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 \[{{120}^{o}}.\]You need to login to perform this action.
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