If ` veca, vecb ` are unit vertors such that `veca -vecb` is also a unit vector, then the angle between ` veca and vecb` , is

Options

A. `pi/6`
B. `pi/3`
C. `pi/4`
D. `pi/2`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the Given Information
We are given that \( \vec{a} \) and \( \vec{b} \) are unit vectors, and \( \vec{a} - \vec{b} \) is also a unit vector.

### Step 2: Set Up the Equation
Since \( \vec{a} \) and \( \vec{b} \) are unit vectors, we know:
\[ |\vec{a}| = 1 \quad \text{and} \quad |\vec{b}| = 1 \]
We need to find the angle \( \theta \) between \( \vec{a} \) and \( \vec{b} \). The magnitude of \( \vec{a} - \vec{b} \) can be expressed as:
\[ |\vec{a} - \vec{b}| = 1 \]

### Step 3: Use the Magnitude Formula
Using the formula for the magnitude of the difference of two vectors:
\[ |\vec{a} - \vec{b}|^2 = |\vec{a}|^2 + |\vec{b}|^2 - 2(\vec{a} \cdot \vec{b}) \]
Substituting the known magnitudes:
\[ 1^2 = 1 + 1 - 2(\vec{a} \cdot \vec{b}) \]
This simplifies to:
\[ 1 = 2 - 2(\vec{a} \cdot \vec{b}) \]

### Step 4: Rearranging the Equation
Rearranging the equation gives:
\[ 2(\vec{a} \cdot \vec{b}) = 2 - 1 \]
\[ 2(\vec{a} \cdot \vec{b}) = 1 \]
\[ \vec{a} \cdot \vec{b} = \frac{1}{2} \]

### Step 5: Relate the Dot Product to the Angle
The dot product of two vectors can also be expressed in terms of the angle between them:
\[ \vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta \]
Since both \( \vec{a} \) and \( \vec{b} \) are unit vectors, this simplifies to:
\[ \vec{a} \cdot \vec{b} = \cos \theta \]
Thus, we have:
\[ \cos \theta = \frac{1}{2} \]

### Step 6: Find the Angle
The angle \( \theta \) for which \( \cos \theta = \frac{1}{2} \) is:
\[ \theta = 60^\circ \]

### Final Answer
The angle between \( \vec{a} \) and \( \vec{b} \) is \( 60^\circ \).

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