If `[(2veca+4vecb,vecc,vecd)]=lamda[(veca,vecc,vecd)]+mu[(vecb,vecc,vecd)]`, then `lamda+mu=`

Options

A. 6
B. -6
C. 10
D. 8

Explanation

To solve the problem, we need to analyze the given equation involving scalar triple products. The equation is:

\[ [(2\vec{a} + 4\vec{b}, \vec{c}, \vec{d)] = \lambda [(\vec{a}, \vec{c}, \vec{d})] + \mu [(\vec{b}, \vec{c}, \vec{d})] \]

### Step-by-Step Solution:

1. **Understanding the Left-Hand Side (LHS)**:
The left-hand side can be expressed using the properties of the scalar triple product. We can rewrite it as:
\[ [(2\vec{a} + 4\vec{b}, \vec{c}, \vec{d})] = (2\vec{a} + 4\vec{b}) \cdot (\vec{c} \times \vec{d}) \]

2. **Distributing the Dot Product**:
We can distribute the dot product over the sum:
\[ = 2\vec{a} \cdot (\vec{c} \times \vec{d}) + 4\vec{b} \cdot (\vec{c} \times \vec{d}) \]

3. **Using the Definition of Scalar Triple Product**:
The terms \( \vec{a} \cdot (\vec{c} \times \vec{d}) \) and \( \vec{b} \cdot (\vec{c} \times \vec{d}) \) can be expressed as scalar triple products:
\[ = 2[(\vec{a}, \vec{c}, \vec{d})] + 4[(\vec{b}, \vec{c}, \vec{d})] \]

4. **Setting the LHS Equal to the RHS**:
Now, we equate this to the right-hand side of the original equation:
\[ 2[(\vec{a}, \vec{c}, \vec{d})] + 4[(\vec{b}, \vec{c}, \vec{d})] = \lambda[(\vec{a}, \vec{c}, \vec{d})] + \mu[(\vec{b}, \vec{c}, \vec{d})] \]

5. **Comparing Coefficients**:
By comparing coefficients from both sides, we find:
- From \( \lambda[(\vec{a}, \vec{c}, \vec{d})] \): \(\lambda = 2\)
- From \( \mu[(\vec{b}, \vec{c}, \vec{d})] \): \(\mu = 4\)

6. **Calculating \(\lambda + \mu\)**:
Finally, we calculate:
\[ \lambda + \mu = 2 + 4 = 6 \]

### Final Answer:
Thus, the value of \( \lambda + \mu \) is \( 6 \).

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