WHAT ARE COPLANAR VECTORS? (WITH SOLVED EXERCISES) - PHYSICAL - 2025

The coplanar vectors or coplanar are those which are contained on the same plane. When there are only two vectors, these are always coplanar, since there are infinite planes, it is always possible to choose one that contains them.

If you have three or more vectors, it may be that some of them are not in the same plane as the others, therefore they could not be considered coplanar. The following figure shows a set of coplanar vectors denoted in bold A, B, C and D:

Figure 1. Four coplanar vectors. Source: self made.

Vectors are related to the behavior and properties of physical quantities relevant to science and engineering; for example velocity, acceleration and force.

A force produces different effects on an object when the way it is applied is varied, for example by changing intensity, direction and direction. Even changing just one of these parameters the results are considerably different.

In many applications, both in statics and dynamics, the forces acting on a body are on the same plane, therefore they are considered coplanar.

Conditions for the vectors to be coplanar

For three vectors to be coplanar they must lie on the same plane and this happens if they meet any of the following conditions:

-Vectors are parallel, therefore their components are proportional and linearly dependent.

-Your mixed product is null.

-If you have three vectors and any of them can be written as a linear combination of the other two, these vectors are coplanar. For example, a vector that results from the sum of two others, the three are all in the same plane.

Alternatively, the coplanarity condition can be set as follows:

Mixed product between three vectors

The mixed product between vectors is defined with three vectors u, v and w, resulting in a scalar that results from performing the following operation:

u · (v x w) = u · (v x w)

First, the cross product that is in parentheses is carried out: v x w , whose result is a normal vector (perpendicular) to the plane in which both v and w lie .

If u is on the same plane as v and w , naturally the scalar product (dot product) between u and said normal vector must be 0. In this way it is verified that the three vectors are coplanar (they lie on the same plane).

When the mixed product is not zero, its result is equal to the volume of the parallelepiped that has the vectors u , v and w as adjacent sides.

Applications

Coplanar, concurrent and non-collinear forces

The concurrent forces are all applied to the same point. If they are also coplanar, they can be replaced by a single one, which is called the resultant force and has the same effect as the original forces.

If a body is in equilibrium thanks to three coplanar forces, concurrent and non-collinear (not parallel), called A , B and C, Lamy's theorem indicates that the relationship between these forces (magnitudes) is the following:

A / sin α = B / sin β = C / sin γ

With α, β and γ as the opposite angles to the applied forces, as shown in the following figure:

Figure 2. Three coplanar forces A, B, and C act on an object. Source: Kiwakwok at English Wikipedia

Solved exercises

-Exercise 1

Find the value of k so that the following vectors are coplanar:

u = <-3, k, 2>

v = <4, 1, 0>

w = <-1, 2, -1>

Solution

Since we have the components of the vectors, the criterion of the mixed product is used, therefore:

u (v x w) = 0

Solve v x w first . The vectors will be expressed in terms of the unit vectors i, j and k that distinguish the three perpendicular directions in space (width, height and depth):

v = 4 i + j + 0 k

w = -1 i + 2 j -1 k

v x w = -4 (ixi) + 8 (ixj) - 4 (ixk) - (jxi) + 2 (jxj) - 2 (jxk) = 8 k + 4 j + k -2 i = -2 i + 4 j + 9 k

Now we consider the scalar product between u and the vector that has resulted from the previous operation, setting the operation equal to 0:

u (v x w) = (-3 i + k j + 2 k) · (-2 i + 4 j + 9 k) = 6 + 4k +18 = 0

24 + 4k = 0

The sought value is: k = - 6

So the vector u is:

u = <-3, -6, 2>

-Exercise 2

The figure shows an object whose weight is W = 600 N, hanging in equilibrium thanks to the cables placed at the angles shown in figure 3. Is it possible to apply Lamy's theorem in this situation? In any case, find the magnitudes of T ₁, T _2, and T ₃ that make equilibrium possible.

Figure 3. A weight hangs in equilibrium under the action of the three stresses shown. Source: self made.

Solution

Lamy's theorem is applicable in this situation if the node on which the three stresses are applied is considered, since they constitute a system of coplanar forces. First, the free-body diagram for the hanging weight is made, in order to determine the magnitude of T _3:

Figure 4. Free-body diagram for hanging weight. Source: self made.

From the equilibrium condition it follows that:

The angles between the forces are marked in red in the following figure, it can be easily verified that their sum is 360º. Now it is possible to apply Lamy's theorem, since one of the forces and the three angles between them are known:

Figure 5.- In red the angles to apply Lamy's theorem. Source: self made.

T ₁ / sin 127º = W / sin 106º

Therefore: T ₁ = sin 127º (W / sin 106º) = 498.5 N

Again Lamy's theorem is applied to solve for T ₂:

T ₂ / sin 127 = T ₁ / sin 127º

T ₂ = T ₁ = 498.5 N

References

1. Figueroa, D. Series: Physics for Sciences and Engineering. Volume 1. Kinematics. 31-68.
2. Physical. Module 8: Vectors. Recovered from: frtl.utn.edu.ar
3. Hibbeler, R. 2006. Mechanics for Engineers. Static 6th Edition. Continental Publishing Company. 28-66.
4. McLean, W. Schaum Series. Mechanics for Engineers: Statics and Dynamics. 3rd Edition. McGraw Hill. 1-15.
5. Wikipedia. Vector. Recovered from: es.wikipedia.org.