Normal vector of the plane, coordinates of the normal vector of the plane. Normal vector of a straight line What is a normal vector

In order to use the coordinate method, you need to know the formulas well. There are three of them:

At first glance, it looks menacing, but just a little practice - and everything will work great.

Task. Find the cosine of the angle between the vectors a = (4; 3; 0) and b = (0; 12; 5).

Decision. Since we are given the coordinates of the vectors, we substitute them into the first formula:

Task. Write an equation for the plane passing through the points M = (2; 0; 1), N = (0; 1; 1) and K = (2; 1; 0), if it is known that it does not pass through the origin.

Decision. The general equation of the plane: Ax + By + Cz + D = 0, but since the desired plane does not pass through the origin - the point (0; 0; 0) - then we set D = 1. Since this plane passes through the points M, N and K, then the coordinates of these points should turn the equation into a true numerical equality.

Let us substitute the coordinates of the point M = (2; 0; 1) instead of x, y and z. We have:
A 2 + B 0 + C 1 + 1 = 0 ⇒ 2A + C + 1 = 0;

Similarly, for the points N = (0; 1; 1) and K = (2; 1; 0) we obtain the equations:
A 0 + B 1 + C 1 + 1 = 0 ⇒ B + C + 1 = 0;
A 2 + B 1 + C 0 + 1 = 0 ⇒ 2A + B + 1 = 0;

So we have three equations and three unknowns. We compose and solve the system of equations:

We got that the equation of the plane has the form: − 0.25x − 0.5y − 0.5z + 1 = 0.

Task. The plane is given by the equation 7x − 2y + 4z + 1 = 0. Find the coordinates of the vector perpendicular to the given plane.

Decision. Using the third formula, we get n = (7; − 2; 4) - that's all!

Calculation of coordinates of vectors

But what if there are no vectors in the problem - there are only points lying on straight lines, and it is required to calculate the angle between these straight lines? It's simple: knowing the coordinates of the points - the beginning and end of the vector - you can calculate the coordinates of the vector itself.

To find the coordinates of a vector, it is necessary to subtract the coordinates of the beginning from the coordinates of its end.

This theorem works equally on the plane and in space. The expression “subtract coordinates” means that the x coordinate of another point is subtracted from the x coordinate of one point, then the same must be done with the y and z coordinates. Here are some examples:

Task. There are three points in space, given by their coordinates: A = (1; 6; 3), B = (3; − 1; 7) and C = (− 4; 3; − 2). Find the coordinates of vectors AB, AC and BC.

Consider the vector AB: its beginning is at point A, and its end is at point B. Therefore, to find its coordinates, it is necessary to subtract the coordinates of point A from the coordinates of point B:
AB = (3 - 1; - 1 - 6; 7 - 3) = (2; - 7; 4).

Similarly, the beginning of the vector AC is still the same point A, but the end is point C. Therefore, we have:
AC = (− 4 − 1; 3 − 6; − 2 − 3) = (− 5; − 3; − 5).

Finally, to find the coordinates of the vector BC, it is necessary to subtract the coordinates of point B from the coordinates of point C:
BC = (− 4 − 3; 3 − (− 1); − 2 − 7) = (− 7; 4; − 9).

Answer: AB = (2; − 7; 4); AC = (−5;−3;−5); BC = (−7; 4; − 9)

Pay attention to the calculation of the coordinates of the last vector BC: a lot of people make mistakes when they work with negative numbers. This applies to the variable y: point B has the coordinate y = − 1, and point C has y = 3. We get exactly 3 − (− 1) = 4, and not 3 − 1, as many people think. Don't make such stupid mistakes!

Computing Direction Vectors for Straight Lines

If you carefully read problem C2, you will be surprised to find that there are no vectors there. There are only straight lines and planes.

Let's start with straight lines. Everything is simple here: on any line there are at least two different points and, conversely, any two different points define a single line...

Does anyone understand what is written in the previous paragraph? I didn’t understand it myself, so I’ll explain it more simply: in problem C2, lines are always given by a pair of points. If we introduce a coordinate system and consider a vector with the beginning and end at these points, we get the so-called directing vector for a straight line:

Why is this vector needed? The point is that the angle between two straight lines is the angle between their direction vectors. Thus, we are moving from incomprehensible straight lines to specific vectors, the coordinates of which are easily calculated. How easy? Take a look at the examples:

Task. Lines AC and BD 1 are drawn in the cube ABCDA 1 B 1 C 1 D 1 . Find the coordinates of the direction vectors of these lines.

Since the length of the edges of the cube is not specified in the condition, we set AB = 1. Let us introduce a coordinate system with the origin at point A and axes x, y, z directed along the lines AB, AD and AA 1, respectively. The unit segment is equal to AB = 1.

Now let's find the coordinates of the direction vector for the straight line AC. We need two points: A = (0; 0; 0) and C = (1; 1; 0). From here we get the coordinates of the vector AC = (1 - 0; 1 - 0; 0 - 0) = (1; 1; 0) - this is the direction vector.

Now let's deal with the straight line BD 1 . It also has two points: B = (1; 0; 0) and D 1 = (0; 1; 1). We get the direction vector BD 1 = (0 − 1; 1 − 0; 1 − 0) = (− 1; 1; 1).

Answer: AC = (1; 1; 0); BD 1 = (− 1; 1; 1)

Task. In a regular triangular prism ABCA 1 B 1 C 1 , all edges of which are equal to 1, straight lines AB 1 and AC 1 are drawn. Find the coordinates of the direction vectors of these lines.

Let us introduce a coordinate system: the origin is at point A, the x-axis coincides with AB, the z-axis coincides with AA 1 , the y-axis forms the OXY plane with the x-axis, which coincides with the ABC plane.

First, let's deal with the straight line AB 1 . Everything is simple here: we have points A = (0; 0; 0) and B 1 = (1; 0; 1). We get the direction vector AB 1 = (1 − 0; 0 − 0; 1 − 0) = (1; 0; 1).

Now let's find the direction vector for AC 1 . Everything is the same - the only difference is that the point C 1 has irrational coordinates. So, A = (0; 0; 0), so we have:

Answer: AB 1 = (1; 0; 1);

A small but very important note about the last example. If the beginning of the vector coincides with the origin, the calculations are greatly simplified: the coordinates of the vector are simply equal to the coordinates of the end. Unfortunately, this is only true for vectors. For example, when working with planes, the presence of the origin of coordinates on them only complicates the calculations.

Calculation of normal vectors for planes

Normal vectors are not vectors that are doing well, or that feel good. By definition, a normal vector (normal) to a plane is a vector perpendicular to the given plane.

In other words, a normal is a vector perpendicular to any vector in a given plane. Surely you have come across such a definition - however, instead of vectors, it was about straight lines. However, just above it was shown that in the C2 problem one can operate with any convenient object - even a straight line, even a vector.

Let me remind you once again that any plane is defined in space by the equation Ax + By + Cz + D = 0, where A, B, C and D are some coefficients. Without diminishing the generality of the solution, we can assume D = 1 if the plane does not pass through the origin, or D = 0 if it does. In any case, the coordinates of the normal vector to this plane are n = (A; B; C).

So, the plane can also be successfully replaced by a vector - the same normal. Any plane is defined in space by three points. How to find the equation of the plane (and hence the normal), we have already discussed at the very beginning of the article. However, this process causes problems for many, so I will give a couple more examples:

Task. The section A 1 BC 1 is drawn in the cube ABCDA 1 B 1 C 1 D 1 . Find the normal vector for the plane of this section if the origin is at point A and the x, y, and z axes coincide with the edges AB, AD, and AA 1, respectively.

Since the plane does not pass through the origin, its equation looks like this: Ax + By + Cz + 1 = 0, i.e. coefficient D \u003d 1. Since this plane passes through points A 1, B and C 1, the coordinates of these points turn the equation of the plane into the correct numerical equality.

A 0 + B 0 + C 1 + 1 = 0 ⇒ C + 1 = 0 ⇒ C = − 1;

Similarly, for points B = (1; 0; 0) and C 1 = (1; 1; 1) we obtain the equations:
A 1 + B 0 + C 0 + 1 = 0 ⇒ A + 1 = 0 ⇒ A = − 1;
A 1 + B 1 + C 1 + 1 = 0 ⇒ A + B + C + 1 = 0;

But the coefficients A = − 1 and C = − 1 are already known to us, so it remains to find the coefficient B:
B = − 1 − A − C = − 1 + 1 + 1 = 1.

We get the equation of the plane: - A + B - C + 1 = 0, Therefore, the coordinates of the normal vector are n = (- 1; 1; - 1).

Task. A section AA 1 C 1 C is drawn in the cube ABCDA 1 B 1 C 1 D 1. Find the normal vector for the plane of this section if the origin is at point A, and the x, y and z axes coincide with the edges AB, AD and AA 1 respectively.

In this case, the plane passes through the origin, so the coefficient D \u003d 0, and the equation of the plane looks like this: Ax + By + Cz \u003d 0. Since the plane passes through points A 1 and C, the coordinates of these points turn the equation of the plane into the correct numerical equality.

Let us substitute the coordinates of the point A 1 = (0; 0; 1) instead of x, y and z. We have:
A 0 + B 0 + C 1 = 0 ⇒ C = 0;

Similarly, for the point C = (1; 1; 0) we obtain the equation:
A 1 + B 1 + C 0 = 0 ⇒ A + B = 0 ⇒ A = − B;

Let B = 1. Then A = − B = − 1, and the equation of the entire plane is: − A + B = 0. Therefore, the coordinates of the normal vector are n = (− 1; 1; 0).

Generally speaking, in the above problems it is necessary to compose a system of equations and solve it. There will be three equations and three variables, but in the second case one of them will be free, i.e. take arbitrary values. That is why we have the right to put B = 1 - without prejudice to the generality of the solution and the correctness of the answer.

Very often in problem C2 it is required to work with points that divide the segment in half. The coordinates of such points are easily calculated if the coordinates of the ends of the segment are known.

So, let the segment be given by its ends - points A \u003d (x a; y a; z a) and B \u003d (x b; y b; z b). Then the coordinates of the middle of the segment - we denote it by the point H - can be found by the formula:

In other words, the coordinates of the middle of a segment are the arithmetic mean of the coordinates of its ends.

Task. The unit cube ABCDA 1 B 1 C 1 D 1 is placed in the coordinate system so that the x, y and z axes are directed along the edges AB, AD and AA 1 respectively, and the origin coincides with point A. Point K is the midpoint of edge A 1 B one . Find the coordinates of this point.

Since the point K is the middle of the segment A 1 B 1 , its coordinates are equal to the arithmetic mean of the coordinates of the ends. Let's write down the coordinates of the ends: A 1 = (0; 0; 1) and B 1 = (1; 0; 1). Now let's find the coordinates of point K:

Task. The unit cube ABCDA 1 B 1 C 1 D 1 is placed in the coordinate system so that the x, y and z axes are directed along the edges AB, AD and AA 1 respectively, and the origin coincides with point A. Find the coordinates of the point L where they intersect diagonals of the square A 1 B 1 C 1 D 1 .

From the course of planimetry it is known that the point of intersection of the diagonals of a square is equidistant from all its vertices. In particular, A 1 L = C 1 L, i.e. point L is the midpoint of the segment A 1 C 1 . But A 1 = (0; 0; 1), C 1 = (1; 1; 1), so we have:

Answer: L = (0.5; 0.5; 1)

Normal vectors are not vectors that are doing well, or that feel good. By definition, a normal vector (normal) to a plane is a vector perpendicular to the given plane.

· Task . The section A 1 BC 1 is drawn in the cube ABCDA 1 B 1 C 1 D 1 . Find the normal vector for the plane of this section if the origin is at point A and the x, y, and z axes coincide with the edges AB, AD, and AA 1, respectively.

Decision. Since the plane does not pass through the origin, its equation looks like this: Ax + By + Cz + 1 = 0, i.e. coefficient D \u003d 1. Since this plane passes through points A 1, B and C 1, the coordinates of these points turn the equation of the plane into the correct numerical equality.

A 0 + B 0 + C 1 + 1 = 0 ⇒ C + 1 = 0 ⇒ C = − 1;

Similarly, for points B = (1; 0; 0) and C 1 = (1; 1; 1) we obtain the equations:
A 1 + B 0 + C 0 + 1 = 0 ⇒ A + 1 = 0 ⇒ A = − 1;
A 1 + B 1 + C 1 + 1 = 0 ⇒ A + B + C + 1 = 0;

But the coefficients A = − 1 and C = − 1 are already known to us, so it remains to find the coefficient B:
B = − 1 − A − B = − 1 + 1 + 1 = 1.

We get the equation of the plane: - A + B - C + 1 = 0, Therefore, the coordinates of the normal vector are n = (- 1; 1; - 1).

Answer: n = (− 1; 1; − 1)

· Task . A section AA 1 C 1 C is drawn in the cube ABCDA 1 B 1 C 1 D 1. Find the normal vector for the plane of this section if the origin is at point A, and the x, y and z axes coincide with the edges AB, AD and AA 1 respectively.

Decision. In this case, the plane passes through the origin, so the coefficient D \u003d 0, and the equation of the plane looks like this: Ax + By + Cz \u003d 0. Since the plane passes through points A 1 and C, the coordinates of these points turn the equation of the plane into the correct numerical equality.

Let us substitute the coordinates of the point A 1 = (0; 0; 1) instead of x, y and z. We have:
A 0 + B 0 + C 1 = 0 ⇒ C = 0;

Similarly, for the point C = (1; 1; 0) we obtain the equation:
A 1 + B 1 + C 0 = 0 ⇒ A + B = 0 ⇒ A = − B;

Let B = 1. Then A = − B = − 1, and the equation of the entire plane is: − A + B = 0. Therefore, the coordinates of the normal vector are n = (− 1; 1; 0).

Answer: n = (− 1; 1; 0)

Normal vector

Planar surface with two normals

In differential geometry, normal- this is a straight line, orthogonal (perpendicular) to a tangent line to some curve or a tangent plane to some surface. They also talk about normal direction.

Normal vector to the surface at a given point is the unit vector applied to the given point and parallel to the direction of the normal. For each point on a smooth surface, you can specify two normal vectors that differ in direction. If a continuous field of normal vectors can be defined on a surface, then this field is said to define orientation surface (that is, selects one of the sides). If this cannot be done, the surface is called non-orientable.

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The normal vector of a plane is a vector that is perpendicular to the given plane. Obviously, any plane has infinitely many normal vectors. But for the solution of problems, one will be enough for us.

If the plane is given by the general equation , then the vector is the normal vector of the given plane. Just to disgrace. All that needs to be done is to "remove" the coefficients from the equation of the plane.

Three screens are waiting for the promised, let's return to Example No. 1 and check it. I remind you that there it was required to construct the equation of the plane using a point and two vectors. As a result of the solution, we got the equation . We check:

First, we substitute the coordinates of the point into the resulting equation:

The correct equality is obtained, which means that the point really lies in the given plane.

Secondly, we remove the normal vector from the equation of the plane: . Since the vectors are parallel to the plane and the vector is perpendicular to the plane, the following facts must hold: . The perpendicularity of vectors can be easily checked using dot product:

Conclusion: the equation of the plane is found correctly.

In the course of testing, I actually quoted the following statement of the theory: vector parallel to the plane if and only if .

Let's solve an important problem that is related to the lesson:

Example 5

Find the unit normal vector of the plane .

Decision: A unit vector is a vector whose length is one. Let's denote this vector by . Basically the landscape looks like this:

It is quite clear that the vectors are collinear.

First, we remove the normal vector from the equation of the plane: .

How to find the unit vector? To find the unit vector , need every vector coordinate divide by the length of the vector .

Let's rewrite the normal vector in the form and find its length:

According to the above:

Answer:

Check: , which was required to check.

Readers who have carefully studied the last paragraph of the lesson Dot product of vectors probably noticed that unit vector coordinates are exactly the direction cosines of the vector :

Let's digress from the disassembled problem: when you are given an arbitrary non-zero vector, and by condition it is required to find its direction cosines (the last tasks of the lesson Dot product of vectors), then you, in fact, also find a unit vector collinear to the given one.

In fact, two tasks in one bottle.

The need to find a unit normal vector arises in some problems of mathematical analysis.

We figured out the fishing of the normal vector, now we will answer the opposite question.