If `vec a,vec b , vec c` are unit vectors such that `veca + vec b+ vecc = 0` and `lambda = veca.vecb + vecb.vecc+vecc.veca `and `d = veca xx vecb + vecb xx vecc + vecc xx veca`, then `(lambda,vecd)` is

Options

A. `(-(3)/(2), 3veccxxvecb)`
B. `((3)/(2), 3vecaxxvecc)`
C. `((3)/(2), 3vecbxxvecc)`
D. `(-(3)/(2), 3vecaxxvecb)`

Answer

To solve the problem, we need to find the values of \( \lambda \) and \( \vec{D} \) given the conditions of the unit vectors \( \vec{a}, \vec{b}, \vec{c} \) such that \( \vec{a} + \vec{b} + \vec{c} = 0 \).

### Step 1: Finding \( \lambda \)

1. Start with the equation \( \vec{a} + \vec{b} + \vec{c} = 0 \).
2. Rearranging gives us \( \vec{c} = -(\vec{a} + \vec{b}) \).
3. We know that \( \vec{a}, \vec{b}, \vec{c} \) are unit vectors, so \( |\vec{a}| = |\vec{b}| = |\vec{c}| = 1 \).
4. Now, square both sides of the equation \( \vec{a} + \vec{b} + \vec{c} = 0 \):
\[ |\vec{a} + \vec{b} + \vec{c}|^2 = 0 \]
Expanding this gives:
\[ |\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 0 \]
5. Since \( |\vec{a}|^2 = |\vec{b}|^2 = |\vec{c}|^2 = 1 \), we have:
\[ 1 + 1 + 1 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 0 \]
Simplifying gives:
\[ 3 + 2(\lambda) = 0 \quad \text{where } \lambda = \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a} \]
6. Rearranging for \( \lambda \):
\[ 2\lambda = -3 \implies \lambda = -\frac{3}{2} \]

### Step 2: Finding \( \vec{D} \)

1. The expression for \( \vec{D} \) is given by:
\[ \vec{D} = \vec{a} \times \vec{b} + \vec{b} \times \vec{c} + \vec{c} \times \vec{a} \]
2. Substitute \( \vec{c} = -(\vec{a} + \vec{b}) \) into the equation:
\[ \vec{D} = \vec{a} \times \vec{b} + \vec{b} \times (-\vec{a} - \vec{b}) + (-\vec{a} - \vec{b}) \times \vec{a} \]
3. Expanding the terms gives:
\[ \vec{D} = \vec{a} \times \vec{b} - \vec{b} \times \vec{a} - \vec{b} \times \vec{b} - \vec{a} \times \vec{a} - \vec{b} \times \vec{a} \]
4. Since the cross product of any vector with itself is zero, we have:
\[ \vec{D} = \vec{a} \times \vec{b} - \vec{b} \times \vec{a} = \vec{a} \times \vec{b} + \vec{a} \times \vec{b} = 2(\vec{a} \times \vec{b}) \]
5. Therefore, we can combine the terms:
\[ \vec{D} = 3(\vec{a} \times \vec{b}) \]

### Final Result

Thus, we have:
\[ (\lambda, \vec{D}) = \left(-\frac{3}{2}, 3(\vec{a} \times \vec{b})\right) \]