Plane equation. How to write an equation for a plane? Mutual arrangement of planes

In the very general case the normal to the surface represents its local curvature, and hence the direction of the specular reflection (Figure 3.5). In relation to our knowledge, we can say that the normal is the vector that determines the orientation of the face (Fig. 3.6).

Rice. 3.5 Fig. 3.6

Many hidden line and surface removal algorithms use only edges and vertices, so in order to combine them with the lighting model, you need to know the approximate value of the normal on the edges and vertices. Let the equations of the planes of polygonal faces be given, then the normal to their common vertex is equal to the average value of the normals to all polygons converging to this vertex. For example, in fig. 3.7 direction of the approximate normal at a point V 1 there is:

n v1 = (a 0 + a 1 + a 4 )i + (b 0 +b 1 +b 4 )j + (c 0 +c 1 +c 4 )k, (3.15)

where a 0 , a 1 , a 4 ,b 0 ,b 1 ,b 4 , c 0 , c 1 , c 4 - coefficients of the equations of the planes of three polygons P 0 , P 1 , P 4 , surrounding V 1 . Note that if you want to find only the direction of the normal, then dividing the result by the number of faces is not necessary.

If the equations of the planes are not given, then the normal to the vertex can be determined by averaging the vector products of all edges intersecting at the vertex. Once again, considering the top V 1 in Fig. 3.7, find the direction of the approximate normal:

n v1 = V 1 V 2 V 1 V 4 +V 1 V 5 V 1 V 2 +V 1 V 4  V 1 V 5 (3.16)

Rice. 3.7 - Approximation of the normal to a polygonal surface

Note that only outer normals are needed. In addition, if the resulting vector is not normalized, then its value depends on the number and area of ​​specific polygons, as well as on the number and length of specific edges. The influence of polygons with larger area and longer ribs.

When the surface normal is used to determine the intensity and a perspective transformation is performed on the image of an object or scene, then the normal should be calculated before the perspective division. Otherwise, the direction of the normal will be distorted, and this will cause the intensity specified by the lighting model to be determined incorrectly.

If the analytical description of the plane (surface) is known, then the normal is calculated directly. Knowing the equation of the plane of each face of the polyhedron, you can find the direction of the outward normal.

If the plane equation is:

then the normal vector to this plane is written as follows:

, (3.18)

where
- unit vectors of axes x,y,z respectively.

Value d is calculated using an arbitrary point belonging to the plane, for example, for a point (
)

Example. Consider a 4-sided flat polygon described by 4 vertices V1(1,0,0), V2(0,1,0), V3(0,0,1) and V4(1,1,1) (see Fig. 3.7).

The plane equation has the form:

x + y + z - 1 = 0.

Let's get the normal to this plane using the vector product of a pair of vectors that are adjacent edges to one of the vertices, for example, V1:

Many hidden line and surface removal algorithms use only edges or vertices, so in order to combine them with the lighting model, you need to know the approximate value of the normal on the edges and vertices.

Let the equations of the planes of the faces of the polyhedron be given, then the normal to their common vertex is equal to the average value of the normals to all faces converging at this vertex.

Namely, about what you see in the title. In essence, this is a "spatial analog" problems of finding a tangent and normals to the graph of a function of one variable, and therefore no difficulties should arise.

Let's start with basic questions: WHAT IS a tangent plane and WHAT IS a normal? Many are aware of these concepts at the level of intuition. The simplest model that comes to mind is a ball on which lies a thin flat cardboard. The cardboard is located as close as possible to the sphere and touches it at a single point. In addition, at the point of contact, it is fixed with a needle sticking straight up.

In theory, there is a rather witty definition of a tangent plane. Imagine an arbitrary surface and the point that belongs to it. It is obvious that a lot passes through the point. spatial lines that belong to this surface. Who has what associations? =) …I personally introduced the octopus. Suppose that each such line has spatial tangent at point .

Definition 1: tangent plane to the surface at a point is plane, containing the tangents to all curves that belong to the given surface and pass through the point .

Definition 2: normal to the surface at a point is straight passing through the given point perpendicular to the tangent plane.

Simple and elegant. By the way, so that you do not die of boredom from the simplicity of the material, a little later I will share with you one elegant secret that allows you to forget about cramming various definitions ONCE AND FOR ALL.

We will get acquainted with the working formulas and the solution algorithm directly on specific example. In the vast majority of problems, it is required to compose both the equation of the tangent plane and the equation of the normal:

Example 1

Decision:if the surface is given by the equation (i.e. implicitly), then the equation of the tangent plane to a given surface at a point can be found by the following formula:

I pay special attention to unusual partial derivatives - their should not be confused with partial derivatives of an implicitly defined function (even though the surface is implicitly defined). When finding these derivatives, one should be guided by rules for differentiating a function of three variables, that is, when differentiating with respect to any variable, the other two letters are considered constants:

Without departing from the cash register, we find the partial derivative at the point:

Similarly:

This was the most unpleasant moment of the decision, in which an error, if not allowed, is constantly imagining. However, there exists effective reception test, which I talked about in the lesson Directional derivative and gradient.

All the “ingredients” have been found, and now it’s up to careful substitution with further simplifications:

– general equation desired tangent plane.

I strongly recommend checking this stage of the decision. First you need to make sure that the coordinates of the touch point really satisfy the found equation:

- true equality.

Now we “remove” the coefficients general equation plane and check them for coincidence or proportionality with the corresponding values. AT this case proportional. As you remember from analytic geometry course, - This normal vector tangent plane, and he - guide vector normal straight line. Let's compose canonical equations normals by point and direction vector:

In principle, the denominators can be reduced by a "two", but there is no particular need for this.

Answer:

It is not forbidden to designate the equations with some letters, however, again - why? Here and so it is very clear what's what.

The next two examples for independent solution. A small "mathematical tongue twister":

Example 2

Find the equations of the tangent plane and the normal to the surface at the point .

And a task interesting from a technical point of view:

Example 3

Compose the equations of the tangent plane and the normal to the surface at a point

At the point.

There is every chance not only to get confused, but also to face difficulties when writing. canonical equations of the line. And the normal equations, as you probably understood, are usually written in this form. Although, due to forgetfulness or ignorance of some nuances, a parametric form is more than acceptable.

Examples of finishing solutions at the end of the lesson.

Is there a tangent plane at any point on the surface? In general, of course not. Classic example- This conical surface and point - the tangents at this point directly form conical surface, and, of course, do not lie in the same plane. It is easy to verify the discord and analytically: .

Another source of problems is the fact non-existence some partial derivative at a point. However, this does not mean that there is no single tangent plane at a given point.

But it was rather popular science than practically significant information, and we return to pressing matters:

How to write the equations of the tangent plane and the normal at a point,
if the surface is given by an explicit function?

Let's rewrite it implicitly:

And by the same principles we find partial derivatives:

Thus, the tangent plane formula is transformed into the following equation:

And correspondingly, canonical equations normals:

As it is easy to guess - it's "real" partial derivatives of a function of two variables at the point , which we used to designate with the letter "Z" and found 100500 times.

Note that in this article it is enough to remember the very first formula, from which, if necessary, it is easy to derive everything else. (of course, having base level training). This is the approach that should be taken when studying exact sciences, i.e. from a minimum of information, one should strive to “pull out” a maximum of conclusions and consequences. "Soobrazhalovka" and already existing knowledge to help! This principle is also useful because it is very likely to save you in a critical situation when you know very little.

Let's work out the "modified" formulas with a couple of examples:

Example 4

Compose the equations of the tangent plane and the normal to the surface at point .

A small overlay here turned out with symbols - now the letter denotes a point of the plane, but what can you do - such a popular letter ....

Decision: we will compose the equation of the desired tangent plane according to the formula:

Let's calculate the value of the function at the point :

Compute partial derivatives of the 1st order at this point:

Thus:

carefully, do not rush:

Let us write the canonical equations of the normal at the point :

Answer:

And a final example for a do-it-yourself solution:

Example 5

Compose the equations of the tangent plane and the normal to the surface at the point.

The final one is because, in fact, I explained all the technical points and there is nothing special to add. Even the functions offered in this task are dull and monotonous - it is almost guaranteed that in practice you will come across a "polynomial", and in this sense, Example No. 2 with the exponent looks like a "black sheep". By the way, it is much more likely to meet the surface, given by the equation and this is another reason why the function was included in the article "second number".

And finally, the promised secret: so how to avoid cramming definitions? (of course, I don’t mean the situation when a student is feverishly cramming something before the exam)

The definition of any concept/phenomenon/object, first of all, gives an answer to next question: WHAT IT IS? (who/such/such/such). Consciously responding to this question, you should try to reflect significant signs, definitely identifying this or that concept/phenomenon/object. Yes, at first it turns out to be somewhat tongue-tied, inaccurate and redundant (the teacher will correct =)), but over time, quite a worthy scientific speech develops.

Practice on the most abstract objects, for example, answer the question: who is Cheburashka? It's not so simple ;-) This is " fairy tale character with big ears, eyes and brown hair"? Far and very far from the definition - you never know there are characters with such characteristics .... But this is much closer to the definition: “Cheburashka is a character invented by the writer Eduard Uspensky in 1966, which ... (listing the main hallmarks)» . Pay attention to how well started

Can be set different ways(one point and a vector, two points and a vector, three points, etc.). It is with this in mind that the equation of the plane can have different kinds. Also, under certain conditions, the planes can be parallel, perpendicular, intersecting, etc. We will talk about this in this article. We will learn how to write the general equation of the plane and not only.

Normal form of the equation

Let's say there is a space R 3 that has a rectangular coordinate system XYZ. We set the vector α, which will be released from the initial point O. Through the end of the vector α we draw the plane P, which will be perpendicular to it.

Denote by P an arbitrary point Q=(x, y, z). We will sign the radius vector of the point Q with the letter p. The length of the vector α is p=IαI and Ʋ=(cosα,cosβ,cosγ).

This is unit vector, which is directed to the side, like the vector α. α, β and γ are the angles that form between the vector Ʋ and the positive directions of the space axes x, y, z, respectively. The projection of some point QϵП onto the vector Ʋ is constant value, which is equal to p: (p,Ʋ) = p(p≥0).

This equation makes sense when p=0. The only thing is that the plane P in this case will intersect the point O (α=0), which is the origin, and the unit vector Ʋ, released from the point O, will be perpendicular to P, regardless of its direction, which means that the vector Ʋ is determined from sign-accurate. The previous equation is the equation of our P plane, expressed in vector form. But in coordinates it will look like this:

P here is greater than or equal to 0. We have found the equation of a plane in space in its normal form.

General Equation

If we multiply the equation in coordinates by any number that is not equal to zero, we get an equation equivalent to the given one, which determines that same plane. It will look like this:

Here A, B, C are numbers that are simultaneously different from zero. This equation is referred to as the general plane equation.

Plane equations. Special cases

Equation in general view may be modified under additional conditions. Let's consider some of them.

Assume that the coefficient A is 0. This means that the given plane is parallel to the given axis Ox. In this case, the form of the equation will change: Ву+Cz+D=0.

Similarly, the form of the equation will change under the following conditions:

• Firstly, if B = 0, then the equation will change to Ax + Cz + D = 0, which will indicate parallelism to the Oy axis.
• Secondly, if С=0, then the equation is transformed into Ах+Ву+D=0, which will indicate parallelism to the given axis Oz.
• Thirdly, if D=0, the equation will look like Ax+By+Cz=0, which will mean that the plane intersects O (the origin).
• Fourth, if A=B=0, then the equation will change to Cz+D=0, which will prove parallel to Oxy.
• Fifth, if B=C=0, then the equation becomes Ax+D=0, which means that the plane to Oyz is parallel.
• Sixth, if A=C=0, then the equation will take the form Ву+D=0, that is, it will report parallelism to Oxz.

Type of equation in segments

In the case when the numbers A, B, C, D are non-zero, the form of equation (0) can be as follows:

x/a + y/b + z/c = 1,

in which a \u003d -D / A, b \u003d -D / B, c \u003d -D / C.

We get as a result It is worth noting that this plane will intersect the Ox axis at a point with coordinates (a,0,0), Oy - (0,b,0), and Oz - (0,0,c).

Taking into account the equation x/a + y/b + z/c = 1, it is easy to visually represent the placement of the plane relative to a given coordinate system.

Normal vector coordinates

The normal vector n to the plane P has coordinates that are the coefficients of the general equation of the given plane, that is, n (A, B, C).

In order to determine the coordinates of the normal n, it is sufficient to know the general equation of a given plane.

When using the equation in segments, which has the form x/a + y/b + z/c = 1, as well as when using the general equation, one can write the coordinates of any normal vector of a given plane: (1/a + 1/b + 1/ with).

It should be noted that the normal vector helps to solve various problems. The most common are tasks that consist in proving the perpendicularity or parallelism of planes, problems in finding angles between planes or angles between planes and lines.

View of the equation of the plane according to the coordinates of the point and the normal vector

A non-zero vector n perpendicular to a given plane is called normal (normal) for a given plane.

Suppose that in the coordinate space (rectangular coordinate system) Oxyz are given:

• point Mₒ with coordinates (xₒ,yₒ,zₒ);
• zero vector n=A*i+B*j+C*k.

It is necessary to compose an equation for a plane that will pass through the point Mₒ perpendicular to the normal n.

In space, we choose any arbitrary point and denote it by M (x y, z). Let the radius vector of any point M (x, y, z) be r=x*i+y*j+z*k, and the radius vector of the point Mₒ (xₒ,yₒ,zₒ) - rₒ=xₒ*i+yₒ *j+zₒ*k. The point M will belong to the given plane if the vector MₒM is perpendicular to the vector n. We write the orthogonality condition using the scalar product:

[MₒM, n] = 0.

Since MₒM \u003d r-rₒ, the vector equation of the plane will look like this:

This equation can take another form. For this, the properties of the scalar product are used, and the left-hand side equations. = - . If denoted as c, then the following equation will be obtained: - c \u003d 0 or \u003d c, which expresses the constancy of the projections onto the normal vector of the radius vectors of the given points that belong to the plane.

Now you can get coordinate view entries of the vector equation of our plane = 0. Since r-rₒ = (x-xₒ)*i + (y-yₒ)*j + (z-zₒ)*k, and n = A*i+B*j+C* k, we have:

It turns out that we have an equation for a plane passing through a point perpendicular to the normal n:

A*(x-xₒ)+B*(y-yₒ)C*(z-zₒ)=0.

View of the plane equation according to the coordinates of two points and a vector collinear to the plane

We define two arbitrary points M′ (x′,y′,z′) and M″ (x″,y″,z″), as well as the vector a (a′,a″,a‴).

Now we can compose an equation for a given plane, which will pass through the available points M′ and M ″, as well as any point M with coordinates (x, y, z) in parallel given vector a.

In this case, the vectors M′M=(x-x′;y-y′;z-z′) and M″M=(x″-x′;y″-y′;z″-z′) must be coplanar with the vector a=(a′,a″,a‴), which means that (M′M, M″M, a)=0.

So, our equation of a plane in space will look like this:

Type of the equation of a plane intersecting three points

Suppose we have three points: (x′, y′, z′), (x″,y″,z″), (x‴,y‴,z‴), which do not belong to the same straight line. It is necessary to write the equation of the plane passing through the given three points. The theory of geometry claims that this kind of plane really exists, only it is the only one and inimitable. Since this plane intersects the point (x′, y′, z′), the form of its equation will be as follows:

Here A, B, C are different from zero at the same time. Also, the given plane intersects two more points: (x″,y″,z″) and (x‴,y‴,z‴). In this regard, the following conditions must be met:

Now we can compose homogeneous system with unknown u, v, w:

In our case x,y or z stands arbitrary point, which satisfies equation (1). Taking into account the equation (1) and the system of equations (2) and (3), the system of equations indicated in the figure above satisfies the vector N (A, B, C), which is non-trivial. That is why the determinant of this system is equal to zero.

Equation (1), which we have obtained, is the equation of the plane. It passes exactly through 3 points, and this is easy to check. To do this, we need to expand our determinant over the elements in the first row. It follows from the existing properties of the determinant that our plane simultaneously intersects three initially given points (x′, y′, z′), (x″,y″,z″), (x‴,y‴,z‴). That is, we have solved the task set before us.

Dihedral angle between planes

A dihedral angle is a spatial geometric figure, formed by two half-planes that emanate from one straight line. In other words, this is the part of space that is limited by these half-planes.

Let's say we have two planes with the following equations:

We know that the vectors N=(A,B,C) and N¹=(A¹,B¹,C¹) are perpendicular according to given planes. In this regard, the angle φ between the vectors N and N¹ is equal to the angle (dihedral), which is between these planes. Scalar product looks like:

NN¹=|N||N¹|cos φ,

precisely because

cosφ= NN¹/|N||N¹|=(AA¹+BB¹+CC¹)/((√(A²+B²+C²))*(√(A¹)²+(B¹)²+(C¹)²)).

It suffices to take into account that 0≤φ≤π.

In fact, two planes that intersect form two (dihedral) angles: φ 1 and φ 2 . Their sum is equal to π (φ 1 + φ 2 = π). As for their cosines, their absolute values ​​are equal, but they differ in signs, that is, cos φ 1 =-cos φ 2. If in equation (0) we replace A, B and C with the numbers -A, -B and -C, respectively, then the equation that we get will determine the same plane, the only angle φ in cos equationφ= NN 1 /|N||N 1 | will be replaced by π-φ.

Perpendicular plane equation

Planes are called perpendicular if the angle between them is 90 degrees. Using the material outlined above, we can find the equation of a plane perpendicular to another. Let's say we have two planes: Ax+By+Cz+D=0 and A¹x+B¹y+C¹z+D=0. We can state that they will be perpendicular if cosφ=0. This means that NN¹=AA¹+BB¹+CC¹=0.

Parallel plane equation

Parallel are two planes that do not contain common points.

The condition (their equations are the same as in the previous paragraph) is that the vectors N and N¹, which are perpendicular to them, are collinear. And this means that the following conditions proportionality:

A/A¹=B/B¹=C/C¹.

If the proportionality conditions are extended - A/A¹=B/B¹=C/C¹=DD¹,

this indicates that these planes coincide. This means that the equations Ax+By+Cz+D=0 and A¹x+B¹y+C¹z+D¹=0 describe one plane.

Distance to plane from point

Let's say we have a plane P, which is given by equation (0). It is necessary to find the distance to it from the point with coordinates (xₒ,yₒ,zₒ)=Qₒ. To do this, you need to bring the equation of the plane P into normal form:

(ρ,v)=p (p≥0).

In this case, ρ(x,y,z) is the radius vector of our point Q located on P, p is the length of the perpendicular to P that was released from the zero point, v is the unit vector that is located in the a direction.

The difference ρ-ρº of the radius vector of some point Q=(x,y,z) belonging to P, as well as the radius vector of a given point Q 0 =(xₒ,yₒ,zₒ) is such a vector, absolute value whose projection on v is equal to the distance d, which must be found from Q 0 \u003d (xₒ, yₒ, zₒ) to P:

D=|(ρ-ρ 0 ,v)|, but

(ρ-ρ 0 ,v)= (ρ,v)-(ρ 0 ,v) =р-(ρ 0 ,v).

So it turns out

d=|(ρ 0 ,v)-p|.

Thus we will find absolute value the resulting expression, that is, the required d.

Using the language of parameters, we get the obvious:

d=|Axₒ+Vuₒ+Czₒ|/√(A²+B²+C²).

If a given point Q 0 is on the other side of the P plane, as well as the origin, then between the vector ρ-ρ 0 and v is therefore:

d=-(ρ-ρ 0 ,v)=(ρ 0 ,v)-p>0.

In the case when the point Q 0, together with the origin, is located on the same side of P, then the angle created is acute, that is:

d \u003d (ρ-ρ 0, v) \u003d p - (ρ 0, v)>0.

As a result, it turns out that in the first case (ρ 0 ,v)> р, in the second (ρ 0 ,v)<р.

Tangent plane and its equation

The tangent plane to the surface at the point of contact Mº is the plane containing all possible tangents to the curves drawn through this point on the surface.

With this form of the surface equation F (x, y, z) \u003d 0, the equation of the tangent plane at the tangent point Mº (xº, yº, zº) will look like this:

F x (xº, yº, zº)(x- xº)+ F x (xº, yº, zº)(y-yº)+ F x (xº, yº, zº)(z-zº)=0.

If you specify the surface in explicit form z=f (x, y), then the tangent plane will be described by the equation:

z-zº = f(xº, yº)(x- xº)+f(xº, yº)(y-yº).

Intersection of two planes

In the coordinate system (rectangular) Oxyz is located, two planes П′ and П″ are given, which intersect and do not coincide. Since any plane located in a rectangular coordinate system is determined by a general equation, we will assume that P′ and P″ are given by the equations A′x+B′y+C′z+D′=0 and A″x+B″y+ С″z+D″=0. In this case, we have the normal n′ (A′, B′, C′) of the P′ plane and the normal n″ (A″, B″, C″) of the P″ plane. Since our planes are not parallel and do not coincide, these vectors are not collinear. Using the language of mathematics, we can write this condition as follows: n′≠ n″ ↔ (A′, B′, C′) ≠ (λ*A″,λ*B″,λ*C″), λϵR. Let the line that lies at the intersection of P′ and P″ be denoted by the letter a, in this case a = P′ ∩ P″.

a is a straight line consisting of the set of all points of (common) planes П′ and П″. This means that the coordinates of any point belonging to the line a must simultaneously satisfy the equations A′x+B′y+C′z+D′=0 and A″x+B″y+C″z+D″=0. This means that the coordinates of the point will be a particular solution of the following system of equations:

As a result, it turns out that the (general) solution of this system of equations will determine the coordinates of each of the points of the straight line, which will act as the intersection point of П′ and П″, and determine the straight line a in the coordinate system Oxyz (rectangular) in space.

What is normal? In simple terms, a normal is a perpendicular. That is, the normal vector of a line is perpendicular to the given line. It is obvious that any straight line has an infinite number of them (as well as directing vectors), and all the normal vectors of the straight line will be collinear (codirectional or not - it does not matter).

Dealing with them will be even easier than with direction vectors:

If a straight line is given by a general equation in a rectangular coordinate system, then the vector is the normal vector of this straight line.

If the coordinates of the direction vector have to be carefully “pulled out” from the equation, then the coordinates of the normal vector are simply “removed”.

The normal vector is always orthogonal to the direction vector of the line. Let's make sure that these vectors are orthogonal using the scalar product:

I will give examples with the same equations as for the direction vector:

Is it possible to write an equation of a straight line, knowing one point and a normal vector? If the normal vector is known, then the direction of the straightest line is also uniquely determined - this is a “rigid structure” with an angle of 90 degrees.

How to write an equation of a straight line given a point and a normal vector?

If some point belonging to the line and the normal vector of this line are known, then the equation of this line is expressed by the formula:

Compose the equation of a straight line given a point and a normal vector. Find the direction vector of the straight line.

Solution: Use the formula:

The general equation of the straight line is obtained, let's check:

1) "Remove" the coordinates of the normal vector from the equation: - yes, indeed, the original vector is obtained from the condition (or the vector should be collinear to the original vector).

2) Check if the point satisfies the equation:

True equality.

After we are convinced that the equation is correct, we will complete the second, easier part of the task. We pull out the direction vector of the straight line:

Answer:

In the drawing, the situation is as follows:

For the purposes of training, a similar task for an independent solution:

Compose the equation of a straight line given a point and a normal vector. Find the direction vector of the straight line.

The final section of the lesson will be devoted to less common, but also important types of equations of a straight line in a plane

Equation of a straight line in segments.
Equation of a straight line in parametric form

The equation of a straight line in segments has the form , where are nonzero constants. Some types of equations cannot be represented in this form, for example, direct proportionality (since the free term is zero and there is no way to get one on the right side).

This is, figuratively speaking, a "technical" type of equation. The usual task is to represent the general equation of a straight line as an equation of a straight line in segments. Why is it convenient? The equation of a straight line in segments allows you to quickly find the points of intersection of a straight line with coordinate axes, which can be very important in some problems of higher mathematics.

Find the point of intersection of the line with the axis. We reset the “y”, and the equation takes the form . The desired point is obtained automatically: .

Same with axis is the point where the line intersects the y-axis.

The actions that I have just explained in detail are performed verbally.

Given a straight line. Compose the equation of a straight line in segments and determine the points of intersection of the graph with the coordinate axes.

Solution: Let's bring the equation to the form . First, we move the free term to the right side:

To get a unit on the right, we divide each term of the equation by -11:

We make fractions three-story:

The points of intersection of the straight line with the coordinate axes surfaced:

Answer:

It remains to attach a ruler and draw a straight line.

It is easy to see that this straight line is uniquely determined by the red and green segments, hence the name - “the equation of a straight line in segments”.

Of course, the points are not so difficult to find from the equation, but the problem is still useful. The considered algorithm will be required to find the points of intersection of the plane with the coordinate axes, to bring the second-order line equation to the canonical form, and in some other problems. Therefore, a couple of straight lines for an independent solution:

Compose the equation of a straight line in segments and determine the points of its intersection with the coordinate axes.

Solutions and answers at the end. Do not forget that if you wish, you can draw everything.

How to write parametric equations for a straight line?

The parametric equations of a straight line are more relevant for straight lines in space, but without them our abstract will be orphaned.

If some point belonging to the line and the direction vector of this line are known, then the parametric equations of this line are given by the system:

Compose parametric equations of a straight line by a point and a direction vector

The solution ended before it could start:

The parameter "te" can take any value from "minus infinity" to "plus infinity", and each parameter value corresponds to a specific point of the plane. For example, if , then we get a point .

Inverse problem: how to check if a condition point belongs to a given line?

Let us substitute the coordinates of the point into the obtained parametric equations:

From both equations it follows that , that is, the system is consistent and has a unique solution.

Let's consider more meaningful tasks:

Compose parametric equations of a straight line

Solution: By condition, the straight line is given in general form. In order to compose the parametric equations of a straight line, you need to know its directing vector and some point belonging to this straight line.

Let's find the direction vector:

Now you need to find some point belonging to the line (any one will do), for this purpose it is convenient to rewrite the general equation in the form of an equation with a slope:

It begs, of course, the point

We compose the parametric equations of the straight line:

And finally, a small creative task for an independent solution.

Compose parametric equations of a straight line if the point belonging to it and the normal vector are known

The task can be done in more than one way. One of the versions of the solution and the answer at the end.

Solutions and answers:

Example 2: Solution: Find the slope:

We compose the equation of a straight line by a point and a slope:

Answer:

Example 4: Solution: We will compose the equation of a straight line according to the formula:

Answer:

Example 6: Solution: Use the formula:

Answer: (y-axis)

Example 8: Decision: Let's make the equation of a straight line on two points:

Multiply both sides by -4:

And divide by 5:

Answer:

Example 10: Decision: Use the formula:

We reduce by -2:

Direction vector direct:
Answer:

Example 12:
a) Decision: Let's transform the equation:

Thus:

Answer:

b) Decision: Let's transform the equation:

Thus:

Answer:

Example 15: Decision: First, we write the general equation of a straight line given a point and the normal vector :

Multiply by 12:

We multiply by 2 more so that after opening the second bracket, get rid of the fraction:

Direction vector direct:
We compose the parametric equations of the straight line by the point and direction vector :
Answer:

The simplest problems with a straight line on a plane.
Mutual arrangement of lines. Angle between lines

We continue to consider these infinite-infinite lines.

How to find the distance from a point to a line?
How to find the distance between two parallel lines?
How to find the angle between two lines?

Mutual arrangement of two straight lines

Consider two straight lines given by equations in general form:

The case when the hall sings along in chorus. Two lines can:

1) match;

2) be parallel: ;

3) or intersect at a single point: .

Please remember the mathematical sign of the intersection , it will occur very often. The entry means that the line intersects with the line at the point.

How to determine the relative position of two lines?

Let's start with the first case:

Two lines coincide if and only if their respective coefficients are proportional, that is, there is such a number of "lambda" that the equalities hold

Let's consider straight lines and compose three equations from the corresponding coefficients: . From each equation it follows that, therefore, these lines coincide.

Indeed, if all the coefficients of the equation multiply by -1 (change signs), and all the coefficients of the equation reduce by 2, you get the same equation: .

The second case when the lines are parallel:

Two lines are parallel if and only if their coefficients at the variables are proportional: , but .

As an example, consider two straight lines. We check the proportionality of the corresponding coefficients for the variables :

However, it is clear that .

And the third case, when the lines intersect:

Two lines intersect if and only if their coefficients at the variables are NOT proportional, that is, there is NOT such a value of "lambda" that the equalities are fulfilled

So, for straight lines we will compose a system:

It follows from the first equation that , and from the second equation: , which means that the system is inconsistent (there are no solutions). Thus, the coefficients at the variables are not proportional.

Conclusion: lines intersect

In practical problems, the solution scheme just considered can be used. By the way, it is very similar to the algorithm for checking vectors for collinearity. But there is a more civilized package:

Find out the relative position of the lines:

The solution is based on the study of directing vectors of straight lines:

a) From the equations we find the direction vectors of the lines: .

, so the vectors are not collinear and the lines intersect.

b) Find the direction vectors of the lines:

The lines have the same direction vector, which means they are either parallel or the same. Here the determinant is not necessary.

Obviously, the coefficients of the unknowns are proportional, while .

Let's find out if the equality is true:

Thus,

c) Find the direction vectors of the lines:

Let's calculate the determinant, composed of the coordinates of these vectors:
, therefore, the direction vectors are collinear. The lines are either parallel or coincide.

The proportionality coefficient "lambda" can be found directly by the ratio of collinear direction vectors. However, it is also possible through the coefficients of the equations themselves: .

Now let's find out if the equality is true. Both free terms are zero, so:

The resulting value satisfies this equation (any number generally satisfies it).

Thus, the lines coincide.

How to draw a line parallel to a given one?

The straight line is given by the equation . Write an equation for a parallel line that passes through the point.

Solution: Denote the unknown straight line by the letter . What does the condition say about it? The line passes through the point. And if the lines are parallel, then it is obvious that the directing vector of the line "ce" is also suitable for constructing the line "te".

We take out the direction vector from the equation:

The geometry of the example looks simple:

Analytical verification consists of the following steps:

1) We check that the lines have the same direction vector (if the equation of the line is not properly simplified, then the vectors will be collinear).

2) Check if the point satisfies the resulting equation.

Analytical verification in most cases is easy to perform orally. Look at the two equations and many of you will quickly figure out how the lines are parallel without any drawing.

Examples for self-solving today will be creative.

Write an equation for a line passing through a point parallel to the line if

The shortest way is at the end.

How to find the point of intersection of two lines?

If straight intersect at the point , then its coordinates are the solution of the system of linear equations

How to find the point of intersection of lines? Solve the system.

So much for the geometric meaning of a system of two linear equations with two unknowns - these are two intersecting (most often) straight lines on a plane.

Find the point of intersection of lines

Solution: There are two ways to solve - graphical and analytical.

The graphical way is to simply draw the given lines and find out the point of intersection directly from the drawing:

Here is our point: . To check, you should substitute its coordinates into each equation of a straight line, they should fit both there and there. In other words, the coordinates of a point are the solution of the system . In fact, we have considered a graphical method for solving a system of linear equations with two equations, two unknowns.

The graphical method, of course, is not bad, but there are noticeable disadvantages. No, the point is not that seventh graders decide this way, the point is that it will take time to make a correct and EXACT drawing. In addition, some lines are not so easy to construct, and the intersection point itself may be somewhere in the thirtieth kingdom outside the notebook sheet.

Therefore, it is more expedient to search for the intersection point by the analytical method. Let's solve the system:

To solve the system, the method of termwise addition of equations was used.

The verification is trivial - the coordinates of the intersection point must satisfy each equation of the system.

Find the point of intersection of the lines if they intersect.

This is a do-it-yourself example. The task can be conveniently divided into several stages. Analysis of the condition suggests that it is necessary:
1) Write the equation of a straight line.
2) Write the equation of a straight line.
3) Find out the relative position of the lines.
4) If the lines intersect, then find the point of intersection.

The development of an action algorithm is typical for many geometric problems, and I will repeatedly focus on this.

Full solution and answer at the end:

Perpendicular lines. The distance from a point to a line.
Angle between lines

How to draw a line perpendicular to a given one?

The straight line is given by the equation . Write an equation for a perpendicular line passing through a point.

Solution: It is known by assumption that . It would be nice to find the direction vector of the straight line. Since the lines are perpendicular, the trick is simple:

From the equation we “remove” the normal vector: , which will be the directing vector of the straight line.

We compose the equation of a straight line by a point and a directing vector:

Answer:

Let's unfold the geometric sketch:

Analytical verification of the solution:

1) Extract the direction vectors from the equations and using the scalar product of vectors, we conclude that the lines are indeed perpendicular: .

By the way, you can use normal vectors, it's even easier.

2) Check if the point satisfies the resulting equation .

Verification, again, is easy to perform verbally.

Find the point of intersection of perpendicular lines, if the equation is known and dot.

This is a do-it-yourself example. There are several actions in the task, so it is convenient to arrange the solution point by point.

Distance from point to line

The distance in geometry is traditionally denoted by the Greek letter "p", for example: - the distance from the point "m" to the straight line "d".

Distance from point to line is expressed by the formula

Find the distance from a point to a line

Solution: all you need to do is carefully plug the numbers into the formula and do the calculations:

Answer:

Let's execute the drawing:

The distance found from the point to the line is exactly the length of the red segment. If you make a drawing on checkered paper on a scale of 1 unit. \u003d 1 cm (2 cells), then the distance can be measured with an ordinary ruler.

Consider another task according to the same drawing:

How to construct a point symmetrical about a straight line?

The task is to find the coordinates of the point , which is symmetrical to the point with respect to the line . I propose to perform the actions on your own, however, I will outline the solution algorithm with intermediate results:

1) Find a line that is perpendicular to a line.

2) Find the point of intersection of the lines: .

In geometry, the angle between two straight lines is taken as the SMALLER angle, from which it automatically follows that it cannot be obtuse. In the figure, the angle indicated by the red arc is not considered to be the angle between intersecting lines. And its “green” neighbor or oppositely oriented “raspberry” corner is considered as such.

If the lines are perpendicular, then any of the 4 angles can be taken as the angle between them.

How are the angles different? Orientation. First, the direction of "scrolling" the corner is fundamentally important. Secondly, a negatively oriented angle is written with a minus sign, for example, if .

Why did I say this? It seems that you can get by with the usual concept of an angle. The fact is that in the formulas by which we will find the angles, a negative result can easily be obtained, and this should not take you by surprise. An angle with a minus sign is no worse, and has a very specific geometric meaning. In the drawing for a negative angle, it is imperative to indicate its orientation (clockwise) with an arrow.

Based on the foregoing, the solution is conveniently formalized in two steps:

1) Calculate the scalar product of directing vectors of straight lines:
so the lines are not perpendicular.

2) We find the angle between the lines by the formula:

Using the inverse function, it is easy to find the angle itself. In this case, we use the oddness of the arc tangent:

Answer:

In the answer, we indicate the exact value, as well as the approximate value (preferably both in degrees and in radians), calculated using a calculator.

Well, minus, so minus, it's okay. Here is a geometric illustration:

It is not surprising that the angle turned out to be of a negative orientation, because in the condition of the problem the first number is a straight line and the “twisting” of the angle began precisely from it.

There is also a third solution. The idea is to calculate the angle between the direction vectors of the lines:

Here we are not talking about an oriented angle, but “just about an angle”, that is, the result will certainly be positive. The catch is that you can get an obtuse angle (not the one you need). In this case, you will have to make a reservation that the angle between the lines is a smaller angle, and subtract the resulting arc cosine from “pi” radians (180 degrees).

Find the angle between the lines.

This is a do-it-yourself example. Try to solve it in two ways.

Solutions and answers:

Example 3: Solution: Find the direction vector of the straight line:

We will compose the equation of the desired straight line using the point and the direction vector

Note: here the first equation of the system is multiplied by 5, then the 2nd is subtracted term by term from the 1st equation.
Answer:

To study the equations of a straight line, it is necessary to have a good understanding of the algebra of vectors. It is important to find the direction vector and the normal vector of the line. This article will consider the normal vector of a straight line with examples and drawings, finding its coordinates if the equations of straight lines are known. A detailed solution will be considered.

Yandex.RTB R-A-339285-1

To make the material easier to digest, you need to understand the concepts of line, plane and definitions that are associated with vectors. First, let's get acquainted with the concept of a straight line vector.

Definition 1

Normal line vector any non-zero vector that lies on any line perpendicular to the given one is called.

It is clear that there is an infinite set of normal vectors located on a given line. Consider the figure below.

We get that the line is perpendicular to one of the two given parallel lines, then its perpendicularity extends to the second parallel line. Hence we obtain that the sets of normal vectors of these parallel lines coincide. When the lines a and a 1 are parallel, and n → is considered a normal vector of the line a , it is also considered a normal vector for the line a 1 . When the line a has a direct vector, then the vector t · n → is non-zero for any value of the parameter t, and is also normal for the line a.

Using the definition of normal and direction vectors, one can conclude that the normal vector is perpendicular to the direction. Consider an example.

If the plane O x y is given, then the set of vectors for O x is the coordinate vector j → . It is considered non-zero and belongs to the coordinate axis O y, perpendicular to O x. The whole set of normal vectors with respect to O x can be written as t · j → , t ∈ R , t ≠ 0 .

The rectangular system O x y z has a normal vector i → related to the line O z . The vector j → is also considered normal. This shows that any non-zero vector located in any plane and perpendicular to O z is considered normal for O z .

Coordinates of the normal vector of the line - finding the coordinates of the normal vector of the line from the known equations of the line

When considering a rectangular coordinate system O x y, we find that the equation of a straight line on a plane corresponds to it, and the determination of normal vectors is made by coordinates. If the equation of the straight line is known, but it is necessary to find the coordinates of the normal vector, then it is necessary to identify the coefficients from the equation A x + B y + C = 0, which correspond to the coordinates of the normal vector of the given straight line.

Example 1

A straight line of the form 2 x + 7 y - 4 = 0 _ is given, find the coordinates of the normal vector.

Decision

By condition, we have that the straight line was given by the general equation, which means that it is necessary to write out the coefficients, which are the coordinates of the normal vector. Hence, the coordinates of the vector have the value 2 , 7 .

Answer: 2 , 7 .

There are times when A or B from an equation is zero. Let's consider the solution of such a task with an example.

Example 2

Specify the normal vector for the given line y - 3 = 0 .

Decision

By condition, we are given the general equation of a straight line, which means we write it in this way 0 · x + 1 · y - 3 = 0. Now we can clearly see the coefficients, which are the coordinates of the normal vector. So, we get that the coordinates of the normal vector are 0 , 1 .

Answer: 0 , 1 .

If an equation is given in segments of the form x a + y b \u003d 1 or an equation with a slope y \u003d k x + b, then it is necessary to reduce to a general equation of a straight line, where you can find the coordinates of the normal vector of this straight line.

Example 3

Find the coordinates of the normal vector if the equation of the straight line x 1 3 - y = 1 is given.

Decision

First you need to move from the equation in the intervals x 1 3 - y = 1 to a general equation. Then we get that x 1 3 - y = 1 ⇔ 3 x - 1 y - 1 = 0 .

This shows that the coordinates of the normal vector have the value 3 , - 1 .

Answer: 3 , - 1 .

If the line is defined by the canonical equation of the line on the plane x - x 1 a x = y - y 1 a y or by the parametric x = x 1 + a x · λ y = y 1 + a y · λ , then getting the coordinates becomes more complicated. According to these equations, it can be seen that the coordinates of the direction vector will be a → = (a x , a y) . The possibility of finding the coordinates of the normal vector n → is possible due to the condition that the vectors n → and a → are perpendicular.

It is possible to obtain the coordinates of a normal vector by reducing the canonical or parametric equations of a straight line to a general one. Then we get:

x - x 1 a x = y - y 1 a y ⇔ a y (x - x 1) = a x (y - y 1) ⇔ a y x - a x y + a x y 1 - a y x 1 x = x 1 + a x λ y = y 1 + a y λ ⇔ x - x 1 a x = y - y 1 a y ⇔ a y x - a x y + a x y 1 - a y x 1 = 0

For the solution, you can choose any convenient way.

Example 4

Find the normal vector of the given line x - 2 7 = y + 3 - 2 .

Decision

From the straight line x - 2 7 = y + 3 - 2 it is clear that the direction vector will have coordinates a → = (7 , - 2) . The normal vector n → = (n x , n y) of the given line is perpendicular to a → = (7 , - 2) .

Let's find out what the scalar product is equal to. To find the scalar product of vectors a → = (7 , - 2) and n → = (n x , n y) we write a → , n → = 7 · n x - 2 · n y = 0 .

The value of n x is arbitrary, you should find n y . If n x = 1, then we get that 7 · 1 - 2 · n y = 0 ⇔ n y = 7 2 .

Hence, the normal vector has coordinates 1 , 7 2 .

The second way of solving comes down to the fact that it is necessary to come to the general form of the equation from the canonical one. For this, we transform

x - 2 7 = y + 3 - 2 ⇔ 7 (y + 3) = - 2 (x - 2) ⇔ 2 x + 7 y - 4 + 7 3 = 0

The result of normal vector coordinates is 2 , 7 .

Answer: 2, 7 or 1 , 7 2 .

Example 5

Specify the coordinates of the normal vector of the line x = 1 y = 2 - 3 · λ .

Decision

First you need to perform a transformation to go to the general form of a straight line. Let's do:

x = 1 y = 2 - 3 λ ⇔ x = 1 + 0 λ y = 2 - 3 λ ⇔ λ = x - 1 0 λ = y - 2 - 3 ⇔ x - 1 0 = y - 2 - 3 ⇔ ⇔ - 3 (x - 1) = 0 (y - 2) ⇔ - 3 x + 0 y + 3 = 0

This shows that the coordinates of the normal vector are - 3 , 0 .

Answer: - 3 , 0 .

Consider ways to find the coordinates of a normal vector in the equation of a straight line in space, given by a rectangular coordinate system O x y z.

When a line is given by the equations of intersecting planes A 1 x + B 1 y + C 1 z + D 1 = 0 and A 2 x + B 2 y + C 2 z + D 2 = 0 , then the normal vector of the plane refers to A 2 x + B 2 y + C 2 z + D 2 = 0 and A 2 x + B 2 y + C 2 z + D 2 = 0, then we get the vectors in the form n 1 → = (A 1 , B 1 , C 1) and n 2 → = (A 2 , B 2 , C 2) .

When the line is defined using the canonical equation of space, having the form x - x 1 a x = y - y 1 a y = z - z 1 a z or parametric, having the form x = x 1 + a x λ y = y 1 + a y λ z = z 1 + a z · λ , hence a x , a y and a z are considered to be the coordinates of the direction vector of the given straight line. Any non-zero vector can be normal for a given line, and be perpendicular to the vector a → = (a x , a y , a z) . It follows that the coordinates of the normal with parametric and canonical equations are found using the coordinates of the vector, which is perpendicular to the given vector a → = (a x, a y, a z) .

If you notice a mistake in the text, please highlight it and press Ctrl+Enter