If a and b are unit vectors, then what is the angle between a and b for `sqrt3a-b` to be a unit vector?

Options

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

Explanation

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

### Step-by-step Solution:

1. **Understanding the Conditions**:
Since \( \mathbf{a} \) and \( \mathbf{b} \) are unit vectors, we have:
\[ |\mathbf{a}| = 1 \quad \text{and} \quad |\mathbf{b}| = 1 \]
We also know that \( |\sqrt{3}\mathbf{a} - \mathbf{b}| = 1 \).

2. **Setting Up the Magnitude Equation**:
We can express the magnitude of \( \sqrt{3}\mathbf{a} - \mathbf{b} \) as follows:
\[ |\sqrt{3}\mathbf{a} - \mathbf{b}|^2 = 1^2 = 1 \]

3. **Expanding the Magnitude**:
Using the properties of magnitudes, we expand the left side:
\[ |\sqrt{3}\mathbf{a} - \mathbf{b}|^2 = (\sqrt{3}\mathbf{a} - \mathbf{b}) \cdot (\sqrt{3}\mathbf{a} - \mathbf{b}) \]
This expands to:
\[ (\sqrt{3}\mathbf{a}) \cdot (\sqrt{3}\mathbf{a}) - 2(\sqrt{3}\mathbf{a}) \cdot \mathbf{b} + (\mathbf{b}) \cdot (\mathbf{b}) \]
Simplifying this, we get:
\[ 3|\mathbf{a}|^2 - 2\sqrt{3}(\mathbf{a} \cdot \mathbf{b}) + |\mathbf{b}|^2 \]

4. **Substituting the Magnitudes**:
Since both \( \mathbf{a} \) and \( \mathbf{b} \) are unit vectors:
\[ 3(1) - 2\sqrt{3}(\mathbf{a} \cdot \mathbf{b}) + 1 = 1 \]
This simplifies to:
\[ 3 - 2\sqrt{3}(\mathbf{a} \cdot \mathbf{b}) + 1 = 1 \]
Therefore:
\[ 4 - 2\sqrt{3}(\mathbf{a} \cdot \mathbf{b}) = 1 \]

5. **Rearranging the Equation**:
Rearranging gives us:
\[ 2\sqrt{3}(\mathbf{a} \cdot \mathbf{b}) = 4 - 1 \]
\[ 2\sqrt{3}(\mathbf{a} \cdot \mathbf{b}) = 3 \]
Dividing both sides by \( 2\sqrt{3} \):
\[ \mathbf{a} \cdot \mathbf{b} = \frac{3}{2\sqrt{3}} = \frac{\sqrt{3}}{2} \]

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

7. **Calculating the Angle**:
The angle \( \theta \) for which \( \cos\theta = \frac{\sqrt{3}}{2} \) is:
\[ \theta = \cos^{-1}\left(\frac{\sqrt{3}}{2}\right) = \frac{\pi}{6} \text{ radians} \quad \text{or} \quad 30^\circ \]

### Final Answer:
The angle between the unit vectors \( \mathbf{a} \) and \( \mathbf{b} \) is \( \frac{\pi}{6} \) radians or \( 30^\circ \).