VECTORS IN SPACE: HOW TO GRAPH, APPLICATIONS, EXERCISES - PHYSICAL - 2025

A vector in space is all that represented by a coordinate system given by x, y, and z. Most of the time the xy plane is the horizontal surface plane and the z axis represents the height (or depth).

The Cartesian coordinate axes shown in Figure 1 divide space into 8 regions called octants, analogous to how the x - y axes divide the plane into 4 quadrants. We will then have 1st octant, 2nd octant and so on.

Figure 1. A vector in space. Source: self made.

Figure 1 contains a representation of a vector v in space. Some perspective is required to create the illusion of three dimensions on the plane of the screen, which is achieved by drawing an oblique view.

To graph a 3D vector, one must use the dotted lines that determine on the grid the coordinates of the projection or "shadow" of v on the xy surface. This projection begins at O ​​and ends at the green point.

Once there, you have to continue along the vertical to the necessary height (or depth) according to the value of z, until you reach P. The vector is drawn starting from O and ending at P, which in the example is in the 1st octant.

Applications

Vectors in space are widely used in mechanics and other branches of physics and engineering, since the structures that surround us require geometry in three dimensions.

Position vectors in space are used to position objects with respect to a reference point called the OR origin. Therefore, they are also necessary tools in navigation, but that is not all.

Forces acting on structures such as bolts, brackets, cables, struts, and more are vector in nature and oriented in space. In order to know its effect, it is necessary to know its address (and also its point of application).

And frequently the direction of a force is known by knowing two points in space that belong to its line of action. In this way the force is:

F = F u

Where F is the magnitude or magnitude of the force and u is the unit vector (module 1) directed along the line of action F.

Notation and 3D Vector Representations

Before we go on to solve some examples, we will briefly review 3D vector notation.

In the example in Figure 1, the vector v, whose point of origin coincides with the origin O and whose end is point P, has positive xyz coordinates, while the y coordinate is negative. These coordinates are: x ₁, y ₁, z ₁, which are precisely the coordinates of P.

So if we have a vector linked to the origin, that is, whose starting point coincides with O, it is very easy to indicate its coordinates, which will be those of the extreme point or P. To distinguish between a point and a vector, we will use to the last bold letters and brackets, like this:

v = <x ₁, y ₁, z ₁ >

While the point P is denoted with parentheses:

P = (x ₁, y ₁, z ₁)

Another representation makes use of the unit vectors i, j and k that define the three directions of space on the x, y and z axes respectively.

These vectors are perpendicular to each other and form an orthonormal basis (see figure 2). This means that a 3D vector can be written in terms of them as:

v = v _x i + v _y j + v _z k

Angles and Director Cosines of a Vector

Figure 2 also shows the director angles γ ₁, γ ₂ and γ ₃ that the vector v makes respectively with the x, y and z axes. Knowing these angles and the magnitude of the vector, it is completely determined. In addition, the cosines of the director angles meet the following relationship:

(cos γ ₁) ² + (cos γ ₂) ² + (cos γ ₃) ² = 1

Figure 2. The unit vectors i, j and k determine the 3 preferential directions of space. Source: self made.

Solved exercises

-Exercise 1

In figure 2 the angles γ ₁, γ ₂ and γ ₃ that the vector v of modulus 50 forms with the coordinate axes are respectively: 75.0º, 60.0º and 34.3º. Find the Cartesian components of this vector and represent it in terms of the unit vectors i, j, and k.

Solution

The projection of the vector v onto the x-axis is v _x = 50. cos 75º = 12,941. In the same way, the projection of v on the y axis is v _y = 50 cos 60 º = 25 and finally on the z axis is v _z = 50. cos 34.3 º = 41.3. Now v can be expressed as:

v = 12.9 i + 25.0 j + 41.3 k

-Exercise 2

Find the tensions in each of the cables that hold the bucket in the figure that is in equilibrium, if its weight is 30 N.

Figure 3. Stress diagram for exercise 2.

Solution

On the bucket, the free-body diagram indicates that T _D (green) offsets the weight W (yellow), therefore T _D = W = 30 N.

At the node, the vector T _D is directed vertically downwards, then:

T _D = 30 (- k) N.

To establish the remaining voltages, follow these steps:

Step 1: Find the Coordinates of All Points

A = (4.5,0,3) (A is on the plane of the wall xz)

B = (1.5,0,0) (B is on the x-axis)

C = (0, 2.5, 3) (C is on the plane of the wall and z)

D = (1.5, 1.5, 0) (D is on the horizontal xy plane)

Step 2: Find the vectors in each direction by subtracting the coordinates of the end and the beginning

DA = <3; -1.5; 3>

DC = <-1.5; one; 3>

DB = <0; -1.5; 0>

Step 3: Calculate modules and unit vectors

A unit vector is obtained by the expression: u = r / r, with r (in bold) being the vector and r (not in bold) being the module of said vector.

DA = (3 ² + (-1.5) ² + 3 ²) ^½ = 4.5; DC = ((-1.5) ² + 1 ² + 3 ²) ^½ = 3.5

u _DA = <3; -1.5; 3> 4.5 = <0.67; -0.33; 0.67>

u _DC = <-1.5; one; 3> 3.5 = <-0.43; 0.29; 0.86>

u _DB = <0; -one; 0>

u _D = <0; 0; -1>

Step 4: Express all stresses as vectors

T _DA = T _DA u _DA = T _DA <0.67; -0.33; 0.67>

T _DC = T _DC u _{DC =} T _DC <-0.43; 0.29; 0.86>

T _DB = T _DB u _DB = T _DB <0; -one; 0>

T _D = 30 <0; 0; -1>

Step 5: Apply the static equilibrium condition and solve the system of equations

Finally, the condition of static equilibrium is applied to the bucket, so that the vector sum of all the forces on the node is zero:

T _DA + T _DC + T _DB + T _D = 0

Since the stresses are in space, it will result in a system of three equations for each component (x, y, and z) of the stresses.

0.67 T _DA -0.43 T _DC + 0 T _DB = 0

-0.33 T _DA + 0.29 T _DC - T _DB = 0

0.67 T _DA + 0.86 T _DC +0 T _DB - 30 = 0

The solution is: T _DA = 14.9 N; T _DA = 23.3 N; T _DB = 1.82 N

References

1. Bedford, 2000. A. Engineering Mechanics: Statics. Addison Wesley. 38-52.
2. Figueroa, D. Series: Physics for Sciences and Engineering. Volume 1. Kinematics. 31-68.
3. Physical. Module 8: Vectors. Recovered from: frtl.utn.edu.ar
4. Hibbeler, R. 2006. Mechanics for Engineers. Static 6th Edition. Continental Publishing Company. 15-53.
5. Vector Addition Calculator. Recovered from: 1728.org