JEE Main 16 March 2021 Shift 1 Solved Paper

Let OA be √3

∧

i

+

∧

j

and OB be α

∧

i

+β

∧

j

.As, we can notice in OA,‌

1

√3

=tan‌30∘. So, it makes an angle of 30∘ with the X-axis.
Now, when OA is rotated further by 45∘ anticlockwise, the resultant vector OB makes an angle of 75∘ with the X-axis.
So, OB =|OA|(cos‌75∘‌

∧

i

+sin‌75∘‌

∧

j

)Let △OBC be the required triangle whose area we have to determine.
Area of △OBC=(1/2)×( Base )×( Height )
=1/2×β×α
=‌

1

2

(2‌sin‌75∘)(2‌cos‌75∘)=2‌sin‌75∘‌cos‌75∘
=sin‌150∘=sin‌30∘
=1/2
Hence, the area is 1/2sq. unit.