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)/(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:

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

### Step 1: Express the cross product and dot product
The magnitude of the cross product \(|\vec{A} \times \vec{B}|\) can be expressed as:

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

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

The dot product \(\vec{A} \cdot \vec{B}\) can be expressed as:

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

### Step 2: Substitute 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 common terms
Assuming \(|\vec{A}| \neq 0\) and \(|\vec{B}| \neq 0\), we can divide both sides by \(|\vec{A}| |\vec{B}|\):

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

### Step 4: Divide both sides by \(\cos \theta\)
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 \(|\vec{A} + \vec{B}|\)
The magnitude of the sum of two vectors can be calculated using the formula:

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

Substituting \(\theta = 60^\circ\) and \(\cos 60^\circ = \frac{1}{2}\):

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

### Step 7: 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:

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

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