If `|vecaxxvecb|=1/sqrt3|veca.vecb|` , find the angle between `vec a and vecb`

Answer

To solve the problem, we will use the relationship between the cross product and the dot product of two vectors.

Given:
\[ |\vec{a} \times \vec{b}| = \frac{1}{\sqrt{3}} |\vec{a} \cdot \vec{b}| \]

### Step 1: Write the expressions for the cross product and dot product

The magnitude of the cross product of two vectors \(\vec{a}\) and \(\vec{b}\) 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 of the two vectors 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:
\[ |\vec{a}| |\vec{b}| \sin \theta = \frac{1}{\sqrt{3}} |\vec{a}| |\vec{b}| \cos \theta \]

### Step 3: Cancel out the common terms

Assuming that neither \(|\vec{a}|\) nor \(|\vec{b}|\) is zero, we can divide both sides by \(|\vec{a}| |\vec{b}|\):
\[ \sin \theta = \frac{1}{\sqrt{3}} \cos \theta \]

### Step 4: Rearrange the equation

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

### Step 5: Use the tangent function

This simplifies to:
\[ \tan \theta = \frac{1}{\sqrt{3}} \]

### Step 6: Find the angle \(\theta\)

The angle \(\theta\) that satisfies \(\tan \theta = \frac{1}{\sqrt{3}}\) is:
\[ \theta = 30^\circ \]

### Conclusion

Thus, the angle between the vectors \(\vec{a}\) and \(\vec{b}\) is:
\[ \theta = 30^\circ \]