If `vec(A) = (2hat(i) + 3hat(j)) and vec(B)= (hat(i)- hat(j))` then component of `vec(A)` perpendicular to vector `vec(B)` and in the same plane is

Options

A. `(5)/(2) (hat(i) + hat(j))`
B. `(5)/(sqrt2) (hat(i) + hat(j))`
C. `(sqrt5)/(2) (hat(i) + hat(j))`
D. `(5)/(sqrt2) (hat(i) + hat(k))`

Answer

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

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

### Step 2: Find the unit vector of \(\vec{B}\)
To find the unit vector \(\hat{B}\) in the direction of \(\vec{B}\), we first calculate the magnitude of \(\vec{B}\):
\[ |\vec{B}| = \sqrt{(1)^2 + (-1)^2} = \sqrt{1 + 1} = \sqrt{2} \]
Now, the unit vector \(\hat{B}\) is:
\[ \hat{B} = \frac{\vec{B}}{|\vec{B}|} = \frac{\hat{i} - \hat{j}}{\sqrt{2}} \]

### Step 3: Calculate the projection of \(\vec{A}\) onto \(\vec{B}\)
The projection of \(\vec{A}\) onto \(\vec{B}\) (denoted as \(\vec{R_1}\)) is given by:
\[ \vec{R_1} = \left(\frac{\vec{A} \cdot \vec{B}}{|\vec{B}|^2}\right) \vec{B} \]
First, we need to calculate \(\vec{A} \cdot \vec{B}\):
\[ \vec{A} \cdot \vec{B} = (2\hat{i} + 3\hat{j}) \cdot (\hat{i} - \hat{j}) = 2 \cdot 1 + 3 \cdot (-1) = 2 - 3 = -1 \]
Now, substituting this into the projection formula:
\[ \vec{R_1} = \left(\frac{-1}{2}\right)(\hat{i} - \hat{j}) = -\frac{1}{2}\hat{i} + \frac{1}{2}\hat{j} \]

### Step 4: Find the component of \(\vec{A}\) perpendicular to \(\vec{B}\)
The component of \(\vec{A}\) that is perpendicular to \(\vec{B}\) (denoted as \(\vec{R}\)) can be found using:
\[ \vec{R} = \vec{A} - \vec{R_1} \]
Substituting the values we have:
\[ \vec{R} = (2\hat{i} + 3\hat{j}) - \left(-\frac{1}{2}\hat{i} + \frac{1}{2}\hat{j}\right) \]
Calculating this gives:
\[ \vec{R} = 2\hat{i} + 3\hat{j} + \frac{1}{2}\hat{i} - \frac{1}{2}\hat{j} \]
Combining the terms:
\[ \vec{R} = \left(2 + \frac{1}{2}\right)\hat{i} + \left(3 - \frac{1}{2}\right)\hat{j} = \frac{5}{2}\hat{i} + \frac{5}{2}\hat{j} \]

### Final Answer
The component of \(\vec{A}\) perpendicular to \(\vec{B}\) is:
\[ \vec{R} = \frac{5}{2}\hat{i} + \frac{5}{2}\hat{j} \]