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

Options

A. `60^(@) `
B. `90^(@)`
C. `105^(@)`
D. `120^(@)`

Explanation

To solve the problem of finding the angle between two vectors of equal magnitude whose resultant is also equal in magnitude to either of the vectors, we can follow these steps:

### Step-by-Step Solution:

1. **Define the Vectors**:
Let the two vectors be \( \vec{A} \) and \( \vec{B} \). Given that both vectors are equal in magnitude, we can denote their magnitudes as:
\[ |\vec{A}| = |\vec{B}| = A \]

2. **Resultant Magnitude**:
According to the problem, the magnitude of the resultant vector \( \vec{R} \) is equal to the magnitude of either vector:
\[ |\vec{R}| = A \]

3. **Use the Formula for Resultant of Two Vectors**:
The magnitude of the resultant vector \( \vec{R} \) when adding two vectors \( \vec{A} \) and \( \vec{B} \) can be calculated using the formula:
\[ |\vec{R}| = \sqrt{|\vec{A}|^2 + |\vec{B}|^2 + 2 |\vec{A}| |\vec{B}| \cos \theta} \]
Substituting the magnitudes:
\[ |\vec{R}| = \sqrt{A^2 + A^2 + 2A \cdot A \cos \theta} \]

4. **Simplify the Equation**:
This simplifies to:
\[ |\vec{R}| = \sqrt{2A^2 + 2A^2 \cos \theta} \]
Since \( |\vec{R}| = A \), we can set the two expressions equal:
\[ A = \sqrt{2A^2 + 2A^2 \cos \theta} \]

5. **Square Both Sides**:
Squaring both sides gives:
\[ A^2 = 2A^2 + 2A^2 \cos \theta \]

6. **Rearranging the Equation**:
Rearranging the equation leads to:
\[ A^2 - 2A^2 = 2A^2 \cos \theta \]
\[ -A^2 = 2A^2 \cos \theta \]

7. **Solve for Cosine**:
Dividing both sides by \( 2A^2 \) gives:
\[ \cos \theta = -\frac{1}{2} \]

8. **Find the Angle**:
The angle \( \theta \) that corresponds to \( \cos \theta = -\frac{1}{2} \) is:
\[ \theta = 120^\circ \]

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

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