The scalar triple product `[veca+vecb-vec c" "vecb+vec c -veca" "vec c+veca-vecb]` is equal to :

Options

A. 0
B. `[vecavecb vec c]`
C. `2[veca vecb vec c]`
D. `4[veca vecb vec c]`

Answer

To solve the problem of finding the scalar triple product \([ \vec{a} + \vec{b} - \vec{c}, \vec{b} + \vec{c} - \vec{a}, \vec{c} + \vec{a} - \vec{b} ]\), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Vectors**:
Let:
\[ \vec{x} = \vec{a} + \vec{b} - \vec{c} \]
\[ \vec{y} = \vec{b} + \vec{c} - \vec{a} \]
\[ \vec{z} = \vec{c} + \vec{a} - \vec{b} \]

2. **Write the Scalar Triple Product**:
The scalar triple product is given by:
\[ \vec{x} \cdot (\vec{y} \times \vec{z}) \]

3. **Calculate \(\vec{y} \times \vec{z}\)**:
We need to compute \(\vec{y} \times \vec{z}\):
\[ \vec{y} \times \vec{z} = (\vec{b} + \vec{c} - \vec{a}) \times (\vec{c} + \vec{a} - \vec{b}) \]
Using the distributive property of the cross product:
\[ = \vec{b} \times \vec{c} + \vec{b} \times \vec{a} - \vec{b} \times \vec{b} + \vec{c} \times \vec{c} + \vec{c} \times \vec{a} - \vec{c} \times \vec{b} - \vec{a} \times \vec{c} - \vec{a} \times \vec{a} + \vec{a} \times \vec{b} \]
Note that any vector crossed with itself is zero:
\[ = \vec{b} \times \vec{c} + \vec{c} \times \vec{a} + \vec{a} \times \vec{b} \]

4. **Substitute Back into the Scalar Triple Product**:
Now substitute \(\vec{y} \times \vec{z}\) back into the scalar triple product:
\[ \vec{x} \cdot (\vec{y} \times \vec{z}) = (\vec{a} + \vec{b} - \vec{c}) \cdot (\vec{b} \times \vec{c} + \vec{c} \times \vec{a} + \vec{a} \times \vec{b}) \]

5. **Distribute the Dot Product**:
Distributing the dot product:
\[ = (\vec{a} + \vec{b} - \vec{c}) \cdot (\vec{b} \times \vec{c}) + (\vec{a} + \vec{b} - \vec{c}) \cdot (\vec{c} \times \vec{a}) + (\vec{a} + \vec{b} - \vec{c}) \cdot (\vec{a} \times \vec{b}) \]

6. **Evaluate Each Term**:
- The first term:
\[ \vec{a} \cdot (\vec{b} \times \vec{c}) + \vec{b} \cdot (\vec{b} \times \vec{c}) - \vec{c} \cdot (\vec{b} \times \vec{c}) = \vec{a} \cdot (\vec{b} \times \vec{c}) + 0 - 0 = \vec{a} \cdot (\vec{b} \times \vec{c}) \]
- The second term:
\[ \vec{a} \cdot (\vec{c} \times \vec{a}) + \vec{b} \cdot (\vec{c} \times \vec{a}) - \vec{c} \cdot (\vec{c} \times \vec{a}) = 0 + \vec{b} \cdot (\vec{c} \times \vec{a}) - 0 = \vec{b} \cdot (\vec{c} \times \vec{a}) \]
- The third term:
\[ \vec{a} \cdot (\vec{a} \times \vec{b}) + \vec{b} \cdot (\vec{a} \times \vec{b}) - \vec{c} \cdot (\vec{a} \times \vec{b}) = 0 + 0 - \vec{c} \cdot (\vec{a} \times \vec{b}) = -\vec{c} \cdot (\vec{a} \times \vec{b}) \]

7. **Combine the Results**:
Combining all the terms:
\[ = \vec{a} \cdot (\vec{b} \times \vec{c}) + \vec{b} \cdot (\vec{c} \times \vec{a}) - \vec{c} \cdot (\vec{a} \times \vec{b}) \]

8. **Final Result**:
The scalar triple product simplifies to:
\[ = 2 \cdot (\vec{a} \cdot (\vec{b} \times \vec{c})) \]
Therefore, the final answer is:
\[ = 2 \cdot [\vec{a}, \vec{b}, \vec{c}] \]