Let `barA=(1)/(sqrt(2))costhetahati+(1)/(sqrt(2)) sin thetahatj` be any vector. What will be the unit vector `hatn` in the direction of `barA` ?

Options

A. `cos theta hati+sin theta hatj`
B. `-cos theta hati-sin theta hatj`
C. `(1)/(sqrt(2))(cos theta hati+sin theta)`
D. `(1)/(sqrt(2))(cos theta hati-sin theta)`

Explanation

To find the unit vector \(\hat{n}\) in the direction of the vector \(\bar{A} = \frac{1}{\sqrt{2}} \cos \theta \hat{i} + \frac{1}{\sqrt{2}} \sin \theta \hat{j}\), we will follow these steps:

### Step 1: Understand the definition of a unit vector
A unit vector is a vector that has a magnitude of 1. The unit vector \(\hat{n}\) in the direction of a vector \(\bar{A}\) can be calculated using the formula:
\[ \hat{n} = \frac{\bar{A}}{|\bar{A}|} \]
where \(|\bar{A}|\) is the magnitude of the vector \(\bar{A}\).

### Step 2: Calculate the magnitude of \(\bar{A}\)
To find the magnitude \(|\bar{A}|\), we use the formula for the magnitude of a vector:
\[ |\bar{A}| = \sqrt{A_x^2 + A_y^2} \]
Here, \(A_x = \frac{1}{\sqrt{2}} \cos \theta\) and \(A_y = \frac{1}{\sqrt{2}} \sin \theta\). Therefore,
\[ |\bar{A}| = \sqrt{\left(\frac{1}{\sqrt{2}} \cos \theta\right)^2 + \left(\frac{1}{\sqrt{2}} \sin \theta\right)^2} \]

### Step 3: Simplify the expression for \(|\bar{A}|\)
Calculating the squares:
\[ |\bar{A}| = \sqrt{\frac{1}{2} \cos^2 \theta + \frac{1}{2} \sin^2 \theta} \]
Factoring out \(\frac{1}{2}\):
\[ |\bar{A}| = \sqrt{\frac{1}{2} (\cos^2 \theta + \sin^2 \theta)} \]
Using the Pythagorean identity \(\cos^2 \theta + \sin^2 \theta = 1\):
\[ |\bar{A}| = \sqrt{\frac{1}{2} \cdot 1} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}} \]

### Step 4: Calculate the unit vector \(\hat{n}\)
Now, substituting \(|\bar{A}|\) back into the formula for the unit vector:
\[ \hat{n} = \frac{\bar{A}}{|\bar{A}|} = \frac{\frac{1}{\sqrt{2}} \cos \theta \hat{i} + \frac{1}{\sqrt{2}} \sin \theta \hat{j}}{\frac{1}{\sqrt{2}}} \]
This simplifies to:
\[ \hat{n} = \cos \theta \hat{i} + \sin \theta \hat{j} \]

### Final Answer
Thus, the unit vector \(\hat{n}\) in the direction of \(\bar{A}\) is:
\[ \hat{n} = \cos \theta \hat{i} + \sin \theta \hat{j} \]

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