If position vector of points A,B and C are respectively `hati,hatj, and hatk` and `AB=CX`, then position vector of point X is

Options

A. `-hati+hatj+hatk`
B. `hati-hatj+hatk`
C. `hati+hatj-hatk`
D. `hati+hatj+hatk`

Explanation

To solve the problem, we need to find the position vector of point X given that the position vectors of points A, B, and C are represented by the unit vectors \( \hat{i} \), \( \hat{j} \), and \( \hat{k} \) respectively, and that \( \overrightarrow{AB} = \overrightarrow{CX} \).

### Step-by-Step Solution:

1. **Identify the Position Vectors**:
- The position vector of point A, \( \overrightarrow{A} = \hat{i} \)
- The position vector of point B, \( \overrightarrow{B} = \hat{j} \)
- The position vector of point C, \( \overrightarrow{C} = \hat{k} \)

2. **Write the Vector Equation**:
- According to the problem, we have \( \overrightarrow{AB} = \overrightarrow{CX} \).
- The vector \( \overrightarrow{AB} \) can be expressed as:
\[ \overrightarrow{AB} = \overrightarrow{B} - \overrightarrow{A} = \hat{j} - \hat{i} \]
- The vector \( \overrightarrow{CX} \) can be expressed as:
\[ \overrightarrow{CX} = \overrightarrow{X} - \overrightarrow{C} = \overrightarrow{X} - \hat{k} \]

3. **Set the Two Vectors Equal**:
- From the equation \( \overrightarrow{AB} = \overrightarrow{CX} \), we have:
\[ \hat{j} - \hat{i} = \overrightarrow{X} - \hat{k} \]

4. **Rearranging the Equation**:
- To find \( \overrightarrow{X} \), we rearrange the equation:
\[ \overrightarrow{X} = (\hat{j} - \hat{i}) + \hat{k} \]

5. **Combine the Vectors**:
- Now, combine the vectors:
\[ \overrightarrow{X} = \hat{j} + \hat{k} - \hat{i} \]
- This can be rearranged as:
\[ \overrightarrow{X} = -\hat{i} + \hat{j} + \hat{k} \]

6. **Final Result**:
- Therefore, the position vector of point X is:
\[ \overrightarrow{X} = -\hat{i} + \hat{j} + \hat{k} \]