If `|vecAxxvecB|=sqrt(3) vecA.vecB`, then the value of `|vecA+vecB|` is

Options

A. a `(A^2+B^2 + "AB"/sqrt3)^(1//2)`
B. b `A+B`
C. c `(A^2+B^2+sqrt3AB)^(1//2)`
D. d `(A^2+B^2+AB)^(1//2)`

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: Use the definitions of cross product and dot product

We know that:

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

### Step 2: Substitute these definitions into the equation

Substituting these into the given equation, we have:

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

### Step 3: Simplify the equation

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: Divide both sides by \( \cos \theta \)

This gives:

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

### Step 5: Determine the angle \( \theta \)

From the tangent function, we know:

\[ \theta = 60^\circ \]

### Step 6: Find the magnitude of \( |\vec{A} + \vec{B}| \)

Using the formula for the magnitude of the sum of two vectors:

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

### Step 7: Substitute \( \cos 60^\circ \)

We know that \( \cos 60^\circ = \frac{1}{2} \). Therefore, we can substitute this into the equation:

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

### 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:

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