Given that A=B. What is the angle between `(vecA+vecB)` and `(vecA-vecB)` ?

Explanation

To find the angle between the vectors \(\vec{A} + \vec{B}\) and \(\vec{A} - \vec{B}\) given that \(\vec{A} = \vec{B}\), we can follow these steps:

### Step 1: Use the formula for the angle between two vectors
The angle \(\theta\) between two vectors \(\vec{P}\) and \(\vec{Q}\) can be calculated using the formula:
\[ \theta = \cos^{-1}\left(\frac{\vec{P} \cdot \vec{Q}}{|\vec{P}| |\vec{Q}|}\right) \]
In our case, we need to find the angle between \(\vec{A} + \vec{B}\) and \(\vec{A} - \vec{B}\).

### Step 2: Identify the vectors
Let:
\[ \vec{P} = \vec{A} + \vec{B} \]
\[ \vec{Q} = \vec{A} - \vec{B} \]

### Step 3: Calculate the dot product \(\vec{P} \cdot \vec{Q}\)
Using the distributive property of the dot product:
\[ \vec{P} \cdot \vec{Q} = (\vec{A} + \vec{B}) \cdot (\vec{A} - \vec{B}) \]
Expanding this:
\[ = \vec{A} \cdot \vec{A} - \vec{A} \cdot \vec{B} + \vec{B} \cdot \vec{A} - \vec{B} \cdot \vec{B} \]
Since \(\vec{A} \cdot \vec{B} = \vec{B} \cdot \vec{A}\), we can simplify:
\[ = |\vec{A}|^2 - |\vec{B}|^2 \]

### Step 4: Substitute the magnitudes
Given that \(|\vec{A}| = |\vec{B}|\), we have:
\[ |\vec{A}|^2 - |\vec{B}|^2 = 0 \]
Thus, the dot product becomes:
\[ \vec{P} \cdot \vec{Q} = 0 \]

### Step 5: Calculate the magnitudes of \(\vec{P}\) and \(\vec{Q}\)
Now we need to find the magnitudes:
\[ |\vec{P}| = |\vec{A} + \vec{B}| = |\vec{A}| + |\vec{B}| \quad (\text{since } \vec{A} = \vec{B}) \]
\[ |\vec{Q}| = |\vec{A} - \vec{B}| = |\vec{A}| - |\vec{B}| = 0 \quad (\text{since } \vec{A} = \vec{B}) \]

### Step 6: Substitute into the angle formula
Now substituting into the angle formula:
\[ \theta = \cos^{-1}\left(\frac{0}{|\vec{P}| \cdot |\vec{Q}|}\right) \]
Since \(|\vec{Q}| = 0\), we have an undefined situation. However, we can deduce that the angle is \(90^\circ\) or \(\frac{\pi}{2}\) radians because the dot product being zero indicates orthogonality.

### Conclusion
Thus, the angle between \(\vec{A} + \vec{B}\) and \(\vec{A} - \vec{B}\) is:
\[ \theta = 90^\circ \quad \text{or} \quad \frac{\pi}{2} \text{ radians} \]