`vec(A) and vec(B)` are two vectors having the same magnitude . What is the condition that (i) the vector sum `vec(A)+vec(B)` and (ii) the vector differece `vec(A) + vec(B)` have the same magnitude as `vec(A) and vec(B)` ?

Explanation

To solve the problem, we need to find the conditions under which the vector sum and vector difference of two vectors \( \vec{A} \) and \( \vec{B} \) (which have the same magnitude) have the same magnitude as \( \vec{A} \) and \( \vec{B} \).

Let’s denote the magnitude of both vectors as \( |\vec{A}| = |\vec{B}| = x \).

### Part (i): Condition for \( |\vec{A} + \vec{B}| = |\vec{A}| \)

1. **Express the magnitude of the vector sum**:
\[ |\vec{A} + \vec{B}| = \sqrt{|\vec{A}|^2 + |\vec{B}|^2 + 2 |\vec{A}| |\vec{B}| \cos \theta} \]
where \( \theta \) is the angle between \( \vec{A} \) and \( \vec{B} \).

2. **Substitute the magnitudes**:
\[ |\vec{A} + \vec{B}| = \sqrt{x^2 + x^2 + 2x^2 \cos \theta} = \sqrt{2x^2(1 + \cos \theta)} \]
This simplifies to:
\[ |\vec{A} + \vec{B}| = x \sqrt{2(1 + \cos \theta)} \]

3. **Set the condition**:
We want \( |\vec{A} + \vec{B}| = |\vec{A}| \), which gives:
\[ x \sqrt{2(1 + \cos \theta)} = x \]

4. **Divide both sides by \( x \) (assuming \( x \neq 0 \))**:
\[ \sqrt{2(1 + \cos \theta)} = 1 \]

5. **Square both sides**:
\[ 2(1 + \cos \theta) = 1 \]

6. **Solve for \( \cos \theta \)**:
\[ 1 + \cos \theta = \frac{1}{2} \implies \cos \theta = -\frac{1}{2} \]

7. **Find \( \theta \)**:
\[ \theta = 120^\circ \]

### Part (ii): Condition for \( |\vec{A} - \vec{B}| = |\vec{A}| \)

1. **Express the magnitude of the vector difference**:
\[ |\vec{A} - \vec{B}| = \sqrt{|\vec{A}|^2 + |\vec{B}|^2 - 2 |\vec{A}| |\vec{B}| \cos \theta} \]

2. **Substitute the magnitudes**:
\[ |\vec{A} - \vec{B}| = \sqrt{x^2 + x^2 - 2x^2 \cos \theta} = \sqrt{2x^2(1 - \cos \theta)} \]
This simplifies to:
\[ |\vec{A} - \vec{B}| = x \sqrt{2(1 - \cos \theta)} \]

3. **Set the condition**:
We want \( |\vec{A} - \vec{B}| = |\vec{A}| \), which gives:
\[ x \sqrt{2(1 - \cos \theta)} = x \]

4. **Divide both sides by \( x \)**:
\[ \sqrt{2(1 - \cos \theta)} = 1 \]

5. **Square both sides**:
\[ 2(1 - \cos \theta) = 1 \]

6. **Solve for \( \cos \theta \)**:
\[ 1 - \cos \theta = \frac{1}{2} \implies \cos \theta = \frac{1}{2} \]

7. **Find \( \theta \)**:
\[ \theta = 60^\circ \]

### Final Conditions:
- For the vector sum \( |\vec{A} + \vec{B}| = |\vec{A}| \), the condition is \( \theta = 120^\circ \).
- For the vector difference \( |\vec{A} - \vec{B}| = |\vec{A}| \), the condition is \( \theta = 60^\circ \).