What should be the angle between `( vec(A) + vec(B)) and ( vec(A) - vec(B))` such that the magnitude of the resultant is `sqrt(3A^(2)+B^(2))` ?

Explanation

To find the angle between the vectors \( \vec{A} + \vec{B} \) and \( \vec{A} - \vec{B} \) such that the magnitude of the resultant is \( \sqrt{3A^2 + B^2} \), we can follow these steps:

### Step 1: Define the vectors
Let:
- \( \vec{X} = \vec{A} + \vec{B} \)
- \( \vec{Y} = \vec{A} - \vec{B} \)

### Step 2: Find the magnitudes of \( \vec{X} \) and \( \vec{Y} \)
The magnitudes of these vectors can be calculated as follows:
- \( |\vec{X}| = |\vec{A} + \vec{B}| = \sqrt{A^2 + B^2 + 2AB \cos \phi} \)
- \( |\vec{Y}| = |\vec{A} - \vec{B}| = \sqrt{A^2 + B^2 - 2AB \cos \phi} \)

Where \( \phi \) is the angle between vectors \( \vec{A} \) and \( \vec{B} \).

### Step 3: Use the resultant magnitude formula
The magnitude of the resultant vector \( \vec{R} \) formed by \( \vec{X} \) and \( \vec{Y} \) is given by:
\[ |\vec{R}| = \sqrt{|\vec{X}|^2 + |\vec{Y}|^2 + 2 |\vec{X}| |\vec{Y}| \cos \theta} \]
where \( \theta \) is the angle between \( \vec{X} \) and \( \vec{Y} \).

### Step 4: Square both sides
We know that the magnitude of the resultant is given as \( \sqrt{3A^2 + B^2} \). Squaring both sides gives:
\[ |\vec{R}|^2 = 3A^2 + B^2 \]

### Step 5: Substitute \( |\vec{X}|^2 \) and \( |\vec{Y}|^2 \)
Now substituting \( |\vec{X}|^2 \) and \( |\vec{Y}|^2 \):
\[ |\vec{X}|^2 = A^2 + B^2 + 2AB \cos \phi \]
\[ |\vec{Y}|^2 = A^2 + B^2 - 2AB \cos \phi \]
Adding these gives:
\[ |\vec{X}|^2 + |\vec{Y}|^2 = 2(A^2 + B^2) \]

### Step 6: Substitute into the resultant equation
Now substituting into the resultant equation:
\[ 2(A^2 + B^2) + 2 |\vec{X}| |\vec{Y}| \cos \theta = 3A^2 + B^2 \]

### Step 7: Simplifying
Rearranging gives:
\[ 2 |\vec{X}| |\vec{Y}| \cos \theta = 3A^2 + B^2 - 2(A^2 + B^2) \]
\[ 2 |\vec{X}| |\vec{Y}| \cos \theta = A^2 - B^2 \]

### Step 8: Calculate \( |\vec{X}| |\vec{Y}| \)
Now, we need to calculate \( |\vec{X}| |\vec{Y}| \):
\[ |\vec{X}| |\vec{Y}| = \sqrt{(A^2 + B^2 + 2AB \cos \phi)(A^2 + B^2 - 2AB \cos \phi)} = \sqrt{(A^2 + B^2)^2 - (2AB \cos \phi)^2} \]
This simplifies to:
\[ |\vec{X}| |\vec{Y}| = \sqrt{(A^2 + B^2)^2 - 4A^2B^2 \cos^2 \phi} \]

### Step 9: Substitute back into the equation
Substituting back gives:
\[ 2 \sqrt{(A^2 + B^2)^2 - 4A^2B^2 \cos^2 \phi} \cos \theta = A^2 - B^2 \]

### Step 10: Solve for \( \cos \theta \)
Now, we can isolate \( \cos \theta \):
\[ \cos \theta = \frac{A^2 - B^2}{2 \sqrt{(A^2 + B^2)^2 - 4A^2B^2 \cos^2 \phi}} \]

### Step 11: Find the angle \( \theta \)
Given that \( \cos \theta = \frac{1}{2} \), we find:
\[ \theta = \cos^{-1}\left(\frac{1}{2}\right) = 60^\circ \]

Thus, the angle between \( \vec{A} + \vec{B} \) and \( \vec{A} - \vec{B} \) is \( 60^\circ \).