A) \[{{({{A}^{2}}+{{B}^{2}}+AB)}^{1/2}}\]
B) \[{{\left( {{A}^{2}}+{{B}^{2}}+\frac{AB}{\sqrt{3}} \right)}^{1/2}}\]
C) \[A+B\]
D) \[{{\left( {{A}^{2}}+{{B}^{2}}+\sqrt{3}AB \right)}^{1/2}}\]
Correct Answer: C
Solution :
\[\vec{p}=\]vector sum\[=\vec{A}+\vec{B}\] \[\vec{Q}=\]vector differences \[=\vec{A}+\vec{B}\] Since \[\vec{P}\]and \[\vec{Q}\]are perpendicular to \[\vec{P}.\vec{Q}=0\] \[\Rightarrow \]\[(\vec{A}+\vec{B}).(\vec{A}-\vec{B})=0\] \[\Rightarrow \]\[{{A}^{2}}={{B}^{2}}\Rightarrow \left| {\vec{A}} \right|=\left| {\vec{B}} \right|\]
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