Three vector a, b and c are such that `abs(a)=1, abs(b)=2, abs(c)=4" and "a+b+c=0`. Then the value of `4a.b+3b.c+3c.a` is equal to

Options

A. 27
B. -68
C. -26
D. -34

Answer

To solve the problem, we need to find the value of \( 4\mathbf{a} \cdot \mathbf{b} + 3\mathbf{b} \cdot \mathbf{c} + 3\mathbf{c} \cdot \mathbf{a} \) given the conditions \( |\mathbf{a}| = 1 \), \( |\mathbf{b}| = 2 \), \( |\mathbf{c}| = 4 \), and \( \mathbf{a} + \mathbf{b} + \mathbf{c} = 0 \).

### Step-by-Step Solution:

1. **Use the condition \( \mathbf{a} + \mathbf{b} + \mathbf{c} = 0 \)**:
\[ \mathbf{c} = -(\mathbf{a} + \mathbf{b}) \]

2. **Dot product with \( \mathbf{a} \)**:
\[ \mathbf{a} \cdot \mathbf{c} = \mathbf{a} \cdot (-\mathbf{a} - \mathbf{b}) = -\mathbf{a} \cdot \mathbf{a} - \mathbf{a} \cdot \mathbf{b} \]
Since \( |\mathbf{a}|^2 = 1 \):
\[ \mathbf{a} \cdot \mathbf{c} = -1 - \mathbf{a} \cdot \mathbf{b} \]

3. **Dot product with \( \mathbf{b} \)**:
\[ \mathbf{b} \cdot \mathbf{c} = \mathbf{b} \cdot (-\mathbf{a} - \mathbf{b}) = -\mathbf{b} \cdot \mathbf{a} - \mathbf{b} \cdot \mathbf{b} \]
Since \( |\mathbf{b}|^2 = 4 \):
\[ \mathbf{b} \cdot \mathbf{c} = -\mathbf{b} \cdot \mathbf{a} - 4 \]

4. **Dot product with \( \mathbf{c} \)**:
\[ \mathbf{c} \cdot \mathbf{c} = |\mathbf{c}|^2 = 16 \]
\[ \mathbf{c} \cdot \mathbf{c} = (-\mathbf{a} - \mathbf{b}) \cdot (-\mathbf{a} - \mathbf{b}) = \mathbf{a} \cdot \mathbf{a} + 2\mathbf{a} \cdot \mathbf{b} + \mathbf{b} \cdot \mathbf{b} \]
So,
\[ 16 = 1 + 2\mathbf{a} \cdot \mathbf{b} + 4 \]
\[ 16 = 5 + 2\mathbf{a} \cdot \mathbf{b} \]
\[ 2\mathbf{a} \cdot \mathbf{b} = 11 \implies \mathbf{a} \cdot \mathbf{b} = \frac{11}{2} \]

5. **Substituting back to find \( \mathbf{b} \cdot \mathbf{c} \)**:
\[ \mathbf{b} \cdot \mathbf{c} = -\frac{11}{2} - 4 = -\frac{11}{2} - \frac{8}{2} = -\frac{19}{2} \]

6. **Substituting back to find \( \mathbf{a} \cdot \mathbf{c} \)**:
\[ \mathbf{a} \cdot \mathbf{c} = -1 - \frac{11}{2} = -\frac{2}{2} - \frac{11}{2} = -\frac{13}{2} \]

7. **Now substitute into \( 4\mathbf{a} \cdot \mathbf{b} + 3\mathbf{b} \cdot \mathbf{c} + 3\mathbf{c} \cdot \mathbf{a} \)**:
\[ 4\mathbf{a} \cdot \mathbf{b} = 4 \cdot \frac{11}{2} = 22 \]
\[ 3\mathbf{b} \cdot \mathbf{c} = 3 \cdot \left(-\frac{19}{2}\right) = -\frac{57}{2} \]
\[ 3\mathbf{c} \cdot \mathbf{a} = 3 \cdot \left(-\frac{13}{2}\right) = -\frac{39}{2} \]

8. **Combine all terms**:
\[ 4\mathbf{a} \cdot \mathbf{b} + 3\mathbf{b} \cdot \mathbf{c} + 3\mathbf{c} \cdot \mathbf{a} = 22 - \frac{57}{2} - \frac{39}{2} \]
\[ = 22 - \frac{96}{2} = 22 - 48 = -26 \]

### Final Answer:
The value of \( 4\mathbf{a} \cdot \mathbf{b} + 3\mathbf{b} \cdot \mathbf{c} + 3\mathbf{c} \cdot \mathbf{a} \) is \( \boxed{-26} \).