Consider the following for the next two items that follow
Let `vec(a).vec(b) and vec(c )` be three vectors such that `vec(a) +vec(b) + vec(c )= vec(0) and |vec(a)|= 10 |vec(b)| =6 and |vec(c )|=14`.
What is `vec(a).vec(b) + vec(b).vec(c ) + vec(c ).vec(a)` equal to ?

Options

A. `-332`
B. `-166`
C. 0
D. 166

Answer

To solve the problem, we need to find the value of \( \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a} \) given the conditions \( \vec{a} + \vec{b} + \vec{c} = \vec{0} \) and the magnitudes of the vectors.

### Step-by-Step Solution:

1. **Understand the given information**:
- We have three vectors \( \vec{a} \), \( \vec{b} \), and \( \vec{c} \).
- The sum of the vectors is \( \vec{a} + \vec{b} + \vec{c} = \vec{0} \).
- The magnitudes are given as:
- \( |\vec{a}| = 10 \)
- \( |\vec{b}| = 6 \)
- \( |\vec{c}| = 14 \)

2. **Use the identity for the square of the sum of vectors**:
\[ |\vec{a} + \vec{b} + \vec{c}|^2 = |\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) \]
Since \( \vec{a} + \vec{b} + \vec{c} = \vec{0} \), we have:
\[ |\vec{0}|^2 = 0 \]

3. **Calculate the squares of the magnitudes**:
- \( |\vec{a}|^2 = 10^2 = 100 \)
- \( |\vec{b}|^2 = 6^2 = 36 \)
- \( |\vec{c}|^2 = 14^2 = 196 \)

4. **Substitute the values into the identity**:
\[ 0 = 100 + 36 + 196 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) \]

5. **Combine the constants**:
\[ 0 = 332 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) \]

6. **Isolate the dot product sum**:
\[ 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = -332 \]
\[ \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a} = -166 \]

### Final Answer:
\[ \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a} = -166 \]