A) \[\overset{\to }{\mathop{\mathbf{A}}}\,\] is parallel to \[\overset{\to }{\mathop{\mathbf{B}}}\,\]
B) \[\overset{\to }{\mathop{\mathbf{A}}}\,\] is anti-parallel to \[\overset{\to }{\mathop{\mathbf{B}}}\,\]
C) \[\overset{\to }{\mathop{\mathbf{A}}}\,\] and \[\overset{\to }{\mathop{\mathbf{B}}}\,\] are equal in magnitude
D) \[\overset{\to }{\mathop{\mathbf{A}}}\,\] and \[\overset{\to }{\mathop{\mathbf{B}}}\,\] 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, \[|\overset{\to }{\mathop{\mathbf{A}}}\,+\overset{\to }{\mathop{\mathbf{B}}}\,|=|\overset{\to }{\mathop{\mathbf{A}}}\,-\overset{\to }{\mathop{\mathbf{B}}}\,|\] Squaring both sides, we get \[|\overset{\to }{\mathop{\mathbf{A}}}\,+\overset{\to }{\mathop{\mathbf{B}}}\,{{|}^{2}}=|\overset{\to }{\mathop{\mathbf{A}}}\,-\overset{\to }{\mathop{\mathbf{B}}}\,{{|}^{2}}\] \[(\overset{\to }{\mathop{\mathbf{A}}}\,+\overset{\to }{\mathop{\mathbf{B}}}\,)\cdot (\overset{\to }{\mathop{\mathbf{A}}}\,+\overset{\to }{\mathop{\mathbf{B}}}\,)=(\overset{\to }{\mathop{\mathbf{A}}}\,-\overset{\to }{\mathop{\mathbf{B}}}\,)\cdot (\overset{\to }{\mathop{\mathbf{A}}}\,-\overset{\to }{\mathop{\mathbf{B}}}\,)\] \[\overset{\to }{\mathop{\mathbf{A}}}\,\cdot \overset{\to }{\mathop{\mathbf{A}}}\,+\overset{\to }{\mathop{\mathbf{A}}}\,\cdot \overset{\to }{\mathop{\mathbf{B}}}\,+\overset{\to }{\mathop{\mathbf{B}}}\,\cdot \overset{\to }{\mathop{\mathbf{A}}}\,+\overset{\to }{\mathop{\mathbf{B}}}\,\cdot \overset{\to }{\mathop{\mathbf{B}}}\,\] \[=\overset{\to }{\mathop{\mathbf{A}}}\,\cdot \overset{\to }{\mathop{\mathbf{A}}}\,-\overset{\to }{\mathop{\mathbf{A}}}\,\cdot \overset{\to }{\mathop{\mathbf{B}}}\,-\overset{\to }{\mathop{\mathbf{B}}}\,\cdot \overset{\to }{\mathop{\mathbf{A}}}\,+\overset{\to }{\mathop{\mathbf{B}}}\,\cdot \overset{\to }{\mathop{\mathbf{B}}}\,\] \[\overset{\to }{\mathop{\mathbf{A}}}\,\cdot \overset{\to }{\mathop{\mathbf{B}}}\,+\overset{\to }{\mathop{\mathbf{B}}}\,\cdot \overset{\to }{\mathop{\mathbf{A}}}\,=-\overset{\to }{\mathop{\mathbf{A}}}\,\cdot \overset{\to }{\mathop{\mathbf{B}}}\,-\overset{\to }{\mathop{\mathbf{B}}}\,\cdot \overset{\to }{\mathop{\mathbf{A}}}\,\] But \[\overset{\to }{\mathop{\mathbf{B}}}\,\cdot \overset{\to }{\mathop{\mathbf{A}}}\,=\overset{\to }{\mathop{\mathbf{A}}}\,\cdot \overset{\to }{\mathop{\mathbf{B}}}\,\] \[\therefore \] \[2(\overset{\to }{\mathop{\mathbf{A}}}\,\cdot \overset{\to }{\mathop{\mathbf{B}}}\,)=-2(\overset{\to }{\mathop{\mathbf{A}}}\,\cdot \overset{\to }{\mathop{\mathbf{B}}}\,)\] \[\Rightarrow \] \[4(\overset{\to }{\mathop{\mathbf{A}}}\,\cdot \overset{\to }{\mathop{\mathbf{B}}}\,)=0\] \[\Rightarrow \] \[\overset{\to }{\mathop{\mathbf{A}}}\,\cdot \overset{\to }{\mathop{\mathbf{B}}}\,=0\] As the scalar product of \[\overset{\to }{\mathop{\mathbf{A}}}\,\] and \[\overset{\to }{\mathop{\mathbf{B}}}\,\] is zero, \[\overset{\to }{\mathop{\mathbf{A}}}\,\] and \[\overset{\to }{\mathop{\mathbf{B}}}\,\] are mutually perpendicular.
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