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) $2\sqrt 2 $

(B) $\frac {1}{2}$

(C) 1

(D) $\frac {1}{\sqrt 2 }$

*Solution*

$\sqrt 3 \hat i + \hat j $ = $2 (\frac {\sqrt 3 }{2} \hat i + \frac {1}{2} \hat j )$

= $2 (cos30^\circ \hat i + sin30^\circ \hat j )$

When the above vector is rotated by $45^\circ $ the obtained vector,

= $\alpha \hat i + \beta \hat j $ = $2 (cos75^\circ \hat i + sin75^\circ \hat j )$

So, $\alpha = 2 cos 75^\circ $ & $\beta = 2 sin75^\circ $

So, area of triangle = $\frac {1}{2} \times \beta \times \alpha $ = $\frac {1}{2} \times 2 sin 75^\circ \times 2 cos 75^\circ = sin 150^\circ = \frac {1}{2}$

Answer: (B)