If O is origin annd the position vector fo A is `4hati+5hatj`, then unit vector parallel to OA is

Options

A. `(4)/(sqrt(41))hati`
B. `(5)/(sqrt(41))hati`
C. `(1)/(sqrt(41))(4hati+5hatj)`
D. `(1)/(sqrt(41))(4hati-5hatj)`

Explanation

To find the unit vector parallel to the vector OA, we will follow these steps:

### Step 1: Identify the position vector of point A
The position vector of point A is given as:
\[ \vec{A} = 4\hat{i} + 5\hat{j} \]

### Step 2: Write the vector OA
Since O is the origin, the vector OA is simply the position vector of A:
\[ \vec{OA} = \vec{A} = 4\hat{i} + 5\hat{j} \]

### Step 3: Calculate the magnitude of vector OA
The magnitude (or modulus) of vector OA can be calculated using the formula:
\[ |\vec{OA}| = \sqrt{(4)^2 + (5)^2} \]
Calculating this gives:
\[ |\vec{OA}| = \sqrt{16 + 25} = \sqrt{41} \]

### Step 4: Find the unit vector parallel to OA
The unit vector parallel to OA, denoted as \(\hat{OA}\), is given by:
\[ \hat{OA} = \frac{\vec{OA}}{|\vec{OA}|} \]
Substituting the values we have:
\[ \hat{OA} = \frac{4\hat{i} + 5\hat{j}}{\sqrt{41}} \]

### Final Answer
Thus, the unit vector parallel to OA is:
\[ \hat{OA} = \frac{4}{\sqrt{41}}\hat{i} + \frac{5}{\sqrt{41}}\hat{j} \]

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