Calculate the component of vector `vecA=(6hati+5hatj)` along the directions of `(hati-hatj) and (hati+hatj)`.

Answer

To calculate the component of the vector \(\vec{A} = 6\hat{i} + 5\hat{j}\) along the directions of \((\hat{i} - \hat{j})\) and \((\hat{i} + \hat{j})\), we can follow these steps:

### Step 1: Define the vectors
Let:
- \(\vec{A} = 6\hat{i} + 5\hat{j}\)
- \(\vec{B_1} = \hat{i} - \hat{j}\)
- \(\vec{B_2} = \hat{i} + \hat{j}\)

### Step 2: Calculate the magnitude of \(\vec{B_1}\) and \(\vec{B_2}\)
The magnitude of \(\vec{B_1}\) is calculated as follows:
\[ |\vec{B_1}| = \sqrt{(1)^2 + (-1)^2} = \sqrt{1 + 1} = \sqrt{2} \]

The magnitude of \(\vec{B_2}\) is calculated as follows:
\[ |\vec{B_2}| = \sqrt{(1)^2 + (1)^2} = \sqrt{1 + 1} = \sqrt{2} \]

### Step 3: Calculate the unit vectors
The unit vector \(\hat{b_1}\) along \(\vec{B_1}\) is:
\[ \hat{b_1} = \frac{\vec{B_1}}{|\vec{B_1}|} = \frac{\hat{i} - \hat{j}}{\sqrt{2}} = \frac{1}{\sqrt{2}}\hat{i} - \frac{1}{\sqrt{2}}\hat{j} \]

The unit vector \(\hat{b_2}\) along \(\vec{B_2}\) is:
\[ \hat{b_2} = \frac{\vec{B_2}}{|\vec{B_2}|} = \frac{\hat{i} + \hat{j}}{\sqrt{2}} = \frac{1}{\sqrt{2}}\hat{i} + \frac{1}{\sqrt{2}}\hat{j} \]

### Step 4: Calculate the component of \(\vec{A}\) along \(\vec{B_1}\)
The component of \(\vec{A}\) along \(\vec{B_1}\) is given by the dot product:
\[ A_{B_1} = \vec{A} \cdot \hat{b_1} = (6\hat{i} + 5\hat{j}) \cdot \left(\frac{1}{\sqrt{2}}\hat{i} - \frac{1}{\sqrt{2}}\hat{j}\right) \]
Calculating the dot product:
\[ A_{B_1} = 6 \cdot \frac{1}{\sqrt{2}} + 5 \cdot \left(-\frac{1}{\sqrt{2}}\right) = \frac{6}{\sqrt{2}} - \frac{5}{\sqrt{2}} = \frac{1}{\sqrt{2}} \]

### Step 5: Calculate the component of \(\vec{A}\) along \(\vec{B_2}\)
The component of \(\vec{A}\) along \(\vec{B_2}\) is given by the dot product:
\[ A_{B_2} = \vec{A} \cdot \hat{b_2} = (6\hat{i} + 5\hat{j}) \cdot \left(\frac{1}{\sqrt{2}}\hat{i} + \frac{1}{\sqrt{2}}\hat{j}\right) \]
Calculating the dot product:
\[ A_{B_2} = 6 \cdot \frac{1}{\sqrt{2}} + 5 \cdot \frac{1}{\sqrt{2}} = \frac{6}{\sqrt{2}} + \frac{5}{\sqrt{2}} = \frac{11}{\sqrt{2}} \]

### Final Result
Thus, the components of vector \(\vec{A}\) along the directions of \((\hat{i} - \hat{j})\) and \((\hat{i} + \hat{j})\) are:
- Along \((\hat{i} - \hat{j})\): \(\frac{1}{\sqrt{2}}\)
- Along \((\hat{i} + \hat{j})\): \(\frac{11}{\sqrt{2}}\)