If a vector `bar(OP) = 3hati + 3hatj` is turned clockwise by an angle of 15°, then the y-component of rotated vector would be

Options

A. `2sqrt9`
B. `(3sqrt2)/2`
C. `2/sqrt9`
D. `(sqrt9)/2`

Answer

To find the y-component of the vector \(\bar{OP} = 3\hat{i} + 3\hat{j}\) after it has been rotated clockwise by an angle of \(15^\circ\), we can follow these steps:

### Step 1: Determine the initial angle of the vector
The vector \(\bar{OP}\) can be represented in terms of its components:
- \(x\)-component = 3 (from \(3\hat{i}\))
- \(y\)-component = 3 (from \(3\hat{j}\))

To find the angle \(\theta\) that the vector makes with the x-axis, we can use the tangent function:
\[ \tan(\theta) = \frac{\text{y-component}}{\text{x-component}} = \frac{3}{3} = 1 \]
This gives:
\[ \theta = \tan^{-1}(1) = 45^\circ \]

### Step 2: Calculate the new angle after rotation
Since the vector is rotated clockwise by \(15^\circ\), the new angle \(\theta'\) with respect to the x-axis will be:
\[ \theta' = \theta - 15^\circ = 45^\circ - 15^\circ = 30^\circ \]

### Step 3: Find the magnitude of the vector
The magnitude of the vector \(\bar{OP}\) can be calculated using the Pythagorean theorem:
\[ |\bar{OP}| = \sqrt{(3^2 + 3^2)} = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2} \]

### Step 4: Calculate the y-component of the rotated vector
To find the y-component of the rotated vector, we can use the sine function:
\[ \text{y-component} = |\bar{OP}| \cdot \sin(\theta') \]
Substituting the values we have:
\[ \text{y-component} = 3\sqrt{2} \cdot \sin(30^\circ) \]
Since \(\sin(30^\circ) = \frac{1}{2}\):
\[ \text{y-component} = 3\sqrt{2} \cdot \frac{1}{2} = \frac{3\sqrt{2}}{2} \]

### Final Answer
Thus, the y-component of the rotated vector is:
\[ \frac{3\sqrt{2}}{2} \]

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