If `[veca xx vecb vecb xx vecc vecc xx veca]=lambda[veca vecb vecc]^2`, then `lambda` is equal to

Options

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

Answer

To solve the problem, we need to find the value of \( \lambda \) in the equation:

\[ [\vec{a} \times \vec{b}, \vec{b} \times \vec{c}, \vec{c} \times \vec{a}] = \lambda [\vec{a}, \vec{b}, \vec{c}]^2 \]

### Step-by-Step Solution:

1. **Understanding the Left-Hand Side (LHS)**:
The left-hand side is the scalar triple product of three cross products:
\[ [\vec{a} \times \vec{b}, \vec{b} \times \vec{c}, \vec{c} \times \vec{a}] \]
This can be rewritten using the property of scalar triple products, which states that \( [\vec{x}, \vec{y}, \vec{z}] = \vec{x} \cdot (\vec{y} \times \vec{z}) \).

2. **Applying the Scalar Triple Product**:
We can express the LHS as:
\[ \vec{a} \times \vec{b} \cdot (\vec{b} \times \vec{c} \times \vec{c} \times \vec{a}) \]
Using the vector identity \( \vec{x} \times (\vec{y} \times \vec{z}) = (\vec{x} \cdot \vec{z}) \vec{y} - (\vec{x} \cdot \vec{y}) \vec{z} \), we can simplify \( \vec{b} \times (\vec{c} \times \vec{a}) \).

3. **Using the Scalar Triple Product Identity**:
We can express this as:
\[ [\vec{b}, \vec{c}, \vec{a}] = \vec{b} \cdot (\vec{c} \times \vec{a}) \]
Therefore, the LHS becomes:
\[ \vec{a} \times \vec{b} \cdot (\vec{c} \cdot \vec{a}) - (\vec{b} \cdot \vec{c}) \vec{a} \]

4. **Rearranging the Terms**:
The result can be rearranged to:
\[ \vec{a} \times \vec{b} \cdot \vec{c} \]
This gives us a scalar value.

5. **Right-Hand Side (RHS)**:
The right-hand side is:
\[ \lambda [\vec{a}, \vec{b}, \vec{c}]^2 \]
Here, \( [\vec{a}, \vec{b}, \vec{c}] \) is the scalar triple product which is equal to \( \vec{a} \cdot (\vec{b} \times \vec{c}) \).

6. **Equating Both Sides**:
Now, we equate the LHS and RHS:
\[ \vec{a} \times \vec{b} \cdot \vec{c} = \lambda [\vec{a}, \vec{b}, \vec{c}]^2 \]

7. **Finding \( \lambda \)**:
Since both sides are equal, we can conclude:
\[ \lambda = 1 \]

### Conclusion:
Thus, the value of \( \lambda \) is:

\[ \lambda = 1 \]