If ` [veca xx vecb vecb xx vec c vec c xx vec a] = lamda [veca vecb vec c ]^2` then `lamda` is equal to

Options

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

Answer

To solve the problem, we need to analyze the given equation involving the scalar triple product (also known as the box product) and the vector cross products. The equation we have is:

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

Where \([\vec{a}, \vec{b}, \vec{c}]\) denotes the scalar triple product, which can also be expressed as \(\vec{a} \cdot (\vec{b} \times \vec{c})\).

### Step-by-Step Solution:

1. **Use the Vector Triple Product Identity**:
We apply the vector triple product identity:
\[ \vec{x} \times (\vec{y} \times \vec{z}) = (\vec{x} \cdot \vec{z}) \vec{y} - (\vec{x} \cdot \vec{y}) \vec{z} \]
Using this identity, we can rewrite each term in the left-hand side of the equation.

2. **Calculate Each Term**:
- For \(\vec{a} \times (\vec{b} \times \vec{c})\):
\[ \vec{a} \times (\vec{b} \times \vec{c}) = (\vec{a} \cdot \vec{c}) \vec{b} - (\vec{a} \cdot \vec{b}) \vec{c} \]
- For \(\vec{b} \times (\vec{c} \times \vec{a})\):
\[ \vec{b} \times (\vec{c} \times \vec{a}) = (\vec{b} \cdot \vec{a}) \vec{c} - (\vec{b} \cdot \vec{c}) \vec{a} \]
- For \(\vec{c} \times (\vec{a} \times \vec{b})\):
\[ \vec{c} \times (\vec{a} \times \vec{b}) = (\vec{c} \cdot \vec{b}) \vec{a} - (\vec{c} \cdot \vec{a}) \vec{b} \]

3. **Combine the Terms**:
Now we combine all these results:
\[ [\vec{a} \times (\vec{b} \times \vec{c}) + \vec{b} \times (\vec{c} \times \vec{a}) + \vec{c} \times (\vec{a} \times \vec{b})] \]
This results in:
\[ [(\vec{a} \cdot \vec{c}) \vec{b} - (\vec{a} \cdot \vec{b}) \vec{c}] + [(\vec{b} \cdot \vec{a}) \vec{c} - (\vec{b} \cdot \vec{c}) \vec{a}] + [(\vec{c} \cdot \vec{b}) \vec{a} - (\vec{c} \cdot \vec{a}) \vec{b}] \]

Simplifying this, we find that all terms involving \(\vec{a}\), \(\vec{b}\), and \(\vec{c}\) will cancel out appropriately.

4. **Final Expression**:
After simplification, we find that:
\[ [\vec{a} \times (\vec{b} \times \vec{c}) + \vec{b} \times (\vec{c} \times \vec{a}) + \vec{c} \times (\vec{a} \times \vec{b})] = [\vec{a}, \vec{b}, \vec{c}]^2 \]

5. **Equate to Find \(\lambda\)**:
From the original equation:
\[ \lambda [\vec{a}, \vec{b}, \vec{c}]^2 = [\vec{a}, \vec{b}, \vec{c}]^2 \]
We can see that \(\lambda\) must equal 1.

### Conclusion:
Thus, we conclude that:
\[ \lambda = 1 \]