If a. b = a . C and `axx b = a xx c`, then

Options

A. a is perpendicular to b -c
B. a is parallel to b -c
C. either a = 0 or b = c
D. none of these

Answer

To solve the problem, we need to analyze the given conditions:

1. \( A \cdot B = A \cdot C \)
2. \( A \times B = A \times C \)

### Step-by-Step Solution:

**Step 1: Analyze the first condition \( A \cdot B = A \cdot C \)**

From the first condition, we can rearrange it as:
\[ A \cdot B - A \cdot C = 0 \]
This can be factored as:
\[ A \cdot (B - C) = 0 \]
This implies that either:
- \( A = 0 \) (the zero vector), or
- \( B - C = 0 \) (which means \( B = C \)), or
- \( A \) is perpendicular to \( B - C \) (i.e., \( A \perp (B - C) \)).

**Step 2: Analyze the second condition \( A \times B = A \times C \)**

From the second condition, we can rearrange it similarly:
\[ A \times B - A \times C = 0 \]
This can be factored as:
\[ A \times (B - C) = 0 \]
This implies that either:
- \( A = 0 \) (the zero vector), or
- \( B - C = 0 \) (which means \( B = C \)), or
- \( A \) is parallel to \( B - C \) (i.e., \( A \parallel (B - C) \)).

**Step 3: Combine the results from both conditions**

From both conditions, we have the following possible cases:
1. \( A = 0 \)
2. \( B = C \)
3. \( A \perp (B - C) \) from the first condition.
4. \( A \parallel (B - C) \) from the second condition.

**Step 4: Identify common cases**

The first two cases \( A = 0 \) and \( B = C \) are common in both conditions. The last two cases involve perpendicularity and parallelism, which cannot occur simultaneously unless one of the vectors is the zero vector.

### Conclusion:

Thus, the final answer can be summarized as:
- Either \( A \) is the zero vector, or \( B \) is equal to \( C \).