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:
Ax = A * cos(θ) and Ay = A * sin(θ),
where θ is the angle of the vector. For vector A, this results in:
Ax = 10 * cos(40°) ≈ 7.7 and Ay = 10 * sin(40°) ≈ 6.4.
For vector B, the components are calculated similarly:
Bx = B * cos(20°) ≈ 2.8 and By = B * sin(20°) ≈ 1.
Now, to find the resultant vector R, we combine the components according to the equation:
Rx = Ax - 2 * Bx and Ry = Ay - 2 * By.
Substituting the values, we find:
Rx = 7.7 - 2 * 2.8 = 2.1 and Ry = 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| = √(Rx2 + Ry2) = √(2.12 + 4.42) ≈ 4.9.
To find the direction, we use the arctangent function:
θR = tan-1(Ry / Rx) = 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.