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 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 the magnitude of the cross product of two vectors can be expressed as:

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

And the dot product is given by:

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

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

Substituting these definitions into the given equation, we have:

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

### Step 3: Cancel out the common terms

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 us:

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

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

From trigonometry, we know:

\[ \tan 60^\circ = \sqrt{3} \]

Thus, we conclude:

\[ \theta = 60^\circ \]

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

The magnitude of the resultant vector \( \vec{A} + \vec{B} \) can be calculated using the formula:

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

### Step 7: Substitute \( \theta = 60^\circ \)

We know that \( \cos 60^\circ = \frac{1}{2} \). Substituting this into our equation gives:

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

### 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:

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