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

Options

A. `2bhat(j)`
B. `a hat(i)`
C. `-2 bhat(j)`
D. None of these

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 relevant vectors
The vector \(\vec{A}\) is given as \(2\hat{i} + 3\hat{j}\). The unit vector along the Y-axis is \(\hat{j}\).

### Step 2: Use the dot product to find the angle
The angle \(\theta\) between two vectors can be found using the formula:
\[ \cos(\theta) = \frac{\vec{A} \cdot \hat{j}}{|\vec{A}| |\hat{j}|} \]
where \(\vec{A} \cdot \hat{j}\) is the dot product of the vectors, and \(|\vec{A}|\) and \(|\hat{j}|\) are the magnitudes of the vectors.

### Step 3: Calculate the dot product
The dot product \(\vec{A} \cdot \hat{j}\) is calculated as follows:
\[ \vec{A} \cdot \hat{j} = (2\hat{i} + 3\hat{j}) \cdot \hat{j} = 2(\hat{i} \cdot \hat{j}) + 3(\hat{j} \cdot \hat{j}) = 0 + 3 = 3 \]

### Step 4: Calculate the magnitude of \(\vec{A}\)
The magnitude of \(\vec{A}\) is given by:
\[ |\vec{A}| = \sqrt{(2^2 + 3^2)} = \sqrt{4 + 9} = \sqrt{13} \]

The magnitude of \(\hat{j}\) is:
\[ |\hat{j}| = 1 \]

### Step 5: Substitute values into the cosine formula
Now substitute the values into the cosine formula:
\[ \cos(\theta) = \frac{3}{\sqrt{13} \cdot 1} = \frac{3}{\sqrt{13}} \]

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

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