If `hata, hatb and hatc ` are mutually perpendicular unit vectors then `|hata+hatb+hatc|` is equal to

Options

A. 3
B. `sqrt3`
C. `(sqrt(a^2+b^2+c^2))/(3) `
D. 1

Answer

To solve the problem, we need to find the modulus of the sum of three mutually perpendicular unit vectors, denoted as \(\hat{a}\), \(\hat{b}\), and \(\hat{c}\).

### Step-by-Step Solution:

1. **Understanding the Vectors**:
- Given that \(\hat{a}\), \(\hat{b}\), and \(\hat{c}\) are mutually perpendicular unit vectors, we know that:
\[ |\hat{a}| = |\hat{b}| = |\hat{c}| = 1 \]
- Additionally, the dot products between any two of these vectors will be zero:
\[ \hat{a} \cdot \hat{b} = 0, \quad \hat{b} \cdot \hat{c} = 0, \quad \hat{c} \cdot \hat{a} = 0 \]

2. **Finding the Modulus of the Sum**:
- We need to calculate \(|\hat{a} + \hat{b} + \hat{c}|\).
- To find the modulus, we square the expression:
\[ |\hat{a} + \hat{b} + \hat{c}|^2 = (\hat{a} + \hat{b} + \hat{c}) \cdot (\hat{a} + \hat{b} + \hat{c}) \]

3. **Expanding the Dot Product**:
- Using the distributive property of the dot product:
\[ |\hat{a} + \hat{b} + \hat{c}|^2 = \hat{a} \cdot \hat{a} + \hat{b} \cdot \hat{b} + \hat{c} \cdot \hat{c} + 2(\hat{a} \cdot \hat{b} + \hat{b} \cdot \hat{c} + \hat{c} \cdot \hat{a}) \]

4. **Substituting Known Values**:
- Since \(\hat{a} \cdot \hat{a} = 1\), \(\hat{b} \cdot \hat{b} = 1\), and \(\hat{c} \cdot \hat{c} = 1\), we have:
\[ |\hat{a} + \hat{b} + \hat{c}|^2 = 1 + 1 + 1 + 2(0 + 0 + 0) = 3 \]

5. **Taking the Square Root**:
- Therefore, we find:
\[ |\hat{a} + \hat{b} + \hat{c}| = \sqrt{3} \]

### Final Answer:
\[ |\hat{a} + \hat{b} + \hat{c}| = \sqrt{3} \]