Let `vec(a)` and `vec(b)` be two unit vectors and `theta` is the angle between them. Then `vec(a)+vec(b)` is a unit vector if :

Options

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

Answer

To determine when the sum of two unit vectors \(\vec{a}\) and \(\vec{b}\) is also a unit vector, we can follow these steps:

### Step 1: Understand the properties of unit vectors
Unit vectors have a magnitude of 1. Therefore, we have:
\[ |\vec{a}| = 1 \quad \text{and} \quad |\vec{b}| = 1 \]

### Step 2: Set up the equation for the magnitude of the sum
We want to find the condition under which the vector \(\vec{a} + \vec{b}\) is also a unit vector. This means:
\[ |\vec{a} + \vec{b}| = 1 \]

### Step 3: Use the formula for the magnitude of the sum of two vectors
The magnitude of the sum of two vectors can be expressed using the cosine of the angle \(\theta\) between them:
\[ |\vec{a} + \vec{b}|^2 = |\vec{a}|^2 + |\vec{b}|^2 + 2|\vec{a}||\vec{b}|\cos(\theta) \]
Substituting the values for unit vectors:
\[ |\vec{a} + \vec{b}|^2 = 1^2 + 1^2 + 2 \cdot 1 \cdot 1 \cdot \cos(\theta) \]
This simplifies to:
\[ |\vec{a} + \vec{b}|^2 = 1 + 1 + 2\cos(\theta) = 2 + 2\cos(\theta) \]

### Step 4: Set the equation equal to 1
Since we want \(|\vec{a} + \vec{b}| = 1\), we square both sides:
\[ |\vec{a} + \vec{b}|^2 = 1^2 \]
Thus, we have:
\[ 2 + 2\cos(\theta) = 1 \]

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

### Step 6: Determine the angle \(\theta\)
The angle \(\theta\) for which \(\cos(\theta) = -\frac{1}{2}\) corresponds to:
\[ \theta = \frac{2\pi}{3} \quad \text{(or } 120^\circ\text{)} \]

### Conclusion
Thus, the angle \(\theta\) between the two unit vectors \(\vec{a}\) and \(\vec{b}\) must be:
\[ \theta = \frac{2\pi}{3} \]

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