Understanding the Magnitude of the Resultant Vector Between Vectors A and B

In this article, we will solve the problem of finding the magnitude of the resultant vector between two given vectors, A and B. We will also discuss the importance of vector components, the concept of resultant vectors, and the application of the Pythagorean theorem. This content is designed to be SEO-friendly, providing valuable insights into solving similar vector problems.

Problem Statement

Vector A has a magnitude of 10 units and makes an angle of 30° with the positive x-axis. Vector B has a magnitude of 20 units and makes an angle of 30° with the negative x-axis. What is the magnitude of the resultant between these two vectors?

Breaking Down the Vectors

To find the magnitude of the resultant vector, we need to first break down each vector into its x and y components.

Vector A

Magnitude: 10 units
Angle with the positive x-axis: 30°

The components of vector A are calculated as:

Ax A · cosθA 10 · cos30° 10 · √3/2 5√3 Ay A · sinθA 10 · sin30° 10 · 1/2 5

Vector B

Magnitude: 20 units
Angle with the negative x-axis: 150° (180° - 30°)

Since B is on the negative x-axis, we can calculate its components as:

Bx B · cosθB 20 · cos150° 20 · -√3/2 -10√3 By B · sinθB 20 · sin150° 20 · 1/2 10

Summing the Components

The components of the resultant vector, R, can be found by summing the x and y components of A and B.

Rx Ax Bx 5√3 - 10√3 -5√3

Ry Ay By 5 10 15

Calculating the Magnitude of the Resultant Vector

The magnitude of the resultant vector, R, can be found using the Pythagorean theorem:

R √(Rx2 Ry2)

Substituting the values:

R √((-5√3)2 152) √(75 225) √300 10√3

Final Answer

The magnitude of the resultant vector between A and B is:

10√3 units

If B is in the third quadrant, the resultant can be 10N at 30° to the -x direction.

Additional Insights

It is clear that the angle between the two vectors A and B is 120°. The sum of the vectors, which is the resultant, AB, can be found using the formula:

AB2 A2 B2 - 2AB·cos(120°)

Substituting the values:

AB2 102 202 - 2·10·20·(-1/2) 100 400 200 700

So, AB √700 10√7.

Note: This is the magnitude of the vector sum of A and B in the third quadrant, which is different from the previous calculation.

When resolving the vectors into x and y components, we get:

A vector: 10cos30° (which is 5√3) and 10sin30° (which is 5) along x and y axis respectively. B vector: -20cos30° (carries a negative sign as it is along the negative X-axis) and 20sin30° (which is 10) along the X and Y axis respectively.

Equating the resolved vectors that are along the x axis we get -5√3, which can be assigned to Rx for convenience. Equating the resolved vectors that are along the y axis we get 15, which can be taken as Ry.

The resultant vector R is found using the Pythagorean theorem:

R √((-5√3)2 152) √(225 75) √300 10√3 17.32.

The magnitude of the resultant vector is 17.32 units.