If `|vecA xx vecB| = sqrt3 vecA *vecB`, then the value of `|vecA + vecB|` is :

Explanation

To solve the problem, we start with the given equation:

\[ |\vec{A} \times \vec{B}| = \sqrt{3} \, \vec{A} \cdot \vec{B} \]

### Step 1: Understand the relationship between cross product and dot product
The magnitude of the cross product can be expressed as:

\[ |\vec{A} \times \vec{B}| = |\vec{A}| |\vec{B}| \sin \theta \]

And the dot product can be expressed as:

\[ \vec{A} \cdot \vec{B} = |\vec{A}| |\vec{B}| \cos \theta \]

### Step 2: Substitute the expressions into the equation
Substituting these expressions into the given equation, we have:

\[ |\vec{A}| |\vec{B}| \sin \theta = \sqrt{3} \, |\vec{A}| |\vec{B}| \cos \theta \]

### Step 3: Cancel out the common terms
Assuming \( |\vec{A}| \) and \( |\vec{B}| \) are not zero, we can divide both sides by \( |\vec{A}| |\vec{B}| \):

\[ \sin \theta = \sqrt{3} \cos \theta \]

### Step 4: Express in terms of tangent
Dividing both sides by \( \cos \theta \) gives:

\[ \tan \theta = \sqrt{3} \]

### Step 5: Find the angle \( \theta \)
The angle \( \theta \) that satisfies \( \tan \theta = \sqrt{3} \) is:

\[ \theta = 60^\circ \]

### Step 6: Calculate the magnitude of \( |\vec{A} + \vec{B}| \)
Now, we can find the magnitude of the sum of the vectors:

\[ |\vec{A} + \vec{B}| = \sqrt{|\vec{A}|^2 + |\vec{B}|^2 + 2 |\vec{A}| |\vec{B}| \cos \theta} \]

### Step 7: Substitute \( \cos 60^\circ \)
Since \( \cos 60^\circ = \frac{1}{2} \):

\[ |\vec{A} + \vec{B}| = \sqrt{|\vec{A}|^2 + |\vec{B}|^2 + 2 |\vec{A}| |\vec{B}| \cdot \frac{1}{2}} \]

### Step 8: Simplify the expression
This simplifies to:

\[ |\vec{A} + \vec{B}| = \sqrt{|\vec{A}|^2 + |\vec{B}|^2 + |\vec{A}| |\vec{B}|} \]

### Final Result
Thus, the value of \( |\vec{A} + \vec{B}| \) is:

\[ |\vec{A} + \vec{B}| = \sqrt{|\vec{A}|^2 + |\vec{B}|^2 + |\vec{A}| |\vec{B}|} \]