If the sum of two unit vectors is also a unit vector. Then magnituce of their difference and angle between the two given unit vectors is

Options

A. `sqrt(3),60^(@)`
B. `sqrt(3), 120^(@)`
C. `sqrt(2), 60^(@)`
D. `sqrt(2),120^(@)`

Answer

To solve the problem, we need to find the magnitude of the difference between two unit vectors \( \mathbf{p} \) and \( \mathbf{q} \) given that their sum is also a unit vector. We will also determine the angle between these two vectors.

### Step 1: Understand the Given Information
We know that both \( \mathbf{p} \) and \( \mathbf{q} \) are unit vectors. This means:
\[ |\mathbf{p}| = 1 \quad \text{and} \quad |\mathbf{q}| = 1 \]
We are also given that the sum of these two vectors is a unit vector:
\[ |\mathbf{p} + \mathbf{q}| = 1 \]

### Step 2: Use the Magnitude Formula for the Sum of Vectors
The magnitude of the sum of two vectors can be expressed as:
\[ |\mathbf{p} + \mathbf{q}|^2 = |\mathbf{p}|^2 + |\mathbf{q}|^2 + 2 |\mathbf{p}| |\mathbf{q}| \cos \theta \]
where \( \theta \) is the angle between \( \mathbf{p} \) and \( \mathbf{q} \).

### Step 3: Substitute Known Values
Since both vectors are unit vectors, we have:
\[ |\mathbf{p} + \mathbf{q}|^2 = 1^2 = 1 \]
Substituting the magnitudes into the equation:
\[ 1 = 1 + 1 + 2 \cdot 1 \cdot 1 \cdot \cos \theta \]
This simplifies to:
\[ 1 = 2 + 2 \cos \theta \]

### Step 4: Solve for \( \cos \theta \)
Rearranging the equation gives:
\[ 2 \cos \theta = 1 - 2 \]
\[ 2 \cos \theta = -1 \]
\[ \cos \theta = -\frac{1}{2} \]

### Step 5: Determine the Angle \( \theta \)
The angle \( \theta \) for which \( \cos \theta = -\frac{1}{2} \) is:
\[ \theta = 120^\circ \quad \text{(in the second quadrant)} \]

### Step 6: Find the Magnitude of the Difference
Now we need to find the magnitude of the difference \( |\mathbf{p} - \mathbf{q}| \):
\[ |\mathbf{p} - \mathbf{q}|^2 = |\mathbf{p}|^2 + |\mathbf{q}|^2 - 2 |\mathbf{p}| |\mathbf{q}| \cos \theta \]
Substituting the known values:
\[ |\mathbf{p} - \mathbf{q}|^2 = 1 + 1 - 2 \cdot 1 \cdot 1 \cdot \left(-\frac{1}{2}\right) \]
This simplifies to:
\[ |\mathbf{p} - \mathbf{q}|^2 = 2 + 1 = 3 \]
Taking the square root gives:
\[ |\mathbf{p} - \mathbf{q}| = \sqrt{3} \]

### Final Answer
Thus, the magnitude of the difference \( |\mathbf{p} - \mathbf{q}| \) is \( \sqrt{3} \) and the angle \( \theta \) between the two unit vectors is \( 120^\circ \).

### Summary of Results
- Magnitude of the difference: \( \sqrt{3} \)
- Angle between the two unit vectors: \( 120^\circ \)