The component of vector `vecA = a_(x) hati+ a_(y) hatj + a_(z) hatk` along th direction of `(hati - hatj)` is

Options

A. `(a_(x) - a_(y) +a_(z))`
B. `(a_(x) - a_(y))`
C. `((a_(x)-a_(y))/sqrt2`
D. `(a_x + a_y + a_z)`

Explanation

To find the component of the vector \(\vec{A} = a_x \hat{i} + a_y \hat{j} + a_z \hat{k}\) along the direction of the vector \((\hat{i} - \hat{j})\), we can follow these steps:

### Step 1: Identify the Direction Vector
The direction vector is given as \(\vec{B} = \hat{i} - \hat{j}\).

### Step 2: Calculate the Magnitude of the Direction Vector
The magnitude of vector \(\vec{B}\) can be calculated using the formula:
\[ |\vec{B}| = \sqrt{(1)^2 + (-1)^2} = \sqrt{1 + 1} = \sqrt{2} \]

### Step 3: Find the Unit Vector of the Direction Vector
The unit vector \(\hat{b}\) in the direction of \(\vec{B}\) is given by:
\[ \hat{b} = \frac{\vec{B}}{|\vec{B}|} = \frac{\hat{i} - \hat{j}}{\sqrt{2}} = \frac{1}{\sqrt{2}} \hat{i} - \frac{1}{\sqrt{2}} \hat{j} \]

### Step 4: Calculate the Dot Product of \(\vec{A}\) and \(\hat{b}\)
Now, we calculate the dot product \(\vec{A} \cdot \hat{b}\):
\[ \vec{A} \cdot \hat{b} = (a_x \hat{i} + a_y \hat{j} + a_z \hat{k}) \cdot \left(\frac{1}{\sqrt{2}} \hat{i} - \frac{1}{\sqrt{2}} \hat{j}\right) \]
\[ = a_x \cdot \frac{1}{\sqrt{2}} + a_y \cdot \left(-\frac{1}{\sqrt{2}}\right) + a_z \cdot 0 \]
\[ = \frac{a_x}{\sqrt{2}} - \frac{a_y}{\sqrt{2}} \]

### Step 5: Find the Component of \(\vec{A}\) along \(\vec{B}\)
The component of vector \(\vec{A}\) along the direction of \(\vec{B}\) is given by:
\[ \text{Component of } \vec{A} \text{ along } \vec{B} = \vec{A} \cdot \hat{b} = \frac{a_x - a_y}{\sqrt{2}} \]

### Final Answer
Thus, the component of vector \(\vec{A}\) along the direction of \((\hat{i} - \hat{j})\) is:
\[ \frac{a_x - a_y}{\sqrt{2}} \]

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