If `|AxxB| = sqrt3 A.B,` then the value of |A+B| is

Options

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

Explanation

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

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

### Step 1: Understand the Cross Product and Dot Product
The magnitude of the cross product \( |A \times B| \) can be expressed as:

\[ |A \times B| = |A| |B| \sin \theta \]

where \( \theta \) is the angle between the vectors \( A \) and \( B \).

The dot product \( A \cdot B \) is given by:

\[ A \cdot B = |A| |B| \cos \theta \]

### Step 2: Substitute into the Given Equation
Substituting the expressions for the magnitudes into the given equation:

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

Assuming \( |A| \) and \( |B| \) are not zero, we can divide both sides by \( |A| |B| \):

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

### Step 3: Use the Tangent Function
Dividing both sides by \( \cos \theta \):

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

### Step 4: Determine the Angle
From trigonometric values, we know:

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

### Step 5: Calculate the Magnitude of \( |A + B| \)
Now, we can find the magnitude of the sum of the vectors \( A \) and \( B \) using the formula:

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

### Step 6: Substitute the Known Values
Substituting \( \cos \theta = \cos 60^\circ = \frac{1}{2} \):

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

This simplifies to:

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

### Final Step: Write the Final Expression
Thus, the final expression for \( |A + B| \) is:

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

### Conclusion
The correct answer is:

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