Question

Let a vector aˆi+βˆj be obtained by rotating the vector √3ˆi+ˆj by an angle 45∘ 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:

A. 1
B. 1/2
C. 1/√2
D. 2/√2

Answer 1

To obtain the vector a, rotate the vector √3i + j by 45° counterclockwise in the first quadrant. The area of the triangle formed by the vertices (α,β), (0,β), and (0,0) is 1.

Step-by-step explanation:

To obtain the vector a, we rotate the vector √3i + j by 45° counterclockwise in the first quadrant. We can use trigonometry to find the components of a. The x-component of a is given by the cosine of the angle multiplied by the magnitude of the original vector, which is √3. So, ax = cos(45°) * √3 = √2.

The y-component of a is given by the sine of the angle multiplied by the magnitude of the original vector, which is 1. So, ay = sin(45°) * 1 = √2.

Now, we can find the area of the triangle formed by the vertices (α,β), (0,β), and (0,0), where α = ax and β = ay. The base of the triangle is β - 0 = √2 - 0 = √2. The height of the triangle is α - 0 = √2 - 0 = √2. Therefore, the area of the triangle is (1/2) * base * height = (1/2) * √2 * √2 = 1.