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

Options

A. `(1)/(2)(hatj+hati)`
B. `(3)/(2)(hatj+hati)`
C. `(5)/(2)(hati+hati)`
D. `(7)/(2)(hatj+hati)`

Explanation

To find the component of the vector \( \mathbf{A} = 3\hat{i} + 4\hat{j} \) along the vector \( \mathbf{B} = \hat{i} + \hat{j} \), we can follow these steps:

### Step 1: Find the unit vector of \( \mathbf{B} \)

The unit vector \( \hat{b} \) in the direction of \( \mathbf{B} \) is calculated using the formula:

\[ \hat{b} = \frac{\mathbf{B}}{|\mathbf{B}|} \]

First, we need to find the magnitude of \( \mathbf{B} \):

\[ |\mathbf{B}| = \sqrt{(1^2 + 1^2)} = \sqrt{2} \]

Now, we can find the unit vector \( \hat{b} \):

\[ \hat{b} = \frac{\hat{i} + \hat{j}}{\sqrt{2}} = \frac{1}{\sqrt{2}}\hat{i} + \frac{1}{\sqrt{2}}\hat{j} \]

### Step 2: Calculate the dot product \( \mathbf{A} \cdot \hat{b} \)

Next, we calculate the dot product of \( \mathbf{A} \) and \( \hat{b} \):

\[ \mathbf{A} \cdot \hat{b} = (3\hat{i} + 4\hat{j}) \cdot \left(\frac{1}{\sqrt{2}}\hat{i} + \frac{1}{\sqrt{2}}\hat{j}\right) \]

Calculating the dot product:

\[ \mathbf{A} \cdot \hat{b} = 3 \cdot \frac{1}{\sqrt{2}} + 4 \cdot \frac{1}{\sqrt{2}} = \frac{3}{\sqrt{2}} + \frac{4}{\sqrt{2}} = \frac{7}{\sqrt{2}} \]

### Step 3: Find the component of \( \mathbf{A} \) along \( \mathbf{B} \)

The component of \( \mathbf{A} \) along \( \mathbf{B} \) is given by:

\[ \text{Component of } \mathbf{A} \text{ along } \mathbf{B} = (\mathbf{A} \cdot \hat{b}) \hat{b} \]

Substituting the values we found:

\[ \text{Component of } \mathbf{A} \text{ along } \mathbf{B} = \frac{7}{\sqrt{2}} \left(\frac{1}{\sqrt{2}}\hat{i} + \frac{1}{\sqrt{2}}\hat{j}\right) \]

Calculating this gives:

\[ = \frac{7}{2} \hat{i} + \frac{7}{2} \hat{j} \]

### Final Answer

Thus, the component of \( (3\hat{i} + 4\hat{j}) \) along \( (\hat{i} + \hat{j}) \) is:

\[ \frac{7}{2} \hat{i} + \frac{7}{2} \hat{j} \]

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