If `sqrt3|vecAxx vecB|= vecA*vecB`, then find the angle between `vecA and vecB`.

Explanation

To solve the problem, we need to find the angle between vectors \(\vec{A}\) and \(\vec{B}\) given the equation:

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

### Step-by-Step Solution:

**Step 1: Understand the Cross Product and Dot Product**

The magnitude of the cross product \(|\vec{A} \times \vec{B}|\) is given by:

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

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

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

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

**Step 2: Substitute the Expressions into the Given Equation**

Substituting the expressions for the cross product and dot product into the given equation:

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

**Step 3: Simplify the Equation**

Assuming \(|\vec{A}| \neq 0\) and \(|\vec{B}| \neq 0\), we can divide both sides by \(|\vec{A}| |\vec{B}|\):

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

**Step 4: Rearrange the Equation**

Rearranging gives us:

\[ \frac{\sin \theta}{\cos \theta} = \frac{1}{\sqrt{3}} \]

This can be rewritten as:

\[ \tan \theta = \frac{1}{\sqrt{3}} \]

**Step 5: Find the Angle \(\theta\)**

The angle whose tangent is \(\frac{1}{\sqrt{3}}\) is:

\[ \theta = 30^\circ \]

### Final Answer:

The angle between vectors \(\vec{A}\) and \(\vec{B}\) is \(30^\circ\).

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