Let a vector $\alpha \hat { \mathrm { i } } + \beta \hat { \mathrm { j } }$ be obtained by rotating the vector $\sqrt { 3 } \hat { \mathrm { i } } + \hat { \mathrm { j } }$ by an angle $45 ^ { \circ }$ about the origin in counterclockwise direction in the first quadrant. Then the area (in sq. units) of triangle having vertices $( \alpha , \beta ) , ( 0 , \beta )$ and $( 0,0 )$ is equal to\\
(1) $\frac { 1 } { 2 }$\\
(2) 1\\
(3) $\frac { 1 } { \sqrt { 2 } }$\\
(4) $2 \sqrt { 2 }$