Let a vector i+ j be obtained by rotating the vector 3 i + j by a… - Vidyadip

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
$2 \sqrt{2}$

✓

Answer

Correct option: A.

$\frac{1}{2}$

a
Area of $\Delta\left( OA ^{\prime} B \right)=\frac{1}{2} OA ^{\prime} \cos 15^{\circ} \times OA ^{\prime} \sin 15^{\circ}$

$=\frac{1}{2}\left( OA ^{\prime}\right)^{2} \frac{\sin 30^{\circ}}{2}$

$=(3+1) \times \frac{1}{8}=\frac{1}{2}$

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