`atimes(btimesc)` is coplanar with

Options

A. b and c
B. a and c
C. a and b are unlike vectors
D. a, b and c

Answer

To determine which vectors are coplanar with the vector \( D = \mathbf{A} \times (\mathbf{B} \times \mathbf{C}) \), we can use the properties of the scalar triple product. The scalar triple product of three vectors \( \mathbf{X}, \mathbf{Y}, \mathbf{Z} \) is given by \( \mathbf{X} \cdot (\mathbf{Y} \times \mathbf{Z}) \), and it is equal to zero if the vectors are coplanar.

### Step-by-Step Solution:

1. **Define the Vector D**:
\[ D = \mathbf{A} \times (\mathbf{B} \times \mathbf{C}) \]
Using the vector triple product identity, we can express \( D \) as:
\[ D = (\mathbf{A} \cdot \mathbf{C}) \mathbf{B} - (\mathbf{A} \cdot \mathbf{B}) \mathbf{C} \]

2. **Check Option A: Vectors B and C**:
To check if \( D \) is coplanar with \( \mathbf{B} \) and \( \mathbf{C} \), we calculate the scalar triple product \( D \cdot (\mathbf{B} \times \mathbf{C}) \):
\[ D \cdot (\mathbf{B} \times \mathbf{C}) = \left((\mathbf{A} \cdot \mathbf{C}) \mathbf{B} - (\mathbf{A} \cdot \mathbf{B}) \mathbf{C}\right) \cdot (\mathbf{B} \times \mathbf{C}) \]
Expanding this gives:
\[ = (\mathbf{A} \cdot \mathbf{C})(\mathbf{B} \cdot (\mathbf{B} \times \mathbf{C})) - (\mathbf{A} \cdot \mathbf{B})(\mathbf{C} \cdot (\mathbf{B} \times \mathbf{C})) \]
Both terms are zero because \( \mathbf{B} \cdot (\mathbf{B} \times \mathbf{C}) = 0 \) and \( \mathbf{C} \cdot (\mathbf{B} \times \mathbf{C}) = 0 \). Thus:
\[ D \cdot (\mathbf{B} \times \mathbf{C}) = 0 \]
Therefore, \( D \) is coplanar with \( \mathbf{B} \) and \( \mathbf{C} \).

3. **Check Option B: Vectors A and C**:
We check if \( D \) is coplanar with \( \mathbf{A} \) and \( \mathbf{C} \) by calculating \( D \cdot (\mathbf{A} \times \mathbf{C}) \):
\[ D \cdot (\mathbf{A} \times \mathbf{C}) = \left((\mathbf{A} \cdot \mathbf{C}) \mathbf{B} - (\mathbf{A} \cdot \mathbf{B}) \mathbf{C}\right) \cdot (\mathbf{A} \times \mathbf{C}) \]
This expands to:
\[ = (\mathbf{A} \cdot \mathbf{C})(\mathbf{B} \cdot (\mathbf{A} \times \mathbf{C})) - (\mathbf{A} \cdot \mathbf{B})(\mathbf{C} \cdot (\mathbf{A} \times \mathbf{C})) \]
The first term is zero since \( \mathbf{B} \cdot (\mathbf{A} \times \mathbf{C}) \) is not guaranteed to be zero, and the second term is zero because \( \mathbf{C} \cdot (\mathbf{A} \times \mathbf{C}) = 0 \). Thus, \( D \) is not coplanar with \( \mathbf{A} \) and \( \mathbf{C} \).

4. **Check Option C: Vectors A and B**:
We check if \( D \) is coplanar with \( \mathbf{A} \) and \( \mathbf{B} \):
\[ D \cdot (\mathbf{A} \times \mathbf{B}) = \left((\mathbf{A} \cdot \mathbf{C}) \mathbf{B} - (\mathbf{A} \cdot \mathbf{B}) \mathbf{C}\right) \cdot (\mathbf{A} \times \mathbf{B}) \]
This expands to:
\[ = (\mathbf{A} \cdot \mathbf{C})(\mathbf{B} \cdot (\mathbf{A} \times \mathbf{B})) - (\mathbf{A} \cdot \mathbf{B})(\mathbf{C} \cdot (\mathbf{A} \times \mathbf{B})) \]
The first term is zero, and the second term is not guaranteed to be zero. Thus, \( D \) is not coplanar with \( \mathbf{A} \) and \( \mathbf{B} \).

5. **Check Option D: Vectors A, B, and C**:
Since we have already established that \( D \) is coplanar with \( \mathbf{B} \) and \( \mathbf{C} \) but not with \( \mathbf{A} \), it follows that \( D \) is not coplanar with all three vectors \( \mathbf{A}, \mathbf{B}, \mathbf{C} \).

### Conclusion:
The only option that is correct is **Option A: Vectors B and C**.