The component of vector `A=a_(x)hati+a_(y)hatj+a_(z)hatk` and the directioin of `hati-hatj` is

Options

A. `a_(x)-a_(y)+a_(z)`
B. `a_(x)-a_(y)`
C. `(a_(X)-a_(y))/(sqrt(2))`
D. `a_(x)+a_(y)+a_(z)`

Answer

To solve the problem, we need to find the component of the vector \( \mathbf{A} = a_x \hat{i} + a_y \hat{j} + a_z \hat{k} \) along the direction of the vector \( \mathbf{B} = \hat{i} - \hat{j} \).

### Step-by-Step Solution:

1. **Identify the Vectors**:
- Given vector \( \mathbf{A} = a_x \hat{i} + a_y \hat{j} + a_z \hat{k} \)
- Direction vector \( \mathbf{B} = \hat{i} - \hat{j} \)

2. **Calculate the Dot Product**:
- The dot product \( \mathbf{A} \cdot \mathbf{B} \) is calculated as follows:
\[ \mathbf{A} \cdot \mathbf{B} = (a_x \hat{i} + a_y \hat{j} + a_z \hat{k}) \cdot (\hat{i} - \hat{j}) \]
- Expanding this gives:
\[ = a_x \cdot 1 + a_y \cdot (-1) + a_z \cdot 0 = a_x - a_y \]

3. **Calculate the Magnitude of Vector \( \mathbf{B} \)**:
- The magnitude of \( \mathbf{B} \) is given by:
\[ |\mathbf{B}| = \sqrt{1^2 + (-1)^2} = \sqrt{1 + 1} = \sqrt{2} \]

4. **Find the Component of \( \mathbf{A} \) along \( \mathbf{B} \)**:
- The component of vector \( \mathbf{A} \) along vector \( \mathbf{B} \) is given by:
\[ \text{Component of } \mathbf{A} \text{ along } \mathbf{B} = \frac{\mathbf{A} \cdot \mathbf{B}}{|\mathbf{B}|} \]
- Substituting the values we found:
\[ = \frac{a_x - a_y}{\sqrt{2}} \]

5. **Conclusion**:
- The component of vector \( \mathbf{A} \) along the direction of \( \hat{i} - \hat{j} \) is:
\[ \frac{a_x - a_y}{\sqrt{2}} \]

### Final Answer:
The component of vector \( \mathbf{A} \) along the direction of \( \hat{i} - \hat{j} \) is \( \frac{a_x - a_y}{\sqrt{2}} \).

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