the angle made by the vector ` sqrt2i + hati` with x-axis body from its initial

Options

A. ` 35.26^(@)`
B. `45^(@)`
C. ` 60^(@)`
D. `90^(@)`

Answer

To find the angle made by the vector \( \sqrt{2} \hat{i} + \hat{j} \) with the x-axis, we can follow these steps:

### Step 1: Identify the components of the vector
The vector given is \( \sqrt{2} \hat{i} + \hat{j} \). Here, the component along the x-axis (i.e., \( \hat{i} \)) is \( \sqrt{2} \) and the component along the y-axis (i.e., \( \hat{j} \)) is \( 1 \).

### Step 2: Use the tangent function to find the angle
The angle \( \theta \) that the vector makes with the x-axis can be found using the tangent function:
\[ \tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} = \frac{\text{y-component}}{\text{x-component}} = \frac{1}{\sqrt{2}} \]

### Step 3: Calculate the angle using the inverse tangent function
To find \( \theta \), we take the arctangent (inverse tangent) of both sides:
\[ \theta = \tan^{-1}\left(\frac{1}{\sqrt{2}}\right) \]

### Step 4: Simplify the expression
Using a calculator or trigonometric tables, we find:
\[ \theta = 35.26^\circ \]

### Step 5: Conclusion
Thus, the angle made by the vector \( \sqrt{2} \hat{i} + \hat{j} \) with the x-axis is approximately \( 35.26^\circ \).

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