In this example, we explore the calculation of the resultant vector from two given vectors, A and B. Vector A has a magnitude of 10 at an angle of 40 degrees, while vector B has a magnitude of 3 at an angle of 20 degrees. The task is to find the resultant vector defined as A - 2B, which involves vector subtraction and scalar multiplication.

To begin, we represent the vectors graphically by drawing them tip to tail. Vector A is drawn from the origin at 40 degrees, and vector B is drawn at 20 degrees. However, since we need to calculate A - 2B, we first need to determine the negative of vector B and then double it. This means we will draw vector -2B in the opposite direction of B with the same length, effectively creating two segments of vector -B.

Next, we break down the vectors into their components using trigonometric functions. The x and y components of vector A can be calculated using:

A_x = A * cos(θ) and A_y = A * sin(θ),

where θ is the angle of the vector. For vector A, this results in:

A_x = 10 * cos(40°) ≈ 7.7 and A_y = 10 * sin(40°) ≈ 6.4.

For vector B, the components are calculated similarly:

B_x = B * cos(20°) ≈ 2.8 and B_y = B * sin(20°) ≈ 1.

Now, to find the resultant vector R, we combine the components according to the equation:

R_x = A_x - 2 * B_x and R_y = A_y - 2 * B_y.

Substituting the values, we find:

R_x = 7.7 - 2 * 2.8 = 2.1 and R_y = 6.4 - 2 * 1 = 4.4.

With the components of the resultant vector determined, we can now calculate its magnitude and direction. The magnitude |R| is found using the Pythagorean theorem:

|R| = √(R_x² + R_y²) = √(2.1² + 4.4²) ≈ 4.9.

To find the direction, we use the arctangent function:

θ_R = tan^-1(R_y / R_x) = tan^-1(4.4 / 2.1) ≈ 64.5°.

In summary, the resultant vector R has a magnitude of approximately 4.9 and is directed at an angle of about 64.5 degrees from the positive x-axis.