Question

Two vectors both are equal in magnitude and have their resultant equal in magnitude of the either. Find the angle between the two vectors.

Answer 1

Hint: The two vectors are equal in magnitude i.e. if the two vectors are $\overrightarrow{A}\text{ and }\overrightarrow{B}$then $\left| \overrightarrow{A} \right|=\left| \overrightarrow{B} \right|$. The resultant of $\overrightarrow{A}\text{ and }\overrightarrow{B}$ is equal in magnitude with them i.e. $\left| \overrightarrow{A}+\overrightarrow{B} \right|=\left| \overrightarrow{A} \right|=\left| \overrightarrow{B} \right|$. Find resultant of $\overrightarrow{A}\text{ and }\overrightarrow{B}$ and then equate the magnitude of resultant with either $\overrightarrow{A}\text{ or }\overrightarrow{B}$. From that value calculate the angle between them.

Formula used: The magnitude of resultant of two vectors \[\overrightarrow{x}\text{ and }\overrightarrow{y}\]with angle $\theta$ in between them is given by \[\left| \overrightarrow{x}+\overrightarrow{y} \right|=\sqrt{{{\left| \overrightarrow{x} \right|}^{2}}+{{\left| \overrightarrow{y} \right|}^{2}}+2\left| \overrightarrow{x} \right|\left| \overrightarrow{y} \right|\cos \theta }\]

Complete step-by-step answer:
$\overrightarrow{A}\text{ and }\overrightarrow{B}$ have equal magnitude so $\left| \overrightarrow{A} \right|=\left| \overrightarrow{B} \right|$. The magnitude of resultant of $\overrightarrow{A}\text{ and }\overrightarrow{B}$ is equal to the magnitude of either of them i.e. $\left| \overrightarrow{A}+\overrightarrow{B} \right|=\left| \overrightarrow{A} \right|=\left| \overrightarrow{B} \right|$.
The magnitude of resultant of $\overrightarrow{A}\text{ and }\overrightarrow{B}$ is given by \[\left| \overrightarrow{A}+\overrightarrow{B} \right|=\sqrt{{{\left| \overrightarrow{A} \right|}^{2}}+{{\left| \overrightarrow{B} \right|}^{2}}+2\left| \overrightarrow{A} \right|\left| \overrightarrow{B} \right|\cos \theta }\]
Where $\theta =\text{ angle between }\overrightarrow{A}\text{ and }\overrightarrow{B}$.
This magnitude is equal to the magnitude of the either of them i.e $\left| \overrightarrow{A} \right|=\left| \overrightarrow{A}+\overrightarrow{B} \right|$
So $\left| \overrightarrow{A} \right|=\sqrt{{{\left| \overrightarrow{A} \right|}^{2}}+{{\left| \overrightarrow{B} \right|}^{2}}+2\left| \overrightarrow{A} \right|\left| \overrightarrow{B} \right|\cos \theta }$
Taking square in both the sides
${{\left| \overrightarrow{A} \right|}^{2}}={{\left| \overrightarrow{A} \right|}^{2}}+{{\left| \overrightarrow{B} \right|}^{2}}+2\left| \overrightarrow{A} \right|\left| \overrightarrow{B} \right|\cos \theta $
Put $\left| \overrightarrow{A} \right|=\left| \overrightarrow{B} \right|$ and ${{\left| \overrightarrow{A} \right|}^{2}}={{\left| \overrightarrow{B} \right|}^{2}}$
So
$\begin{align}
  & {{\left| \overrightarrow{A} \right|}^{2}}={{\left| \overrightarrow{A} \right|}^{2}}+{{\left| \overrightarrow{A} \right|}^{2}}+2\left| \overrightarrow{A} \right|\left| \overrightarrow{A} \right|\cos \theta \\
 & \Rightarrow 2{{\left| \overrightarrow{A} \right|}^{2}}\cos \theta +{{\left| \overrightarrow{A} \right|}^{2}}=0 \\
 & \Rightarrow 2\cos \theta =-1 \\
 & \Rightarrow \cos \theta =\dfrac{-1}{2} \\
 & \Rightarrow \theta ={{\cos }^{-1}}\left( \dfrac{-1}{2} \right)=\dfrac{2\pi }{3}\text{ or 120}{}^\circ \\
\end{align}$
So the required angle between $\overrightarrow{A}\text{ and }\overrightarrow{B}$ is $\dfrac{2\pi }{3}$ or $\text{120}{}^\circ $.

Note: During addition of vectors both the magnitude and direction should be taken into consideration. If the direction is not taken correctly then then there would be errors in calculation. Also the angle between the vectors also plays an important role.
If $\theta =90{}^\circ $then $\left| \overrightarrow{A}+\overrightarrow{B} \right|=\sqrt{{{\left| \overrightarrow{A}\, \right|}^{2}}+{{\left| \overrightarrow{B} \right|}^{2}}}$