If `vec(a)` and `vec(b)` are unit vectors, then the angle between `vec(a)` and `vec(b)` for `sqrt( 3) vec( a) - vec(b)` to be a unit vector is

Options

A. `30^(@)`
B. `45^(@)`
C. `60^(@)`
D. `90^(@)`

Answer

To solve the problem, we need to find the angle between two unit vectors \( \vec{a} \) and \( \vec{b} \) such that the vector \( \sqrt{3} \vec{a} - \vec{b} \) is also a unit vector.

### Step-by-Step Solution:

1. **Understanding the Problem**:
We know that both \( \vec{a} \) and \( \vec{b} \) are unit vectors. This means that \( |\vec{a}| = 1 \) and \( |\vec{b}| = 1 \). We need to find the angle \( \theta \) between these two vectors.

2. **Setting Up the Equation**:
Since \( \sqrt{3} \vec{a} - \vec{b} \) is a unit vector, its magnitude must equal 1:
\[ |\sqrt{3} \vec{a} - \vec{b}| = 1 \]

3. **Squaring Both Sides**:
Squaring the magnitude gives us:
\[ |\sqrt{3} \vec{a} - \vec{b}|^2 = 1^2 \]
Expanding this, we have:
\[ (\sqrt{3} \vec{a} - \vec{b}) \cdot (\sqrt{3} \vec{a} - \vec{b}) = 1 \]

4. **Using the Dot Product**:
Expanding the dot product:
\[ (\sqrt{3} \vec{a}) \cdot (\sqrt{3} \vec{a}) - 2(\sqrt{3} \vec{a}) \cdot \vec{b} + \vec{b} \cdot \vec{b} = 1 \]
Since \( |\vec{a}|^2 = 1 \) and \( |\vec{b}|^2 = 1 \):
\[ 3 - 2\sqrt{3} (\vec{a} \cdot \vec{b}) + 1 = 1 \]

5. **Simplifying the Equation**:
This simplifies to:
\[ 4 - 2\sqrt{3} (\vec{a} \cdot \vec{b}) = 1 \]
Rearranging gives:
\[ 2\sqrt{3} (\vec{a} \cdot \vec{b}) = 3 \]
Thus:
\[ \vec{a} \cdot \vec{b} = \frac{3}{2\sqrt{3}} = \frac{\sqrt{3}}{2} \]

6. **Finding the Angle**:
The dot product of two vectors is also given by:
\[ \vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta \]
Since both are unit vectors:
\[ \vec{a} \cdot \vec{b} = \cos \theta \]
Therefore:
\[ \cos \theta = \frac{\sqrt{3}}{2} \]

7. **Determining the Angle**:
The angle \( \theta \) that corresponds to \( \cos \theta = \frac{\sqrt{3}}{2} \) is:
\[ \theta = 30^\circ \]

### Final Answer:
The angle between \( \vec{a} \) and \( \vec{b} \) is \( 30^\circ \).