Let `hatu, hatv, hatw` be three unit vectors such that `hatu+hatv+hatw=hata, hata.hatu=3/2, hata.hatv=7/4` & `|hata|=2`, then

Options

A. `hatu. hatv=3/4`
B. `hatu.hatw=0`
C. `hatu. hatv=1/4`
D. `hatu.hatw=-1/4`

Explanation

To solve the problem, we will follow a systematic approach using the given information about the unit vectors \( \hat{u}, \hat{v}, \hat{w} \) and the vector \( \hat{a} \).

### Step-by-Step Solution

1. **Given Information**:
- \( \hat{u} + \hat{v} + \hat{w} = \hat{a} \)
- \( \hat{a} \cdot \hat{u} = \frac{3}{2} \)
- \( \hat{a} \cdot \hat{v} = \frac{7}{4} \)
- \( |\hat{a}| = 2 \)

2. **Squaring the Equation**:
We start by squaring both sides of the equation \( \hat{u} + \hat{v} + \hat{w} = \hat{a} \):
\[ |\hat{u} + \hat{v} + \hat{w}|^2 = |\hat{a}|^2 \]
Expanding the left-hand side:
\[ |\hat{u}|^2 + |\hat{v}|^2 + |\hat{w}|^2 + 2(\hat{u} \cdot \hat{v} + \hat{v} \cdot \hat{w} + \hat{w} \cdot \hat{u}) = |\hat{a}|^2 \]
Since \( \hat{u}, \hat{v}, \hat{w} \) are unit vectors, we have \( |\hat{u}|^2 = |\hat{v}|^2 = |\hat{w}|^2 = 1 \):
\[ 1 + 1 + 1 + 2(\hat{u} \cdot \hat{v} + \hat{v} \cdot \hat{w} + \hat{w} \cdot \hat{u}) = 2^2 \]
This simplifies to:
\[ 3 + 2(\hat{u} \cdot \hat{v} + \hat{v} \cdot \hat{w} + \hat{w} \cdot \hat{u}) = 4 \]
Rearranging gives:
\[ 2(\hat{u} \cdot \hat{v} + \hat{v} \cdot \hat{w} + \hat{w} \cdot \hat{u}) = 1 \]
Thus,
\[ \hat{u} \cdot \hat{v} + \hat{v} \cdot \hat{w} + \hat{w} \cdot \hat{u} = \frac{1}{2} \tag{1} \]

3. **Using \( \hat{a} \cdot \hat{u} \)**:
Multiply the original equation \( \hat{u} + \hat{v} + \hat{w} = \hat{a} \) by \( \hat{u} \):
\[ \hat{u} \cdot \hat{u} + \hat{u} \cdot \hat{v} + \hat{u} \cdot \hat{w} = \hat{a} \cdot \hat{u} \]
Since \( \hat{u} \cdot \hat{u} = 1 \):
\[ 1 + \hat{u} \cdot \hat{v} + \hat{u} \cdot \hat{w} = \frac{3}{2} \]
Rearranging gives:
\[ \hat{u} \cdot \hat{v} + \hat{u} \cdot \hat{w} = \frac{3}{2} - 1 = \frac{1}{2} \tag{2} \]

4. **Using \( \hat{a} \cdot \hat{v} \)**:
Multiply the original equation by \( \hat{v} \):
\[ \hat{v} \cdot \hat{u} + \hat{v} \cdot \hat{v} + \hat{v} \cdot \hat{w} = \hat{a} \cdot \hat{v} \]
Since \( \hat{v} \cdot \hat{v} = 1 \):
\[ \hat{u} \cdot \hat{v} + 1 + \hat{v} \cdot \hat{w} = \frac{7}{4} \]
Rearranging gives:
\[ \hat{u} \cdot \hat{v} + \hat{v} \cdot \hat{w} = \frac{7}{4} - 1 = \frac{3}{4} \tag{3} \]

5. **Substituting into Equation (1)**:
From equations (2) and (3), we can substitute:
\[ \hat{u} \cdot \hat{w} = \frac{1}{2} - \hat{u} \cdot \hat{v} \]
Substituting \( \hat{u} \cdot \hat{v} \) from equation (2) into equation (1):
\[ \frac{1}{2} - \hat{u} \cdot \hat{v} + \hat{v} \cdot \hat{w} = \frac{1}{2} \]
This gives:
\[ \hat{v} \cdot \hat{w} = 0 \]

6. **Conclusion**:
Since \( \hat{v} \cdot \hat{w} = 0 \), the vectors \( \hat{v} \) and \( \hat{w} \) are orthogonal.