If a and b are two unit vectors inclined at an angle `2theta` to each other, then `|a + b| lt 1` if

Options

A. `(pi)/(3) lt thetalt (2pi)/(3)`
B. `theta lt (pi)/(3)`
C. `theta lt (2pi)/(3)`
D. `theta = (pi)/(2)`

Answer

To solve the problem, we need to determine the conditions under which the magnitude of the sum of two unit vectors \( \mathbf{a} \) and \( \mathbf{b} \), which are inclined at an angle \( 2\theta \) to each other, is less than 1.

### Step-by-Step Solution:

1. **Understand the Magnitude of the Sum of Vectors**:
Since \( \mathbf{a} \) and \( \mathbf{b} \) are unit vectors, we have:
\[ |\mathbf{a}| = 1 \quad \text{and} \quad |\mathbf{b}| = 1 \]
The magnitude of the sum of two vectors can be expressed using the formula:
\[ |\mathbf{a} + \mathbf{b}|^2 = |\mathbf{a}|^2 + |\mathbf{b}|^2 + 2(\mathbf{a} \cdot \mathbf{b}) \]

2. **Substituting the Magnitudes**:
Since both vectors are unit vectors, we substitute \( |\mathbf{a}|^2 = 1 \) and \( |\mathbf{b}|^2 = 1 \):
\[ |\mathbf{a} + \mathbf{b}|^2 = 1 + 1 + 2(\mathbf{a} \cdot \mathbf{b}) = 2 + 2(\mathbf{a} \cdot \mathbf{b}) \]

3. **Expressing the Dot Product**:
The dot product \( \mathbf{a} \cdot \mathbf{b} \) can be expressed in terms of the angle \( 2\theta \):
\[ \mathbf{a} \cdot \mathbf{b} = |\mathbf{a}| |\mathbf{b}| \cos(2\theta) = 1 \cdot 1 \cdot \cos(2\theta) = \cos(2\theta) \]

4. **Substituting the Dot Product**:
Substitute \( \mathbf{a} \cdot \mathbf{b} \) back into the equation for the magnitude:
\[ |\mathbf{a} + \mathbf{b}|^2 = 2 + 2\cos(2\theta) \]

5. **Setting Up the Inequality**:
We want to find when \( |\mathbf{a} + \mathbf{b}| < 1 \):
\[ |\mathbf{a} + \mathbf{b}|^2 < 1^2 \]
Thus, we have:
\[ 2 + 2\cos(2\theta) < 1 \]

6. **Simplifying the Inequality**:
Rearranging gives:
\[ 2\cos(2\theta) < 1 - 2 \]
\[ 2\cos(2\theta) < -1 \]
Dividing both sides by 2:
\[ \cos(2\theta) < -\frac{1}{2} \]

7. **Finding the Angles**:
The cosine function is less than \(-\frac{1}{2}\) in the intervals:
\[ 2\theta \in \left( \frac{2\pi}{3}, \frac{4\pi}{3} \right) \]
Dividing by 2 gives:
\[ \theta \in \left( \frac{\pi}{3}, \frac{2\pi}{3} \right) \]

### Final Result:
Thus, the condition for \( |\mathbf{a} + \mathbf{b}| < 1 \) is:
\[ \theta \in \left( \frac{\pi}{3}, \frac{2\pi}{3} \right) \]