Let `vec(a) and vec(b)` be two unit vectors and α be the angle between them. If `(vec(a) + vec(b))` is also the unit vectors, then what is the value of α ?

Options

A. `(pi)/(4)`
B. `(pi)/(3)`
C. `(2pi)/(3)`
D. `(pi)/(2)`

Answer

To solve the problem, we need to find the angle \( \alpha \) between two unit vectors \( \vec{a} \) and \( \vec{b} \) given that \( \vec{a} + \vec{b} \) is also a unit vector. Let's break down the solution step by step.

### Step 1: Understand the Given Information
We are given:
- \( \vec{a} \) and \( \vec{b} \) are unit vectors.
- The angle between \( \vec{a} \) and \( \vec{b} \) is \( \alpha \).
- \( \vec{a} + \vec{b} \) is also a unit vector.

### Step 2: Write the Magnitude Condition
Since \( \vec{a} \) and \( \vec{b} \) are unit vectors, we have:
\[ |\vec{a}| = 1 \quad \text{and} \quad |\vec{b}| = 1 \]
We also know that:
\[ |\vec{a} + \vec{b}| = 1 \]

### Step 3: Use the Magnitude Formula
The magnitude of the sum of two vectors can be expressed as:
\[ |\vec{a} + \vec{b}|^2 = |\vec{a}|^2 + |\vec{b}|^2 + 2|\vec{a}||\vec{b}|\cos(\alpha) \]
Substituting the known values:
\[ |\vec{a} + \vec{b}|^2 = 1^2 + 1^2 + 2 \cdot 1 \cdot 1 \cdot \cos(\alpha) \]
This simplifies to:
\[ |\vec{a} + \vec{b}|^2 = 1 + 1 + 2\cos(\alpha) = 2 + 2\cos(\alpha) \]

### Step 4: Set the Magnitude Equal to 1
Since \( |\vec{a} + \vec{b}| = 1 \), we have:
\[ 2 + 2\cos(\alpha) = 1 \]

### Step 5: Solve for \( \cos(\alpha) \)
Rearranging the equation gives:
\[ 2\cos(\alpha) = 1 - 2 \]
\[ 2\cos(\alpha) = -1 \]
\[ \cos(\alpha) = -\frac{1}{2} \]

### Step 6: Find the Angle \( \alpha \)
To find \( \alpha \), we take the inverse cosine:
\[ \alpha = \cos^{-1}\left(-\frac{1}{2}\right) \]
The angle whose cosine is \( -\frac{1}{2} \) is:
\[ \alpha = \frac{2\pi}{3} \quad \text{(or } 120^\circ\text{)} \]

### Conclusion
Thus, the value of \( \alpha \) is:
\[ \alpha = \frac{2\pi}{3} \]