Find the angle between two vectors if their resultant is equal to either of them.

Hint: Here, we will assume two different variables for two different vectors. We will then find the resultant of two vectors by using the resultant formula and by using the given condition. Then we will use the trigonometric identity to find the angle between two vectors.

Formula Used:
Resultant of two vectors is given by the formula \[R = \sqrt {{a^2} + {b^2} + 2ab\cos \theta } \] where \[a,b\] are the two vectors and \[\theta \] be the angle between two vectors.

Complete step-by-step answer:
We are given that their resultant is equal to either of them.
Let \[P,Q\] be the two vectors.
We are given that their resultant is equal to either of them so that the two vectors are equal.
We know that the resultant of two vectors is equal to either of them if and only if the two vectors are of same magnitude i.e., \[\left| P \right| = \left| Q \right|\] , so \[P = Q\]
Let \[R\] be the resultant of two vectors.
So, we get \[R = \left| P \right| = \left| Q \right|\]
Resultant of two vectors is given by the formula \[R = \sqrt {{a^2} + {b^2} + 2ab\cos \theta } \] where \[a,b\] are the two vectors and \[\theta \] be the angle between two vectors.
Let \[\theta \] be the angle between two vectors.
Now, by using the Resultant of two vectors formula, we get
\[ \Rightarrow R = \sqrt {{P^2} + {Q^2} + 2PQ\cos \theta } \]
Since \[P = Q\] , we get
\[ \Rightarrow R = \sqrt {{P^2} + {P^2} + 2PP\cos \theta } \]
By simplifying, we get
\[ \Rightarrow R = \sqrt {2{P^2} + 2{P^2}\cos \theta } \]
Since \[R = \left| P \right| = \left| Q \right|\] , we get
\[ \Rightarrow \left| P \right| = \sqrt {2{P^2} + 2{P^2}\cos \theta } \]
By squaring on both the sides, we get
\[ \Rightarrow {P^2} = 2{P^2} + 2{P^2}\cos \theta \]
By rewriting the equation, we get
\[ \Rightarrow 2{P^2}\cos \theta = - 2{P^2} + {P^2}\]
\[ \Rightarrow 2{P^2}\cos \theta = - {P^2}\]
By cancelling out the common terms, we get
\[ \Rightarrow 2\cos \theta = - 1\]
By rewriting the equation, we get
\[ \Rightarrow \cos \theta = \dfrac{{ - 1}}{2}\]
\[ \Rightarrow \theta = {\cos ^{ - 1}}\left( {\dfrac{{ - 1}}{2}} \right)\]
\[ \Rightarrow \theta = 120^\circ \]
Therefore, the angle between two vectors is \[120^\circ \] if their resultant is equal to either of them.

Note: We know that a vector is an object which has both a magnitude and a direction. We should know that the resultant of two vectors is equal to either of them if and only if the two vectors are of same magnitude and the vectors are inclined to each other at an angle.
\[ \Rightarrow {P^2} = 2{P^2}\left( {1 + \cos \theta } \right)\]
By rewriting the equation, we get
\[ \Rightarrow \left( {1 + \cos \theta } \right) = \dfrac{{{P^2}}}{{2{P^2}}}\]
By using the trigonometric formula \[1 + \cos \theta = 2{\cos ^2}\dfrac{\theta }{2}\]
\[ \Rightarrow 2{\cos ^2}\dfrac{\theta }{2} = \dfrac{1}{2}\]
By rewriting the equation, we get
\[ \Rightarrow {\cos ^2}\dfrac{\theta }{2} = \dfrac{1}{4}\]
By taking square root on both the sides, we get
\[ \Rightarrow \cos \dfrac{\theta }{2} = \dfrac{1}{2}\]
Now taking inverse, we get
\[ \Rightarrow \dfrac{\theta }{2} = {\cos ^{ - 1}}\dfrac{1}{2}\]
\[ \Rightarrow \dfrac{\theta }{2} = 60^\circ \]
Multiplying both sides by 2, we get
\[ \Rightarrow \theta = 2 \times 60^\circ \]
\[ \Rightarrow \theta = 120^\circ \]