If `vecA=(1)/sqrt(2)cos theta hati+(1)/sqrt(2)sin thetahatj`, what will be the unit vector perpendiuclar to `vecA`.

Options

A. `cos theta hati+sin thetahatj`
B. `-cos theta hati+sin thetahatj`
C. `(cos thetahati+sin thetahatj)/sqrt(2)`
D. `sin thetahati-cos theta hatj`

Explanation

To find the unit vector perpendicular to the vector \(\vec{A} = \frac{1}{\sqrt{2}} \cos \theta \hat{i} + \frac{1}{\sqrt{2}} \sin \theta \hat{j}\), we can follow these steps:

### Step 1: Define the vector \(\vec{A}\)
The vector \(\vec{A}\) is given as:
\[ \vec{A} = \frac{1}{\sqrt{2}} \cos \theta \hat{i} + \frac{1}{\sqrt{2}} \sin \theta \hat{j} \]

### Step 2: Assume a perpendicular vector \(\vec{B}\)
Let \(\vec{B}\) be a vector that is perpendicular to \(\vec{A}\). We can express \(\vec{B}\) in terms of its components:
\[ \vec{B} = A \hat{i} + B \hat{j} \]

### Step 3: Use the dot product to find the relationship
Since \(\vec{A}\) and \(\vec{B}\) are perpendicular, their dot product must equal zero:
\[ \vec{A} \cdot \vec{B} = 0 \]
Substituting the components of \(\vec{A}\) and \(\vec{B}\):
\[ \left(\frac{1}{\sqrt{2}} \cos \theta\right) A + \left(\frac{1}{\sqrt{2}} \sin \theta\right) B = 0 \]

### Step 4: Rearranging the equation
Rearranging the equation gives us:
\[ A \cos \theta + B \sin \theta = 0 \]
From this, we can express \(B\) in terms of \(A\):
\[ B = -A \frac{\cos \theta}{\sin \theta} \]

### Step 5: Substitute \(B\) back into \(\vec{B}\)
Now substituting \(B\) back into the expression for \(\vec{B}\):
\[ \vec{B} = A \hat{i} - A \frac{\cos \theta}{\sin \theta} \hat{j} \]
This simplifies to:
\[ \vec{B} = A \hat{i} - A \cot \theta \hat{j} \]

### Step 6: Find the magnitude of \(\vec{B}\)
The magnitude of \(\vec{B}\) is given by:
\[ |\vec{B}| = \sqrt{A^2 + \left(-A \cot \theta\right)^2} = \sqrt{A^2 + A^2 \cot^2 \theta} \]
Factoring out \(A^2\):
\[ |\vec{B}| = A \sqrt{1 + \cot^2 \theta} = A \csc \theta \]

### Step 7: Find the unit vector \(\hat{B}\)
The unit vector in the direction of \(\vec{B}\) is given by:
\[ \hat{B} = \frac{\vec{B}}{|\vec{B}|} = \frac{A \hat{i} - A \cot \theta \hat{j}}{A \csc \theta} \]
This simplifies to:
\[ \hat{B} = \sin \theta \hat{i} - \cos \theta \hat{j} \]

### Final Result
Thus, the unit vector perpendicular to \(\vec{A}\) is:
\[ \hat{B} = \sin \theta \hat{i} - \cos \theta \hat{j} \]