If `veca=hati+hatj+hatk, vecb=hati+hatj,vecc=hati` and `(vecaxxvecb)xxvecc=lamda veca+mu vecb`, then `lamda+mu=`

Options

A. 0
B. 1
C. 2
D. 3

Answer

To solve the problem, we need to compute the vector triple product \( \vec{a} \times (\vec{b} \times \vec{c}) \) and express it in terms of the vectors \( \vec{a} \) and \( \vec{b} \).

### Step-by-Step Solution:

1. **Identify the vectors**:
\[ \vec{a} = \hat{i} + \hat{j} + \hat{k}, \quad \vec{b} = \hat{i} + \hat{j}, \quad \vec{c} = \hat{i} \]

2. **Use the vector triple product identity**:
The vector triple product can be expressed as:
\[ \vec{a} \times (\vec{b} \times \vec{c}) = (\vec{a} \cdot \vec{c}) \vec{b} - (\vec{b} \cdot \vec{c}) \vec{a} \]

3. **Calculate \( \vec{a} \cdot \vec{c} \)**:
\[ \vec{a} \cdot \vec{c} = (\hat{i} + \hat{j} + \hat{k}) \cdot \hat{i} = 1 + 0 + 0 = 1 \]

4. **Calculate \( \vec{b} \cdot \vec{c} \)**:
\[ \vec{b} \cdot \vec{c} = (\hat{i} + \hat{j}) \cdot \hat{i} = 1 + 0 = 1 \]

5. **Substitute the dot products into the triple product formula**:
\[ \vec{a} \times (\vec{b} \times \vec{c}) = (1) \vec{b} - (1) \vec{a} = \vec{b} - \vec{a} \]

6. **Express \( \vec{b} - \vec{a} \)**:
\[ \vec{b} - \vec{a} = (\hat{i} + \hat{j}) - (\hat{i} + \hat{j} + \hat{k}) = -\hat{k} \]

7. **Set the expression equal to \( \lambda \vec{a} + \mu \vec{b} \)**:
We have:
\[ -\hat{k} = \lambda (\hat{i} + \hat{j} + \hat{k}) + \mu (\hat{i} + \hat{j}) \]

8. **Equate coefficients**:
Since the left side has no \( \hat{i} \) or \( \hat{j} \) components, we can equate the coefficients:
- For \( \hat{i} \): \( \lambda + \mu = 0 \)
- For \( \hat{j} \): \( \lambda + \mu = 0 \)
- For \( \hat{k} \): \( \lambda = -1 \)

9. **Solve for \( \lambda \) and \( \mu \)**:
From \( \lambda = -1 \), substituting into \( \lambda + \mu = 0 \):
\[ -1 + \mu = 0 \implies \mu = 1 \]

10. **Calculate \( \lambda + \mu \)**:
\[ \lambda + \mu = -1 + 1 = 0 \]

### Final Answer:
\[ \lambda + \mu = 0 \]