JEE Main · 2021 · 16 Mar 2021 · Shift 1 · Vector Algebra

Let a vector

α \hat{i} + β \hat{j}

be obtained by rotating the vector

3 \hat{i} + \hat{j}

by an angle

4 5^{\circ}

about the origin in counterclockwise direction in the first quadrant. Then the area of triangle having vertices

(α, β), (0, β)

and

(0, 0)

is equal to :

Answer: (B)

Solution

The vector

3 \hat{i} + \hat{j}

has magnitude

2

and makes

3 0^{\circ}

with the

x

-axis, so after a counterclockwise rotation through

4 5^{\circ}

it makes

7 5^{\circ}

and

α = 2 cos 7 5^{\circ}, β = 2 sin 7 5^{\circ}

The triangle with vertices

(α, β)

(0, β)

and

(0, 0)

is right angled at

(0, β)

, with legs of lengths

α

and

β

, so its area is

\frac{1}{2} α β = \frac{1}{2} \cdot 4 sin 7 5^{\circ} cos 7 5^{\circ} = sin 15 0^{\circ} = \frac{1}{2}

The area of the triangle is

\frac{1}{2}

square units.

JEE Main · 2021 · 16 Mar 2021 · Shift 1 · Single correct option