{B}\] \[=\vec{A}.\vec{A}-\vec{A}.\vec{B}-\vec{B}.\vec{A}+\vec{B}.\vec{B}\] \[\vec{A}.\vec{B}+\vec{B}.\vec{A}=-\vec{A}.\vec{B}-\vec{B}.\vec{A}\] But \[\vec{B}.\vec{A}=\vec{A}.\vec{B}\] \[\therefore \] \[2(\vec{A}.\vec{B})=-2(\vec{A}.\vec{B})\] \[\Rightarrow \] \[4(\vec{A}.\vec{B})=0\] \[\Rightarrow \] \[\vec{A}.\vec{B}=0.\] As the scalar product of \[\vec{A}\]and \[\vec{B}\]is zero,\[\vec{A}\] and \[\vec{B}\]are mutually perpendicular.
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