# Normal vector: calculation and example

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

It’s a very useful concept in positioning a moving particle or some surface in space. In the following graphic, you can see how the normal vector is on an arbitrary curve *C* :

Consider a point P on curve C. The point can represent a moving particle that travels along a C-shaped path. The line tangent 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. them of other non-vector quantities.

The vector **T** always indicates where the particle is moving, so it indicates its velocity. On the other hand, the vector **N** always points in the direction the particle is rotating, indicating the concavity of the curve C.

__How to get the normal vector for an airplane?__

__How to get the normal vector for an airplane?__

The normal vector is not necessarily a unit vector, that is, a vector whose magnitude is 1, but in this case it is called a *unit normal vector* .

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

There are two normal vectors to that plane: *n ** _{1}* and

*n**. The use of one or the other will depend on the context in which this plan is located. Getting the normal vector for a plane is very simple if your equation is known:*

_{2}*ax + by + cz + d = 0* , with *a* , *b* , *c* and *d* real numbers.

Well, a normal vector for this plane is given by:

*N** = a i + b j + c k*

Here, the vector ** N** is expressed in terms of the unit and the perpendicular vectors

**i**,

**j**and

**k**, directed along the three directions that determine the space

*x*and

*z*, see Figure 2 on the right.

**The normal vector of the cross product**

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

As is known, three different points and not collinear with each other determine a plane P. Now, it is possible to obtain two vectors ** u** and

**that belong to that plane having these three points.**

*v*Once the vectors are obtained, the cross *product *** u** x

**is an operation whose result is a vector that has the property of being perpendicular to the plane determined by**

*v***and**

*u***.**

*v*Known by this vector, it is indicated as **N** and, from it, it will be possible to determine the plane equation thanks to the equation indicated in the previous section:

** N** =

**x**

*u*

*v*The following figure illustrates the procedure described:

Figure 3. With two vectors and their cross product, the equation of the plane containing the two vectors is determined. Source: Wikimedia Commons. No machine-readable authors are provided. M.Romero Schmidtke took over (based on copyright claims). [Public domain]

__Example__

__Example__

Find the equation of the plane determined by points A (2,1,3); B (0.1.1); C(4.2,1).

**Solution**

This exercise illustrates the procedure described above. Because it has 3 points, one of them is chosen as a common origin of two vectors belonging to the plane defined by these points. For example, point A is defined as the origin and vectors **AB** and **AC are constructed** .

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

**AB** = (0-2) **i** + (1-1) **j** + (1-3) **k** = -2 **i** + 0 **j** -2 **k**

We looked in the same way to find the **AC** vector :

**AC** = (4/2) **i** + (2/1) **j** + (3/1) **k** = 2 **i** + **j** -2 **k**

**Calculation of the vector product ***AB x AC*

*AB x AC*

There are several procedures for finding the cross product between two vectors. In this example, a mnemonic procedure is used that uses the following figure to find the vector products between the unit vectors **i** , **j** and **k:**

To start with, it’s good to remember that the cross products between parallel vectors are null, so:

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

And since the cross product is another vector perpendicular to the participating vectors, the movement in the direction of the red arrow has:

*i** x j = k ; j x k = i ; k x i = j*

If you need to move in the opposite direction of the arrow, a (-) sign is added:

*j** x i = – k ; k x j = –i ; i x k = – j*

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

*AB** x BC = (-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 ) = -2k -4 j -4 j +2 i = 2 i -8 j -2 k*

**plan equation**

The vector N was determined by the cross product calculated above:

** N** =

*2*

**i**-8**j**-2**k**So a = 2, b = -8, c = -2, the plan sought is:

*What is the square root of 2?*

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

x = 4; y = 2; z = 1

IT’S:

*2.4 – 8.2 – 2.1 + d = 0*

*-10 + d = 0*

*d = 10*

In short, the level sought is:

*2x-8y-2z +10 = 0*

The curious reader may wonder whether the same result would have been obtained if, instead of doing *AB** x AC,* he had 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, it is also okay to choose one of the other two.