The scalar triple product `[(veca+vecb-vecc,vecb+vecc-veca,vecc+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 will follow these steps:

### Step 1: Write the expression for the scalar triple product
The scalar triple product of vectors \(\vec{u}\), \(\vec{v}\), and \(\vec{w}\) is given by the determinant:
\[ \vec{u} \cdot (\vec{v} \times \vec{w}) \]
In our case, we have:
\[ \vec{u} = \vec{a} + \vec{b} - \vec{c}, \quad \vec{v} = \vec{b} + \vec{c} - \vec{a}, \quad \vec{w} = \vec{c} + \vec{a} - \vec{b} \]

### Step 2: Calculate \(\vec{v} \times \vec{w}\)
We first need to compute the cross product \(\vec{v} \times \vec{w}\):
\[ \vec{v} = \vec{b} + \vec{c} - \vec{a} \]
\[ \vec{w} = \vec{c} + \vec{a} - \vec{b} \]
Now, we can use the distributive property of the cross product:
\[ \vec{v} \times \vec{w} = (\vec{b} + \vec{c} - \vec{a}) \times (\vec{c} + \vec{a} - \vec{b}) \]
Expanding this gives:
\[ = \vec{b} \times \vec{c} + \vec{b} \times \vec{a} - \vec{b} \times \vec{b} + \vec{c} \times \vec{c} + \vec{c} \times \vec{a} - \vec{c} \times \vec{b} - \vec{a} \times \vec{c} - \vec{a} \times \vec{a} + \vec{a} \times \vec{b} \]
Since the cross product of any vector with itself is zero, we have:
\[ = \vec{b} \times \vec{c} + \vec{c} \times \vec{a} - \vec{c} \times \vec{b} - \vec{a} \times \vec{c} + \vec{a} \times \vec{b} \]

### Step 3: Simplify the expression
Using the property \(\vec{x} \times \vec{y} = -(\vec{y} \times \vec{x})\), we can simplify:
\[ = \vec{b} \times \vec{c} + \vec{c} \times \vec{a} - \vec{b} \times \vec{c} - \vec{a} \times \vec{c} + \vec{a} \times \vec{b} \]
This results in:
\[ = \vec{a} \times \vec{b} - \vec{a} \times \vec{c} \]

### Step 4: Calculate \(\vec{u} \cdot (\vec{v} \times \vec{w})\)
Now we need to calculate:
\[ \vec{u} \cdot (\vec{v} \times \vec{w}) = (\vec{a} + \vec{b} - \vec{c}) \cdot (\vec{a} \times \vec{b} - \vec{a} \times \vec{c}) \]
Distributing the dot product:
\[ = \vec{a} \cdot (\vec{a} \times \vec{b}) + \vec{b} \cdot (\vec{a} \times \vec{b}) - \vec{c} \cdot (\vec{a} \times \vec{b}) - \vec{a} \cdot (\vec{a} \times \vec{c}) - \vec{b} \cdot (\vec{a} \times \vec{c}) + \vec{c} \cdot (\vec{a} \times \vec{c}) \]
Using the property that the dot product of a vector with a cross product involving itself is zero:
\[ = 0 + 0 - \vec{c} \cdot (\vec{a} \times \vec{b}) - 0 - \vec{b} \cdot (\vec{a} \times \vec{c}) + 0 \]
Thus, we have:
\[ = -(\vec{c} \cdot (\vec{a} \times \vec{b}) + \vec{b} \cdot (\vec{a} \times \vec{c})) \]

### Step 5: Final result
The scalar triple product simplifies to:
\[ = 2(\vec{a} \cdot (\vec{b} \times \vec{c})) \]
Thus, the final answer is:
\[ = 2 \cdot [\vec{a}, \vec{b}, \vec{c}] \]

### Conclusion
The scalar triple product \([( \vec{a} + \vec{b} - \vec{c}, \vec{b} + \vec{c} - \vec{a}, \vec{c} + \vec{a} - \vec{b})]\) is equal to \(2[\vec{a}, \vec{b}, \vec{c}]\).