If sum of two unit vectors is a unit vector; prove that the magnitude of their difference is `sqrt3`

Options

A. `sqrt(2)`
B. `sqrt(3)`
C. 1
D. none of these

Explanation

To prove that the magnitude of the difference of two unit vectors \( \mathbf{a} \) and \( \mathbf{b} \) is \( \sqrt{3} \) given that their sum is also a unit vector, we can follow these steps:

### Step 1: Define the unit vectors
Let \( \mathbf{a} \) and \( \mathbf{b} \) be two unit vectors. This means:
\[ |\mathbf{a}| = 1 \quad \text{and} \quad |\mathbf{b}| = 1 \]

### Step 2: Express the condition of their sum
Given that the sum of the two unit vectors is also a unit vector, we have:
\[ |\mathbf{a} + \mathbf{b}| = 1 \]

### Step 3: Use the property of magnitudes
Using the property of magnitudes, we can square both sides:
\[ |\mathbf{a} + \mathbf{b}|^2 = 1^2 \]
Expanding the left side using the dot product:
\[ |\mathbf{a}|^2 + |\mathbf{b}|^2 + 2(\mathbf{a} \cdot \mathbf{b}) = 1 \]
Since both \( |\mathbf{a}| \) and \( |\mathbf{b}| \) are 1, we can substitute:
\[ 1 + 1 + 2(\mathbf{a} \cdot \mathbf{b}) = 1 \]
This simplifies to:
\[ 2 + 2(\mathbf{a} \cdot \mathbf{b}) = 1 \]

### Step 4: Solve for \( \mathbf{a} \cdot \mathbf{b} \)
Rearranging gives:
\[ 2(\mathbf{a} \cdot \mathbf{b}) = 1 - 2 \]
\[ 2(\mathbf{a} \cdot \mathbf{b}) = -1 \]
\[ \mathbf{a} \cdot \mathbf{b} = -\frac{1}{2} \]

### Step 5: Use the dot product to find the magnitude of the difference
Now, we want to find the magnitude of the difference \( |\mathbf{a} - \mathbf{b}| \):
\[ |\mathbf{a} - \mathbf{b}|^2 = |\mathbf{a}|^2 + |\mathbf{b}|^2 - 2(\mathbf{a} \cdot \mathbf{b}) \]
Substituting the known values:
\[ |\mathbf{a} - \mathbf{b}|^2 = 1 + 1 - 2\left(-\frac{1}{2}\right) \]
This simplifies to:
\[ |\mathbf{a} - \mathbf{b}|^2 = 2 + 1 = 3 \]

### Step 6: Take the square root
Taking the square root gives:
\[ |\mathbf{a} - \mathbf{b}| = \sqrt{3} \]

Thus, we have proved that the magnitude of the difference of the two unit vectors is \( \sqrt{3} \).