Let `vecu,vecv and vecw` be vectors such that `vecu+ vecv + vecw =0` if `|vecu|= 3, |vecv|=4 and |vecw|=5` then `vecu.vecv + vecv .vecw + vecw .vecu ` is (a) 47 (b) -25 (c) 0 (d) 25

Options

A. 47
B. `-25`
C. 0
D. 25

Explanation

To solve the problem, we need to find the value of \( \vec{u} \cdot \vec{v} + \vec{v} \cdot \vec{w} + \vec{w} \cdot \vec{u} \) given that \( \vec{u} + \vec{v} + \vec{w} = 0 \) and the magnitudes of the vectors are \( |\vec{u}| = 3 \), \( |\vec{v}| = 4 \), and \( |\vec{w}| = 5 \).

### Step-by-Step Solution:

1. **Use the identity for the square of a sum of vectors**:
We know that:
\[ (\vec{u} + \vec{v} + \vec{w})^2 = \vec{u}^2 + \vec{v}^2 + \vec{w}^2 + 2(\vec{u} \cdot \vec{v} + \vec{v} \cdot \vec{w} + \vec{w} \cdot \vec{u}) \]

2. **Substituting the condition \( \vec{u} + \vec{v} + \vec{w} = 0 \)**:
Since \( \vec{u} + \vec{v} + \vec{w} = 0 \), we have:
\[ 0^2 = \vec{u}^2 + \vec{v}^2 + \vec{w}^2 + 2(\vec{u} \cdot \vec{v} + \vec{v} \cdot \vec{w} + \vec{w} \cdot \vec{u}) \]
This simplifies to:
\[ 0 = \vec{u}^2 + \vec{v}^2 + \vec{w}^2 + 2(\vec{u} \cdot \vec{v} + \vec{v} \cdot \vec{w} + \vec{w} \cdot \vec{u}) \]

3. **Calculate the squares of the magnitudes**:
We know:
\[ |\vec{u}|^2 = 3^2 = 9, \quad |\vec{v}|^2 = 4^2 = 16, \quad |\vec{w}|^2 = 5^2 = 25 \]
Therefore:
\[ \vec{u}^2 + \vec{v}^2 + \vec{w}^2 = 9 + 16 + 25 = 50 \]

4. **Substituting back into the equation**:
Substituting this into the equation gives:
\[ 0 = 50 + 2(\vec{u} \cdot \vec{v} + \vec{v} \cdot \vec{w} + \vec{w} \cdot \vec{u}) \]

5. **Solving for the dot products**:
Rearranging the equation:
\[ 2(\vec{u} \cdot \vec{v} + \vec{v} \cdot \vec{w} + \vec{w} \cdot \vec{u}) = -50 \]
Dividing by 2:
\[ \vec{u} \cdot \vec{v} + \vec{v} \cdot \vec{w} + \vec{w} \cdot \vec{u} = -25 \]

6. **Final answer**:
Thus, the value of \( \vec{u} \cdot \vec{v} + \vec{v} \cdot \vec{w} + \vec{w} \cdot \vec{u} \) is \( -25 \).

### Conclusion:
The correct answer is (b) -25.