The expression `(1/(sqrt(2))hat(i)+1/(sqrt(2))hat(j))` is a

Options

A. unit vector
B. unll vector
C. vector of magnitude`sqrt(2)`
D. Scalar

Explanation

To determine the nature of the expression \((\frac{1}{\sqrt{2}} \hat{i} + \frac{1}{\sqrt{2}} \hat{j})\), we will analyze it step by step.

### Step 1: Identify the components of the vector
The expression consists of two components:
- The \(\hat{i}\) component is \(\frac{1}{\sqrt{2}}\).
- The \(\hat{j}\) component is \(\frac{1}{\sqrt{2}}\).

### Step 2: Calculate the magnitude of the vector
The magnitude \( |V| \) of a vector \( V = a \hat{i} + b \hat{j} \) is given by the formula:
\[ |V| = \sqrt{a^2 + b^2} \]
In our case, \( a = \frac{1}{\sqrt{2}} \) and \( b = \frac{1}{\sqrt{2}} \).

Substituting these values into the formula:
\[ |V| = \sqrt{\left(\frac{1}{\sqrt{2}}\right)^2 + \left(\frac{1}{\sqrt{2}}\right)^2} \]

### Step 3: Simplify the expression
Calculating the squares:
\[ |V| = \sqrt{\frac{1}{2} + \frac{1}{2}} = \sqrt{1} \]

### Step 4: Conclude the magnitude
Thus, the magnitude of the vector is:
\[ |V| = 1 \]

### Step 5: Determine the type of vector
A vector is classified as a unit vector if its magnitude is equal to 1. Since we have found that the magnitude of the vector is 1, we conclude that:
\[ \frac{1}{\sqrt{2}} \hat{i} + \frac{1}{\sqrt{2}} \hat{j} \text{ is a unit vector.} \]

### Final Answer
The expression \((\frac{1}{\sqrt{2}} \hat{i} + \frac{1}{\sqrt{2}} \hat{j})\) is a **unit vector**.

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