Question

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:

• A. $1$
• B. $\frac{1}{2}$
• C. $\frac{1}{\sqrt{2}}$
• D. $2\sqrt{2}$

Answer 1

The correct option is B $\frac{1}{2}$


$(\alpha ,\beta )\equiv (2\cos {75}^{\circ },2\sin {75}^{\circ })\text{Area}=\frac{1}{2}\left(2\cos {75}^{\circ }\right)\left(2\sin {75}^{\circ }\right)$
$=\sin \left({150}^{\circ }\right)=\frac{1}{2}$ sq. unit