Question

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

Answer 1

Hint : Vector is defined as a quantity that has both magnitude and direction. Vector is generally represented by an arrow whose direction is the same as that of the quantity and also while the length is proportional to the quantity’s magnitude.

Complete Step by step solution:
The magnitude of resultant of two vectors $\overrightarrow x $ and $\overrightarrow y $with an angle $\theta $ is given by
\[\left| {\overrightarrow x + \overrightarrow y } \right| = \sqrt {{{\left| {\overrightarrow x } \right|}^2} + {{\left| {\overrightarrow y } \right|}^2} + 2\left| {\overrightarrow x } \right|\left| {\overrightarrow y } \right|\cos \theta } \]
Let \[\left| {\overrightarrow A } \right|\]and \[\left| {\overrightarrow B } \right|\]are the two vectors which are having equal magnitude and it is given that two vectors are equal in magnitude and hence \[\left| {\overrightarrow A } \right| = \left| {\overrightarrow B } \right|\]
Thus the magnitude of the resultant vectors of \[\left| {\overrightarrow A } \right|\]and \[\left| {\overrightarrow B } \right|\]will be equal to the magnitude of any of them
Hence \[\left| {\overrightarrow A + \overrightarrow B } \right| = \left| {\overrightarrow A } \right| = \left| {\overrightarrow B } \right|\]
Thus the magnitude of two vectors is given by \[\left| {\overrightarrow A + \overrightarrow B } \right| = \sqrt {{{\left| {\overrightarrow A } \right|}^2} + {{\left| {\overrightarrow B } \right|}^2} + 2\left| {\overrightarrow A } \right|\left| {\overrightarrow B } \right|\cos \theta } \_\_\_\_\_\left( 1 \right)\]
Where $\theta $ is the angle between the two vectors \[\left| {\overrightarrow A } \right|\]and \[\left| {\overrightarrow B } \right|\]
Substituting \[\left| {\overrightarrow A + \overrightarrow B } \right| = \left| {\overrightarrow A } \right|\]in equation 1, we get
\[\left| {\overrightarrow A } \right| = \sqrt {{{\left| {\overrightarrow A } \right|}^2} + {{\left| {\overrightarrow B } \right|}^2} + 2\left| {\overrightarrow A } \right|\left| {\overrightarrow B } \right|\cos \theta } \]
Squaring on both sides we get,
\[{\left| {\overrightarrow A } \right|^2} = {\left| {\overrightarrow A } \right|^2} + {\left| {\overrightarrow B } \right|^2} + 2\left| {\overrightarrow A } \right|\,\left| {\overrightarrow B } \right|\cos \theta \]
Substituting \[\left| {\overrightarrow A } \right| = \left| {\overrightarrow B } \right|\] and \[{\left| {\overrightarrow A } \right|^2} = {\left| {\overrightarrow B } \right|^2}\], we get
\[{\left| {\overrightarrow A } \right|^2} = {\left| {\overrightarrow A } \right|^2} + {\left| {\overrightarrow A } \right|^2} + 2\left| {\overrightarrow A } \right|\,\left| {\overrightarrow A } \right|\cos \theta \]



\[ \Rightarrow 2{\left| {\overrightarrow A } \right|^2}\cos \theta + {\left| {\overrightarrow A } \right|^2} = 0\]
\[ \Rightarrow 2\cos \theta = - 1\]
\[ \Rightarrow \cos \theta = - \dfrac{1}{2}\]
\[ \Rightarrow \theta = co{s^{ - 1}}\left( {\dfrac{{ - 1}}{2}} \right)\]
\[ \Rightarrow \theta = \dfrac{{2\pi }}{3} = 120^\circ \]

Thus the required angle is \[120^\circ \]

Note:
1. Magnitude is generally defined as the quantity or the distance concerning the movement. We also relate it to the size and speed of the object, when it is traveling.
2. Inorder to find the magnitude of the vector, we have to calculate the length of the vector. In general, this quantity is the length between the initial point and endpoint of the vector. Examples of vector quantities are displacement, velocity, force, momentum, etc.
3. The direction of a vector is the measurement of the angle which is made with the horizontal line.
4. Scalar quantities have only magnitude. Examples of scalar are speed, mass, temperature, distance, volume, etc.