If `|vec(A)xxvec(B)|=sqrt(3)vec(A).vec(B)`, then the value of `|vec(A)+vec(B)|` is

Options

A. `(A^(2)+B^(2)+AB)^(1//2)`
B. `(A^(2)+B^(2)+(AB)/(sqrt(3)))^(1//2)`
C. `A+B`
D. `(A^(2)+B^(2)sqrt(3)+AB)^(1//2)`

Answer

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

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

### Step 1: Understand the relationship between cross product and dot product
The magnitude of the cross product of two vectors \(\vec{A}\) and \(\vec{B}\) is given by:

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

where \(A = |\vec{A}|\), \(B = |\vec{B}|\), and \(\theta\) is the angle between the two vectors.

The dot product of the vectors is given by:

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

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

\[ AB \sin \theta = \sqrt{3} (AB \cos \theta) \]

### Step 3: Simplify the equation
Assuming \(AB \neq 0\), we can divide both sides by \(AB\):

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

### Step 4: Divide both sides by \(\cos \theta\)
This gives us:

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

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

\[ \theta = 60^\circ \quad \text{or} \quad \theta = \frac{\pi}{3} \text{ radians} \]

### Step 6: Use the parallelogram law to find \(|\vec{A} + \vec{B}|\)
The magnitude of the sum of two vectors can be calculated using the parallelogram law:

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

Substituting \(\theta = 60^\circ\):

\[ |\vec{A} + \vec{B}| = \sqrt{A^2 + B^2 + 2AB \cos(60^\circ)} \]

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

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

This simplifies to:

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

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

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