JEE MAIN - Mathematics (2021 - 16th March Morning Shift - No. 1)
Let a vector $$\alpha \widehat i + \beta \widehat j$$ be obtained by rotating the vector $$\sqrt 3 \widehat i + \widehat j$$ 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 :
$${1 \over {\sqrt 2 }}$$
$${1 \over 2}$$
1
2$${\sqrt 2 }$$
คำอธิบาย
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($$\alpha$$, $$\beta$$) $$ \equiv $$ (2 cos 75$$^\circ$$, 2 sin 75$$^\circ$$)
Area = $${1 \over 2}$$ (2 cos 75$$^\circ$$) (2 sin 75$$^\circ$$)
= sin(150$$^\circ$$) = $${1 \over 2}$$ square unit