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

Options

A. `(A^2+B^2 +AB)^(1//2)`
B. `(A^2 +B^2 + (AB)/(sqrt3))^(1//2)`
C. `(A+B)`
D. `(A^2 +B^2 +sqrt3 AB)^(1//2)`

Explanation

To solve the problem, we need to find the magnitude of the vector sum |A + B| given that |A × B| = √3 (A · B).

### Step-by-Step Solution:

1. **Understanding the Given Condition**:
We are given that the magnitude of the cross product of vectors A and B is equal to √3 times the dot product of A and B.
\[ |A \times B| = \sqrt{3} (A \cdot B) \]

2. **Using the Definitions of Cross Product and Dot Product**:
The magnitude of the cross product can be expressed as:
\[ |A \times B| = |A||B| \sin \theta \]
The dot product can be expressed as:
\[ A \cdot B = |A||B| \cos \theta \]
where θ is the angle between vectors A and B.

3. **Substituting the Definitions into the Given Condition**:
Substitute the expressions for the cross product and dot product into the given equation:
\[ |A||B| \sin \theta = \sqrt{3} |A||B| \cos \theta \]

4. **Canceling Common Terms**:
Assuming |A| and |B| are not zero, we can divide both sides by |A||B|:
\[ \sin \theta = \sqrt{3} \cos \theta \]

5. **Dividing Both Sides by cos θ**:
This gives us:
\[ \tan \theta = \sqrt{3} \]

6. **Finding the Angle θ**:
The angle θ that satisfies this equation is:
\[ \theta = 60^\circ \]

7. **Finding the Magnitude of A + B**:
We can now find the magnitude of the vector sum |A + B| using the formula:
\[ |A + B| = \sqrt{|A|^2 + |B|^2 + 2|A||B| \cos \theta} \]
Substituting θ = 60° (where cos 60° = 1/2):
\[ |A + B| = \sqrt{|A|^2 + |B|^2 + 2|A||B| \cdot \frac{1}{2}} \]
Simplifying this:
\[ |A + B| = \sqrt{|A|^2 + |B|^2 + |A||B|} \]

8. **Final Expression**:
Thus, the magnitude of the vector sum |A + B| is:
\[ |A + B| = \sqrt{|A|^2 + |B|^2 + |A||B|} \]

### Final Answer:
The value of |A + B| is \(\sqrt{|A|^2 + |B|^2 + |A||B|}\).