The angle made by the vector `vecA=2hati+3hatj` with Y-axis is

Options

A. `tan^(-1)((3)/(2))`
B. `tan^(-1)((2)/(3))`
C. `sin^(-1)((2)/(3))`
D. `cos^(-1)((3)/(2))`

Explanation

To find the angle made by the vector \(\vec{A} = 2\hat{i} + 3\hat{j}\) with the Y-axis, we can follow these steps:

### Step 1: Identify the components of the vector
The vector \(\vec{A}\) has components:
- \(A_x = 2\) (the coefficient of \(\hat{i}\))
- \(A_y = 3\) (the coefficient of \(\hat{j}\))

### Step 2: Find the magnitude of the vector \(\vec{A}\)
The magnitude of the vector \(\vec{A}\) is given by the formula:
\[ |\vec{A}| = \sqrt{A_x^2 + A_y^2} \]
Substituting the values:
\[ |\vec{A}| = \sqrt{2^2 + 3^2} = \sqrt{4 + 9} = \sqrt{13} \]

### Step 3: Identify the unit vector along the Y-axis
The unit vector along the Y-axis is:
\[ \hat{j} \]
The magnitude of \(\hat{j}\) is 1.

### Step 4: Calculate the dot product of \(\vec{A}\) and \(\hat{j}\)
The dot product \(\vec{A} \cdot \hat{j}\) is calculated as follows:
\[ \vec{A} \cdot \hat{j} = A_x \cdot 0 + A_y \cdot 1 = 0 + 3 = 3 \]

### Step 5: Use the dot product to find the cosine of the angle \(\theta\)
The cosine of the angle \(\theta\) between the vector \(\vec{A}\) and the Y-axis is given by:
\[ \cos \theta = \frac{\vec{A} \cdot \hat{j}}{|\vec{A}| \cdot |\hat{j}|} \]
Substituting the values:
\[ \cos \theta = \frac{3}{\sqrt{13} \cdot 1} = \frac{3}{\sqrt{13}} \]

### Step 6: Find the angle \(\theta\)
To find the angle \(\theta\), we take the inverse cosine:
\[ \theta = \cos^{-1}\left(\frac{3}{\sqrt{13}}\right) \]

### Step 7: Conclusion
The angle made by the vector \(\vec{A}\) with the Y-axis is:
\[ \theta = \cos^{-1}\left(\frac{3}{\sqrt{13}}\right) \]