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Let a vector $α \hat{i} + β \hat{j}$ be obtained by rotating the vector $\sqrt{3} \hat{i} + \hat{j}$ by an angle $45^{0}$ about the origin in counterclockwise direction in the first quadrant. Then the area of triangle having vertices $(α, β), (0, β)$ and $(0, 0)$ is equal to:

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$\frac{1}{\sqrt{2}}$

$2 \sqrt{2}$

$\frac{1}{2}$

answer is D.

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Detailed Solution

$α = 2 \cos 75^{o} = \frac{\sqrt{3} - 1}{\sqrt{2}}, β = 2 \sin 75^{0} = \frac{\sqrt{3} + 1}{\sqrt{2}}$

Area $= \frac{1}{2} | α . β | = \frac{1}{2} (\frac{3 - 1}{2}) = \frac{1}{2}$

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