If vectors `vecA and vecB` are such that `|vecA+vecB|= |vecA|= |vecB|` then `|vecA-vecB|` may be equated to

Answer

To solve the problem, we need to analyze the given condition and apply the relevant vector formulas step by step.

### Step-by-Step Solution:

1. **Given Condition**: We have two vectors \(\vec{A}\) and \(\vec{B}\) such that \(|\vec{A} + \vec{B}| = |\vec{A}| = |\vec{B}|\). Let's denote the magnitude of both vectors as \(a\). Thus, we can write:
\[ |\vec{A}| = |\vec{B}| = a \]

2. **Magnitude of the Sum**: From the given condition, we can express the magnitude of the sum of the vectors:
\[ |\vec{A} + \vec{B}| = \sqrt{|\vec{A}|^2 + |\vec{B}|^2 + 2|\vec{A}||\vec{B}|\cos\theta} \]
where \(\theta\) is the angle between the vectors \(\vec{A}\) and \(\vec{B}\).

3. **Substituting Values**: Since \(|\vec{A}| = |\vec{B}| = a\), we substitute into the equation:
\[ |\vec{A} + \vec{B}| = \sqrt{a^2 + a^2 + 2a^2\cos\theta} = \sqrt{2a^2(1 + \cos\theta)} = a\sqrt{2(1 + \cos\theta)} \]

4. **Equating Magnitudes**: According to the problem, we have:
\[ |\vec{A} + \vec{B}| = a \]
Therefore, we can set the two expressions equal:
\[ a\sqrt{2(1 + \cos\theta)} = a \]

5. **Dividing by \(a\)**: Assuming \(a \neq 0\), we can divide both sides by \(a\):
\[ \sqrt{2(1 + \cos\theta)} = 1 \]

6. **Squaring Both Sides**: Squaring both sides gives:
\[ 2(1 + \cos\theta) = 1 \]
Simplifying this, we find:
\[ 1 + \cos\theta = \frac{1}{2} \]
\[ \cos\theta = -\frac{1}{2} \]

7. **Finding the Angle**: The angle \(\theta\) that satisfies \(\cos\theta = -\frac{1}{2}\) is:
\[ \theta = 120^\circ \quad \text{or} \quad \theta = 240^\circ \]

8. **Magnitude of the Difference**: Now we need to find \(|\vec{A} - \vec{B}|\):
\[ |\vec{A} - \vec{B}| = \sqrt{|\vec{A}|^2 + |\vec{B}|^2 - 2|\vec{A}||\vec{B}|\cos\theta} \]
Substituting the values:
\[ |\vec{A} - \vec{B}| = \sqrt{a^2 + a^2 - 2a^2\cos(120^\circ)} \]
Since \(\cos(120^\circ) = -\frac{1}{2}\):
\[ |\vec{A} - \vec{B}| = \sqrt{a^2 + a^2 + a^2} = \sqrt{3a^2} = a\sqrt{3} \]

9. **Final Result**: Therefore, the magnitude \(|\vec{A} - \vec{B}|\) is:
\[ |\vec{A} - \vec{B}| = a\sqrt{3} \]

### Conclusion:
The final answer is:
\[ |\vec{A} - \vec{B}| = \sqrt{3} |\vec{A}| = \sqrt{3} |\vec{B}| \]