What is the component of `(3hati+4hatj)` along `(hati+hatj)` ?

Explanation

To find the component of the vector \( \vec{A} = 3\hat{i} + 4\hat{j} \) along the vector \( \vec{B} = \hat{i} + \hat{j} \), we will use the formula for the component of one vector along another.

### Step-by-Step Solution:

1. **Identify the vectors**:
- Let \( \vec{A} = 3\hat{i} + 4\hat{j} \)
- Let \( \vec{B} = \hat{i} + \hat{j} \)

2. **Calculate the dot product \( \vec{A} \cdot \vec{B} \)**:
\[ \vec{A} \cdot \vec{B} = (3\hat{i} + 4\hat{j}) \cdot (\hat{i} + \hat{j}) = 3 \cdot 1 + 4 \cdot 1 = 3 + 4 = 7 \]

3. **Calculate the magnitude of vector \( \vec{B} \)**:
\[ |\vec{B}| = \sqrt{(\hat{i})^2 + (\hat{j})^2} = \sqrt{1^2 + 1^2} = \sqrt{2} \]

4. **Calculate the unit vector \( \hat{B} \)**:
\[ \hat{B} = \frac{\vec{B}}{|\vec{B}|} = \frac{\hat{i} + \hat{j}}{\sqrt{2}} \]

5. **Calculate the component of \( \vec{A} \) along \( \vec{B} \)** using the formula:
\[ \text{Component of } \vec{A} \text{ along } \vec{B} = \frac{\vec{A} \cdot \vec{B}}{|\vec{B}|^2} \hat{B} \]
- Here, \( |\vec{B}|^2 = (\sqrt{2})^2 = 2 \)
- Therefore:
\[ \text{Component} = \frac{7}{2} \hat{B} = \frac{7}{2} \left( \frac{\hat{i} + \hat{j}}{\sqrt{2}} \right) \]

6. **Simplify the expression**:
\[ \text{Component} = \frac{7}{2\sqrt{2}} (\hat{i} + \hat{j}) = \frac{7\hat{i}}{2\sqrt{2}} + \frac{7\hat{j}}{2\sqrt{2}} \]

### Final Answer:
The component of \( (3\hat{i} + 4\hat{j}) \) along \( (\hat{i} + \hat{j}) \) is:
\[ \frac{7\hat{i}}{2\sqrt{2}} + \frac{7\hat{j}}{2\sqrt{2}} \]