The three vectors a, b and c with magnitude 3, 4 and 5 respectively and `a+b+c=0,` then the value of `a.b+b.c+c.a` is

Options

A. 47
B. 25
C. 50
D. `-25`

Answer

To solve the problem, we need to find the value of \( a \cdot b + b \cdot c + c \cdot a \) given that the vectors \( a, b, c \) have magnitudes 3, 4, and 5 respectively, and that \( a + b + c = 0 \).

### Step-by-Step Solution:

1. **Understanding the Given Information:**
We are given:
- Magnitudes: \( |a| = 3 \), \( |b| = 4 \), \( |c| = 5 \)
- Vector equation: \( a + b + c = 0 \)

2. **Finding the Magnitude of the Sum:**
Since \( a + b + c = 0 \), we can say:
\[ |a + b + c| = 0 \]

3. **Squaring the Magnitude:**
Squaring both sides gives:
\[ |a + b + c|^2 = 0^2 = 0 \]
Using the property of magnitudes, we can expand this:
\[ |a + b + c|^2 = |a|^2 + |b|^2 + |c|^2 + 2(a \cdot b + b \cdot c + c \cdot a) \]

4. **Substituting the Magnitudes:**
Now substitute the magnitudes:
\[ |a|^2 = 3^2 = 9, \quad |b|^2 = 4^2 = 16, \quad |c|^2 = 5^2 = 25 \]
Therefore:
\[ 0 = 9 + 16 + 25 + 2(a \cdot b + b \cdot c + c \cdot a) \]

5. **Calculating the Sum of Squares:**
Calculate the sum:
\[ 9 + 16 + 25 = 50 \]
So we have:
\[ 0 = 50 + 2(a \cdot b + b \cdot c + c \cdot a) \]

6. **Rearranging the Equation:**
Rearranging gives:
\[ 2(a \cdot b + b \cdot c + c \cdot a) = -50 \]

7. **Dividing by 2:**
Dividing both sides by 2:
\[ a \cdot b + b \cdot c + c \cdot a = -25 \]

### Final Answer:
Thus, the value of \( a \cdot b + b \cdot c + c \cdot a \) is:
\[ \boxed{-25} \]