If 'a' and 'b' are unit vectors and `|a+b|=1`, then `|a-b|` is equal to

Options

A. `sqrt(2)`
B. `1`
C. `sqrt(5)`
D. `sqrt(3)`

Explanation

To solve the problem, we need to find the value of \(|a-b|\) given that \(a\) and \(b\) are unit vectors and \(|a+b|=1\).

### Step-by-Step Solution:

1. **Understanding Unit Vectors**:
Since \(a\) and \(b\) are unit vectors, we have:
\[ |a| = 1 \quad \text{and} \quad |b| = 1 \]

2. **Using the Given Information**:
We know that:
\[ |a+b| = 1 \]
Squaring both sides gives:
\[ |a+b|^2 = 1^2 = 1 \]

3. **Expanding the Magnitude**:
The magnitude squared of the sum of two vectors can be expressed as:
\[ |a+b|^2 = |a|^2 + |b|^2 + 2(a \cdot b) \]
Substituting the magnitudes of \(a\) and \(b\):
\[ |a+b|^2 = 1^2 + 1^2 + 2(a \cdot b) = 1 + 1 + 2(a \cdot b) = 2 + 2(a \cdot b) \]

4. **Setting Up the Equation**:
From the previous step, we have:
\[ 2 + 2(a \cdot b) = 1 \]
Rearranging gives:
\[ 2(a \cdot b) = 1 - 2 \]
\[ 2(a \cdot b) = -1 \]
Thus, we find:
\[ a \cdot b = -\frac{1}{2} \]

5. **Finding \(|a-b|\)**:
Now we need to find \(|a-b|\). Squaring the magnitude gives:
\[ |a-b|^2 = |a|^2 + |b|^2 - 2(a \cdot b) \]
Substituting the known values:
\[ |a-b|^2 = 1^2 + 1^2 - 2(a \cdot b) = 1 + 1 - 2\left(-\frac{1}{2}\right) \]
Simplifying this:
\[ |a-b|^2 = 1 + 1 + 1 = 3 \]

6. **Taking the Square Root**:
Finally, we take the square root to find \(|a-b|\):
\[ |a-b| = \sqrt{3} \]

### Final Answer:
\[ |a-b| = \sqrt{3} \]