If a,b,c be three non-coplanar vectors, then
`(i) [a-b,b-c,c-a]=`

Options

A. `[abc]`
B. `2[abc]`
C. 0
D. `[abc]^(2)`

Answer

To solve the problem, we need to find the scalar triple product of the vectors \( (a-b), (b-c), (c-a) \). The scalar triple product of three vectors \( \mathbf{u}, \mathbf{v}, \mathbf{w} \) is given by \( \mathbf{u} \cdot (\mathbf{v} \times \mathbf{w}) \).

### Step-by-step Solution:

1. **Identify the vectors:**
We have three vectors:
- \( \mathbf{u} = \mathbf{a} - \mathbf{b} \)
- \( \mathbf{v} = \mathbf{b} - \mathbf{c} \)
- \( \mathbf{w} = \mathbf{c} - \mathbf{a} \)

2. **Write the scalar triple product:**
The scalar triple product can be expressed as:
\[ [\mathbf{u}, \mathbf{v}, \mathbf{w}] = (\mathbf{a} - \mathbf{b}) \cdot ((\mathbf{b} - \mathbf{c}) \times (\mathbf{c} - \mathbf{a})) \]

3. **Calculate the cross product:**
We need to compute \( (\mathbf{b} - \mathbf{c}) \times (\mathbf{c} - \mathbf{a}) \):
\[ (\mathbf{b} - \mathbf{c}) \times (\mathbf{c} - \mathbf{a}) = \mathbf{b} \times \mathbf{c} - \mathbf{b} \times \mathbf{a} + \mathbf{c} \times \mathbf{a} - \mathbf{c} \times \mathbf{c} \]
Since \( \mathbf{c} \times \mathbf{c} = \mathbf{0} \), this simplifies to:
\[ \mathbf{b} \times \mathbf{c} - \mathbf{b} \times \mathbf{a} + \mathbf{c} \times \mathbf{a} \]

4. **Substitute back into the scalar triple product:**
Now substitute the result of the cross product back into the scalar triple product:
\[ [\mathbf{u}, \mathbf{v}, \mathbf{w}] = (\mathbf{a} - \mathbf{b}) \cdot [(\mathbf{b} \times \mathbf{c}) - (\mathbf{b} \times \mathbf{a}) + (\mathbf{c} \times \mathbf{a})] \]

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

6. **Evaluate each term:**
- The first term \( (\mathbf{a} - \mathbf{b}) \cdot (\mathbf{b} \times \mathbf{c}) \) gives us \( \mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) - \mathbf{b} \cdot (\mathbf{b} \times \mathbf{c}) = \mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) \) (the second term is zero).
- The second term \( -(\mathbf{a} - \mathbf{b}) \cdot (\mathbf{b} \times \mathbf{a}) = -(\mathbf{a} \cdot (\mathbf{b} \times \mathbf{a}) - \mathbf{b} \cdot (\mathbf{b} \times \mathbf{a})) \) results in zero since \( \mathbf{b} \times \mathbf{a} \) is perpendicular to \( \mathbf{b} \).
- The third term \( (\mathbf{a} - \mathbf{b}) \cdot (\mathbf{c} \times \mathbf{a}) = \mathbf{a} \cdot (\mathbf{c} \times \mathbf{a}) - \mathbf{b} \cdot (\mathbf{c} \times \mathbf{a}) \) also results in zero for the first part.

7. **Final result:**
Therefore, the scalar triple product simplifies to:
\[ [\mathbf{a} - \mathbf{b}, \mathbf{b} - \mathbf{c}, \mathbf{c} - \mathbf{a}] = \mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) - 0 + 0 = \mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) \]

### Conclusion:
The scalar triple product \( [\mathbf{a} - \mathbf{b}, \mathbf{b} - \mathbf{c}, \mathbf{c} - \mathbf{a}] \) equals \( \mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) \).