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)         \[~150{}^\circ \]

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 vector\[(\vec{A}=\vec{B})\] 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 it 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 \[\vec{A}\]and \[\vec{B}.\] When\[\vec{R}=\vec{A}=B,\] 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 \[{{120}^{o}}.\]