If `vecA = (2hati + 3hatj)` and `vecB =(hati - hatj)` then component of `vecA` perpendicular to vector `vecB` and in the same plane is:

Options

A. `5/2(hati + hatj)`
B. `5/sqrt(2)(hati + hatj)`
C. `sqrt(5)/2(hati + hatj)`
D. `5/sqrt(2)(hati + hatk)`

Answer

To find the component of vector \(\vec{A}\) that is perpendicular to vector \(\vec{B}\) in the same plane, we can follow these steps:

### Step 1: Define the vectors
Given:
\[ \vec{A} = 2\hat{i} + 3\hat{j} \]
\[ \vec{B} = \hat{i} - \hat{j} \]

### Step 2: Find the dot product of \(\vec{A}\) and \(\vec{B}\)
The dot product \(\vec{A} \cdot \vec{B}\) is calculated as follows:
\[ \vec{A} \cdot \vec{B} = (2\hat{i} + 3\hat{j}) \cdot (\hat{i} - \hat{j}) = 2(\hat{i} \cdot \hat{i}) + 3(\hat{j} \cdot -\hat{j}) = 2(1) + 3(-1) = 2 - 3 = -1 \]

### Step 3: Calculate the magnitude of vector \(\vec{B}\)
The magnitude of \(\vec{B}\) is given by:
\[ |\vec{B}| = \sqrt{(1)^2 + (-1)^2} = \sqrt{1 + 1} = \sqrt{2} \]

### Step 4: Find the component of \(\vec{A}\) along \(\vec{B}\)
The component of \(\vec{A}\) along \(\vec{B}\) is given by:
\[ \text{Component of } \vec{A} \text{ along } \vec{B} = \frac{\vec{A} \cdot \vec{B}}{|\vec{B}|^2} \vec{B} \]
Substituting the values:
\[ \text{Component of } \vec{A} \text{ along } \vec{B} = \frac{-1}{2} (\hat{i} - \hat{j}) = -\frac{1}{2}\hat{i} + \frac{1}{2}\hat{j} \]

### Step 5: Find the component of \(\vec{A}\) perpendicular to \(\vec{B}\)
Using the relation:
\[ \vec{A} = \vec{A}_{\parallel} + \vec{A}_{\perpendicular} \]
We can rearrange it to find \(\vec{A}_{\perpendicular}\):
\[ \vec{A}_{\perpendicular} = \vec{A} - \vec{A}_{\parallel} \]
Substituting the values:
\[ \vec{A}_{\perpendicular} = (2\hat{i} + 3\hat{j}) - \left(-\frac{1}{2}\hat{i} + \frac{1}{2}\hat{j}\right) \]
\[ = 2\hat{i} + 3\hat{j} + \frac{1}{2}\hat{i} - \frac{1}{2}\hat{j} \]
\[ = \left(2 + \frac{1}{2}\right)\hat{i} + \left(3 - \frac{1}{2}\right)\hat{j} \]
\[ = \frac{5}{2}\hat{i} + \frac{5}{2}\hat{j} \]

### Final Result
Thus, the component of \(\vec{A}\) perpendicular to \(\vec{B}\) is:
\[ \vec{A}_{\perpendicular} = \frac{5}{2} \hat{i} + \frac{5}{2} \hat{j} \]

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