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Physics Lesson 3.2.3 - Position in the system of coordinates

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Welcome to our Physics lesson on Position in the system of coordinates, this is the third lesson of our suite of physics lessons covering the topic of Position, Reference Point, you can find links to the other lessons within this tutorial and access additional physics learning resources below this lesson.

Position in the system of coordinates

It is known that in a system of coordinates we can assign a letter to each direction available. Thus, if there is only one direction available (1-D) as shown in the above figure, we denote the axis by the letter x and the object's position by the vector x or Ox). We can write

xblue bicycle = -1 m

and

xgreen bicycle = +4 m

When two directions of motion (2-D) are available, we can express the position of an object using a pair or coordinates (one for each direction). We must use two letters (usually x and y) to label the directions. Therefore, we need to know both coordinates to determine the location of an object in 2-D. Look at the figure below:

Physics Tutorials: This image shows a grid with point A marked as grid position 2y,4x

From the above graph, we can see that the object A is 4m on the right and 2m above the reference point. Therefore, we say the position of the object A is at (4m, 2m) and by this, we understand that the position of object A is represented through the vector

OA = 4m/2m

and the object A is

|OA| = √OA2x + OA2y
= √4m2 + 2m2
= √20m2
= 4.47m

away from the origin in the direction of the vector OA.

Likewise, we can use the same approaches in 3-D (in space) as well. We have another direction added in this case. Usually, it is denoted by the letter z. Therefore, we must write all three coordinates to determine the position of an object. Look at the graph below.

Physics Tutorials: This image shows the grid position 5y,6x and 5z to indicate the three dimensional grid coordinates for position reference on a three dimensional plain

The object A is in the 3 dimensional space. It is diverted 6m from the origin according the x-direction, 5 m from the origin in the y-direction and 6m from the origin in the z-direction, all in the positive direction. This means the components of the vector OA which represents geometrically the linear distance from the origin, are OAx = 6m, OAy = 5m and OAz = 6m respectively.

From the concept of vector's magnitude, we know that

|OA| = √OA2x + OA2y + OA2z

Therefore, substituting the values, we obtain for the magnitude of the vector OA

|OA| = √(6m)2 + (5m)2 + (6m)2
= √36m2 + 25m2 + 36m2
= √97m2
= 9.85m
≈ 10m

Thus, the object A is nearly 10m away from the origin (reference point) in the direction of the vector OA.

Example 1

Write the position of the objects A, B and C shown in the figure. How far are they from the origin?

Physics Tutorials: This image expands on the prvious chart to short tthree refence points (A, B and C) to visually illustrate the coordinates on a three dimensional plain

Solution 1

The object A is in the xOy plane. It has no z-coordinate, so we need to know only two coordinates to show its position.

From the figure, we can see that Ax = 4m, Ay = 3m (and Az = 0m). Therefore, the position OA of the object A is

OA = 4m3m0

The distance from the origin of the Object A is found by calculating the magnitude of the vector OA. Therefore, using the known procedure explained in the article "Vectors and Scalars" for calculating the magnitude of a vector, we can write

|OA| = √OA2x + OA2y + OA2z

Substituting the values, we obtain for the magnitude of the vector OA

|OA| = √(4m)2 + (3m)2 + (0m)2
= √16m2 + 9m2
= √25m2
= 5m

This means the point A is 5m away from the origin in the direction of the vector OA.

The same procedure is used for the other two objects. Thus, for the object B we have only one coordinate Bz = 5m as it lies on the z-axis only. It is not necessary to calculate the magnitude of the vector OB as it is clear that |OB|= 5m.

As for the object C, we can see from the figure that it contains all three coordinates. Thus, Cx = 3m, Cy = 6m and Cz = 4m. Therefore, the magnitude of the vector OC which represents the position of the object C (its linear distance from the origin), is

|OC| = √OC2x + OC2y + OC2z

Substituting the values, we obtain for the magnitude of the vector (OC)

|OC| = √(3m)2 + (6m)2 + (4m)2
= √9m2 + 36m2 + 16m2
= √61m2
= 7.81m

Therefore, the object C is 7.81m away from the origin in the direction of the vector OC.

The figure below shows the position vectors for the three objects.

Physics Tutorials: This image shows the same chart as in the previous example with the addition of verctor arrows from the datam point in the direction of the final reference point

Remark! The 1 and 2 dimensional motions are special cases of the 3 dimensional motion. We can either write 0 in the place of the missing coordinates or simply represent the position in as many coordinates as given. We can illustrate this aspect using the position of the vector B. This position can be mathematically represented in three ways:

In one dimension (according z only)

OB = 5m

In two dimensions (according x and z, or y and z)

OB = 05m

In three dimensions (according x, y and z)

OB = 005m

All these three presentations show the same thing: the vector OB. Hence, they are all equivalent.

You have reach the end of Physics lesson 3.2.3 Position in the system of coordinates. There are 3 lessons in this physics tutorial covering Position, Reference Point, you can access all the lessons from this tutorial below.

More Position, Reference Point Lessons and Learning Resources

Kinematics Learning Material
Tutorial IDPhysics Tutorial TitleTutorialVideo
Tutorial
Revision
Notes
Revision
Questions
3.2Position, Reference Point
Lesson IDPhysics Lesson TitleLessonVideo
Lesson
3.2.1The meaning of reference point
3.2.2What is Position? How does it differ from Location?
3.2.3Position in the system of coordinates

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