`alpha hati + beta hatj` is obtained by rotating `sqrt3hati+hatj` by `45^@` in counter direction about only `45^@`. Find area of `Delta` made by `(0,0), (0,beta)and (alpha , beta)`

Answer

To solve the problem step by step, we need to follow the geometric transformations and calculations outlined in the video transcript. Here’s the detailed solution:

### Step 1: Understand the Rotation
We start with the vector \( \sqrt{3} \hat{i} + \hat{j} \). We need to rotate this vector by \( 45^\circ \) in the counterclockwise direction.

### Step 2: Use Rotation Matrix
The rotation of a vector \( \begin{pmatrix} x \\ y \end{pmatrix} \) by an angle \( \theta \) can be done using the rotation matrix:
\[ R(\theta) = \begin{pmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{pmatrix} \]
For \( \theta = 45^\circ \):
\[ R(45^\circ) = \begin{pmatrix} \cos 45^\circ & -\sin 45^\circ \\ \sin 45^\circ & \cos 45^\circ \end{pmatrix} = \begin{pmatrix} \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \end{pmatrix} \]

### Step 3: Apply the Rotation
Now, we apply this rotation matrix to the vector \( \begin{pmatrix} \sqrt{3} \\ 1 \end{pmatrix} \):
\[ \begin{pmatrix} \alpha \\ \beta \end{pmatrix} = R(45^\circ) \begin{pmatrix} \sqrt{3} \\ 1 \end{pmatrix} = \begin{pmatrix} \frac{1}{\sqrt{2}} \sqrt{3} - \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} \sqrt{3} + \frac{1}{\sqrt{2}} \end{pmatrix} \]
This simplifies to:
\[ \begin{pmatrix} \alpha \\ \beta \end{pmatrix} = \begin{pmatrix} \frac{\sqrt{3} - 1}{\sqrt{2}} \\ \frac{\sqrt{3} + 1}{\sqrt{2}} \end{pmatrix} \]

### Step 4: Calculate Area of Triangle
The area \( A \) of the triangle formed by the points \( (0, 0) \), \( (0, \beta) \), and \( (\alpha, \beta) \) can be calculated using the formula:
\[ A = \frac{1}{2} \times \text{base} \times \text{height} \]
Here, the base is \( \alpha \) and the height is \( \beta \):
\[ A = \frac{1}{2} \times \alpha \times \beta \]

### Step 5: Substitute Values of \( \alpha \) and \( \beta \)
Substituting the values of \( \alpha \) and \( \beta \):
\[ A = \frac{1}{2} \times \frac{\sqrt{3} - 1}{\sqrt{2}} \times \frac{\sqrt{3} + 1}{\sqrt{2}} = \frac{1}{2} \times \frac{(\sqrt{3} - 1)(\sqrt{3} + 1)}{2} \]
Using the difference of squares:
\[ A = \frac{1}{2} \times \frac{3 - 1}{2} = \frac{1}{2} \times \frac{2}{2} = \frac{1}{2} \]

### Final Answer
Thus, the area of the triangle is:
\[ \boxed{\frac{1}{2}} \]