MCQ
Let a vector $\alpha \hat{i}+\beta \hat{j}$ be obtained by rotating the vector $\sqrt{3} \hat{ i }+\hat{ j }$ by an angle $45^{\circ}$ about the origin in counterclockwise direction in the first quadrant. Then the area of triangle having vertices $(\alpha, \beta),(0, \beta)$ and $(0,0)$ is equal to
- ✓$\frac{1}{2}$
- B$1$
- C$\frac{1}{\sqrt{2}}$
- D$2 \sqrt{2}$
