question_answer

Two equal vectors have a resultant equal to either of them, then the angle between them will be

A)  \[{{120}^{o}}\]

B)  \[{{110}^{o}}\]

C)  \[{{60}^{o}}\]

D)  \[{{150}^{o}}\]

Correct Answer: A

Solution :

Key Idea: Two vectors (inclined at any angle) and their sum vector form a triangle. It is given that two vectors have a resultant equal to either of them, hence these three vectors form an equilateral triangle each angle of \[{{60}^{o}}\]. In the figure \[\vec{A}\] and \[\vec{B}\] are two vectors \[(\vec{A}=\vec{B})\] having their sum vectors \[\vec{R}\] such that.                 \[\vec{R}=\vec{A}=\vec{B}\] Thus, the vectors \[\vec{A}\] and \[\vec{B}\] of same magnitude have the resultant vectors \[\vec{R}\] of the same magnitude. In this case angle between \[\vec{A}\] and \[\vec{B}\] is \[{{120}^{o}}\]. Alternative: Let there be two vectors\[\vec{A}\]and\[\vec{B}\]. where, \[A=B\] Their sum is                 \[\vec{R}=\vec{A}+\vec{B}\] Taking self product of both sides, we get                 \[\vec{R}.\,\,\vec{R}=(\vec{A}+\vec{B}).\,\,(\vec{A}+\vec{B})\]                 \[=\vec{A}\,\,.\,\,\vec{A}+2\vec{A}\,\,.\,\,\vec{B}+\vec{B}\,\,.\,\,\vec{B}\]                 \[={{A}^{2}}+2AB\cos \theta +{{B}^{2}}\] where \[\theta \] is angle between A and B. When \[\vec{R}=\vec{A}=\vec{B}\], then we have                 \[{{A}^{2}}={{A}^{2}}+2{{A}^{2}}\cos \theta +{{A}^{2}}\] \[\Rightarrow \] \[2{{A}^{2}}\cos \theta =-{{A}^{2}}\] \[\Rightarrow \] \[\cos \theta =-\frac{1}{2}\] \[\Rightarrow \] \[\theta ={{120}^{o}}\] In this condition angle between given vectors should be \[{{120}^{o}}\].