NORMAL VECTOR: CALCULATION AND EXAMPLE - PHYSICAL - 2025

The normal vector is one that defines the direction perpendicular to some geometric entity under consideration, which can be by a curve, a plane or a surface, for example.

It is a very useful concept in the positioning of a moving particle or some surface in space. In the following graph it is possible to see what the normal vector to an arbitrary curve C is like:

Figure 1. A curve C with the vector normal to the curve at point P. Source: Svjo

Consider a point P on curve C. The point can represent a moving particle that is moving along a C-shaped path. The tangent line to the curve at point P is drawn in red.

Note that vector T is tangent to C at each point, while vector N is perpendicular to T and points to the center of an imaginary circle whose arc is a segment of C. Vectors are denoted in bold type in printed text, for distinguish them from other non-vector quantities.

The vector T always indicates where the particle is moving, therefore it indicates the speed of the particle. On the other hand, the vector N always points in the direction in which the particle is rotating, in this way it indicates the concavity of the curve C.

How to get the normal vector to a plane?

The normal vector is not necessarily a unit vector, that is, a vector whose modulus is 1, but if so, it is called a normal unit vector.

Figure 2. On the left a plane P and the two vectors normal to said plane. On the right the unit vectors in the three directions that determine space. Source: Wikimedia Commons. See page for author

In many applications it is necessary to know the vector normal to a plane rather than a curve. This vector reveals the orientation of said plane in space. For example, consider the plane P (yellow) of the figure:

There are two normal vectors to this plane: n ₁ and n ₂. The use of one or the other will depend on the context in which said plane is found. Obtaining the normal vector to a plane is very simple if the equation of the plane is known:

Here the vector N is expressed in terms of the perpendicular unit vectors i, j and k, directed along the three directions that determine the xyz space, see figure 2 right.

The normal vector from the vector product

A very simple procedure to find the normal vector makes use of the properties of the vector product between two vectors.

As is known, three different points, not collinear with each other, determine a plane P. Now, it is possible to obtain two vectors u and v that belong to said plane having these three points.

Once the vectors are obtained, the vector product u x v is an operation whose result is in turn a vector, which has the property of being perpendicular to the plane determined by u and v.

Known this vector, it is denoted as N, and from it it will be possible to determine the equation of the plane thanks to the equation indicated in the preceding section:

N = u x v

The following figure illustrates the procedure described:

Figure 3. With two vectors and their vector product or cross, the equation of the plane that contains the two vectors is determined. Source: Wikimedia Commons. No machine-readable author provided. M.Romero Schmidtke assumed (based on copyright claims).

Example

Find the equation of the plane determined by the points A (2,1,3); B (0,1,1); C (4.2.1).

Solution

This exercise illustrates the procedure described above. By having 3 points, one of them is chosen as the common origin of two vectors that belong to the plane defined by these points. For example, point A is set as the origin and vectors AB and AC are constructed.

Vector AB is the vector whose origin is point A and whose endpoint is point B. The coordinates of vector AB are determined by respectively subtracting the coordinates of B from the coordinates of A:

We proceed in the same way to find the vector AC:

Calculation of the vector product

There are several procedures to find the cross product between two vectors. This example uses a mnemonic procedure that uses the following figure to find the vector products between the unit vectors i, j and k:

Figure 4. Graph to determine the vector product between the unit vectors. Source: self made.

To begin, it is good to remember that the vector products between parallel vectors are null, therefore:

i x i = 0; j x j = 0; k x k = 0

And since the vector product is another vector perpendicular to the participating vectors, moving in the direction of the red arrow we have:

If you have to move in the opposite direction to the arrow then add a sign (-):

In total it is possible to make 9 vector products with the unit vectors i, j and k, of which 3 will be null.

AB x AC = (-2 i + 0 j -2 k) x (2 i + j -2 k) = -4 (i x i) -2 (i x j) +4 (i x k) +0 (j x i) + 0 (j x j) - 0 (j x k) - 4 (k x i) -2 (k x j) + 4 (k x k) = -2 k -4j -4 j +2 i = 2 i -8 j -2 k

Equation of the plane

The vector N has been determined by the vector product previously calculated:

N = 2 i -8 j -2 k

Therefore a = 2, b = -8, c = -2, the plane sought is:

The value of d remains to be determined. This is easy if the values ​​of any of the points A, B or C that are available are substituted in the equation of the plane. Choosing C for example:

x = 4; y = 2; z = 1

Remains:

In short, the map sought is:

The inquisitive reader may wonder if the same result would have been obtained if instead of doing AB x AC it had been chosen to do AC x AB. The answer is yes, the plane determined by these three points is unique and has two normal vectors, as shown in figure 2.

As for the point selected as the origin of the vectors, there is no problem in choosing any of the other two.

References

1. Figueroa, D. (2005). Series: Physics for Science and Engineering. Volume 1. Kinematics. Edited by Douglas Figueroa (USB). 31- 62.
2. Finding the normal to a plane. Recovered from: web.ma.utexas.edu.
3. Larson, R. (1986). Calculus and Analytical Geometry. Mc Graw Hill. 616-647.
4. Lines and planes in R 3. Recovered from: math.harvard.edu.
5. Normal vector. Recovered from mathworld.wolfram.com.