Two vectors of equal magnitude are acting through a point. The magnitude of resultant is equal to the magnitude of either vectoe. Find the angle between the vectors.

Options

A. 120°
B. 90°
C. 60°
D. 0°

Answer

To solve the problem of finding the angle between two vectors of equal magnitude when the resultant vector's magnitude is equal to the magnitude of either vector, we can follow these steps:

### Step-by-Step Solution:

1. **Define the Vectors**:
Let the two vectors be \( \vec{A} \) and \( \vec{B} \) with equal magnitudes. Let the magnitude of each vector be \( a \). Therefore, we have:
\[ |\vec{A}| = |\vec{B}| = a \]

2. **Resultant Vector Magnitude**:
The magnitude of the resultant vector \( \vec{R} \) when two vectors are acting at an angle \( \theta \) is given by the formula:
\[ R = \sqrt{A^2 + B^2 + 2AB \cos \theta} \]
Since \( A = B = a \), we can substitute:
\[ R = \sqrt{a^2 + a^2 + 2a^2 \cos \theta} \]
This simplifies to:
\[ R = \sqrt{2a^2(1 + \cos \theta)} \]

3. **Given Condition**:
According to the problem, the magnitude of the resultant \( R \) is equal to the magnitude of either vector \( a \):
\[ R = a \]
Therefore, we can equate the two expressions:
\[ a = \sqrt{2a^2(1 + \cos \theta)} \]

4. **Square Both Sides**:
To eliminate the square root, we square both sides:
\[ a^2 = 2a^2(1 + \cos \theta) \]

5. **Simplify the Equation**:
Dividing both sides by \( a^2 \) (assuming \( a \neq 0 \)):
\[ 1 = 2(1 + \cos \theta) \]
This simplifies to:
\[ 1 = 2 + 2\cos \theta \]

6. **Rearranging the Equation**:
Rearranging gives:
\[ 2\cos \theta = 1 - 2 \]
\[ 2\cos \theta = -1 \]
\[ \cos \theta = -\frac{1}{2} \]

7. **Finding the Angle**:
The angle \( \theta \) for which \( \cos \theta = -\frac{1}{2} \) is:
\[ \theta = 120^\circ \quad \text{(or equivalently, } 240^\circ \text{, but we consider the acute angle)} \]

### Final Answer:
The angle between the two vectors is \( \theta = 120^\circ \).