Lines ab and bc are parallel intersecting and crossing. Definition

Theorem. If one line lies in a given plane, and another line intersects this plane at a point that does not belong to the first line, then these two lines intersect. Sign of intersecting lines Proof. Let line a lie in a plane, and line b intersect the plane at a point B not belonging to line a. If the lines a and b lie in the same plane, then the point B would also lie in this plane. Since there is only one plane passing through the line and a point outside this line, this plane must be a plane. But then the line b would lie in a plane, which contradicts the condition. Therefore, the lines a and b do not lie in the same plane, i.e. interbreed.

How many pairs of skew lines are there that contain the edges of a regular triangular prism? Solution: For each base edge, there are three edges intersecting with it. For each side edge, there are two edges intersecting with it. Therefore, the desired number of pairs of skew lines is Exercise 5

How many pairs of skew lines are there that contain the edges of a regular hexagonal prism? Solution: Each base edge participates in 8 pairs of skew lines. Each side edge participates in 8 pairs of intersecting lines. Therefore, the desired number of pairs of skew lines is Exercise 6

If two lines in space have a common point, then these two lines are said to intersect. In the following figure, lines a and b intersect at point A. Lines a and c do not intersect.

Any two lines either have only one common point, or do not have common points.

Parallel lines

Two lines in space are called parallel if they lie in the same plane and do not intersect. To designate parallel lines use a special icon - ||.

The notation a||b means that line a is parallel to line b. In the figure above, lines a and c are parallel.

Parallel line theorem

Through any point in space that does not lie on a given line, there passes a line parallel to the given line and, moreover, only one.

Crossed lines

Two lines that lie in the same plane can either intersect or be parallel. But in space, two straight lines do not have to belong to the same plane. They can be located in two different planes.

Obviously, lines located in different planes do not intersect and are not parallel lines. Two lines that do not lie in the same plane are called crossing lines.

The following figure shows two intersecting lines a and b that lie in different planes.

Sign and the skew lines theorem

If one of two lines lies in a certain plane, and the other line intersects this plane at a point not lying on the first line, then these lines are skew.

Crossing lines theorem: through each of the two intersecting lines there passes a plane parallel to the other line, and moreover, only one.

Thus, we have considered all possible cases of mutual arrangement of lines in space. There are only three of them.

1. The lines intersect. (That is, they have only one common point.)

2. Lines are parallel. (That is, they do not have common points and lie in the same plane.)

3. Straight lines intersect. (That is, they are located in different planes.)

In less than a minute, I created a new Verdov file and continued on such an exciting topic. You need to catch the moments of the working mood, so there will be no lyrical introduction. There will be prosaic spanking =)

The two straight spaces can:

1) interbreed;

2) intersect at the point ;

3) be parallel;

4) match.

Case #1 is fundamentally different from the other cases. Two lines intersect if they do not lie in the same plane.. Raise one arm up and stretch the other arm forward - here is an example of intersecting lines. In points 2-4, the lines necessarily lie in one plane.

How to find out the relative position of lines in space?

Consider two straight spaces:

is a straight line given by a point and a directing vector ;
is a straight line defined by a point and a direction vector .

For a better understanding, let's make a schematic drawing:

The drawing shows skew lines as an example.

How to deal with these lines?

Since the points are known, it is easy to find the vector.

If straight interbreed, then the vectors not coplanar(see lesson Linear (non) dependence of vectors. Vector basis), which means that the determinant composed of their coordinates is nonzero. Or, which is actually the same, will be different from zero: .

In cases No. 2-4, our construction “falls” into one plane, while the vectors coplanar, and the mixed product of linearly dependent vectors is equal to zero: .

We expand the algorithm further. Let's pretend that , therefore, the lines either intersect, or are parallel, or coincide.

If the direction vectors collinear, then the lines are either parallel or coincide. As a final nail, I propose the following technique: we take any point of one straight line and substitute its coordinates into the equation of the second straight line; if the coordinates “approached”, then the lines coincide, if they “did not approach”, then the lines are parallel.

The course of the algorithm is unpretentious, but practical examples still do not interfere:

Example 11

Find out the relative position of two lines

Solution: as in many problems of geometry, it is convenient to arrange the solution point by point:

1) We extract points and direction vectors from the equations:

2) Find the vector:

Thus, the vectors are coplanar, which means that the lines lie in the same plane and can intersect, be parallel or coincide.

4) Check the direction vectors for collinearity.

Let's compose a system from the corresponding coordinates of these vectors:

From everyone The equation implies that , therefore, the system is consistent, the corresponding coordinates of the vectors are proportional, and the vectors are collinear.

Conclusion: lines are parallel or coincide.

5) Find out if the lines have common points. Let's take a point belonging to the first straight line and substitute its coordinates into the equations of the straight line:

Thus, the lines have no common points, and there is nothing left for them but to be parallel.

Answer:

An interesting example to solve on your own:

Example 12

Find out the relative position of the lines

This is a do-it-yourself example. Note that the second line has the letter as a parameter. Logically. In the general case, these are two different lines, so each line has its own parameter.

And again I urge you not to skip examples, I will smack the tasks I propose are far from random ;-)

Problems with a straight line in space

In the final part of the lesson, I will try to consider the maximum number of different problems with spatial lines. In this case, the started order of the story will be respected: first we will consider problems with intersecting lines, then with intersecting lines, and at the end we will talk about parallel lines in space. However, I must say that some of the tasks of this lesson can be formulated for several cases of straight lines at once, and in this regard, the division of the section into paragraphs is somewhat arbitrary. There are simpler examples, there are more complex examples, and hopefully everyone will find what they need.

Crossed lines

I remind you that lines intersect if there is no plane in which they both lie. When I was thinking about the practice, a monster task came to mind, and now I am glad to present to your attention a dragon with four heads:

Example 13

Given are straight lines. Required:

a) prove that the lines intersect;

b) find the equations of the line passing through the point perpendicular to the given lines;

c) compose the equations of a straight line that contains common perpendicular intersecting lines;

d) find the distance between the lines.

Solution: The road will be mastered by the walking one:

a) Let us prove that the lines intersect. Let's find the points and direction vectors of these straight lines:

Let's find the vector:

Compute mixed product of vectors:

So the vectors not coplanar, which means that the lines intersect, which was to be proved.

Probably, everyone has long noticed that for skew lines, the verification algorithm turns out to be the shortest.

b) Let's find the equations of the line that passes through the point and is perpendicular to the lines. Let's make a schematic drawing:

For variety, I posted a direct BEHIND straight lines, see how it is slightly erased at the crossing points. Crossbreeds? Yes, in the general case, the line "de" will intersect with the original lines. Although we are not interested in this moment, we just need to build a perpendicular line and that's it.

What is known about the direct "de"? The point belonging to it is known. The direction vector is missing.

By condition, the line must be perpendicular to the lines, which means that its direction vector will be orthogonal to the direction vectors. The motif already familiar from Example No. 9, let's find the vector product:

Let's compose the equations of the straight line "de" by the point and the directing vector:

Ready. In principle, one can change the signs in the denominators and write the answer in the form , but there is no need for this.

To check, it is necessary to substitute the coordinates of the point into the obtained equations of the straight line, then using dot product of vectors make sure that the vector is really orthogonal to the direction vectors "pe one" and "pe two".

How to find the equations of a line containing a common perpendicular?

c) This problem is more difficult. I recommend that dummies skip this paragraph, I don’t want to cool your sincere sympathy for analytic geometry =) By the way, it might be better for more prepared readers to wait too, the fact is that in terms of complexity the example should be put last in the article, but according to the logic of presentation it should be located here.

So, it is required to find the equations of the straight line, which contains the common perpendicular of the skew lines.

is a line segment that connects the given lines and is perpendicular to the given lines:

Here is our handsome man: - common perpendicular of intersecting lines. He is the only one. There is no other like it. We also need to compose the equations of a straight line that contains a given segment.

What is known about the direct "uh"? Its direction vector is known, found in the previous paragraph. But, unfortunately, we do not know a single point belonging to the straight line "em", we do not know the ends of the perpendicular - points. Where does this perpendicular line intersect the two original lines? Africa, Antarctica? From the initial review and analysis of the condition, it is not at all clear how to solve the problem .... But there is a tricky move associated with the use of parametric equations of a straight line.

Let's make a decision point by point:

1) Let's rewrite the equations of the first straight line in parametric form:

Let's consider a point. We do not know the coordinates. BUT. If a point belongs to a given line, then its coordinates correspond to , denote it by . Then the coordinates of the point will be written as:

Life is getting better, one unknown - after all, not three unknowns.

2) The same outrage must be carried out on the second point. Let us rewrite the equations of the second straight line in parametric form:

If a point belongs to a given line, then with a very specific meaning its coordinates must satisfy the parametric equations:

Or:

3) The vector , like the previously found vector , will be the directing vector of the line . How to compose a vector from two points was considered in time immemorial in the lesson Vectors for dummies. Now the difference is that the coordinates of the vectors are written with unknown parameter values. So what? Nobody forbids subtracting the corresponding coordinates of the beginning of the vector from the coordinates of the end of the vector.

There are two points: .

Finding a vector:

4) Since the direction vectors are collinear, then one vector is linearly expressed through the other with some proportionality coefficient "lambda":

Or coordinatewise:

It turned out to be the most ordinary system of linear equations with three unknowns , which is standard solvable, for example, Cramer's method. But here there is an opportunity to get off with little blood, from the third equation we will express "lambda" and substitute it into the first and second equations:

Thus: , and "lambda" we do not need. The fact that the values ​​of the parameters turned out to be the same is pure chance.

5) The sky clears up completely, substitute the found values to our locations:

The direction vector is not particularly needed, since its counterpart has already been found.

After a long journey, it is always interesting to perform a check.

:

The correct equalities are obtained.

Substitute the coordinates of the point into the equations :

The correct equalities are obtained.

6) The final chord: we will compose the equations of a straight line for a point (you can take) and a directing vector:

In principle, you can pick up a “good” point with integer coordinates, but this is cosmetic.

How to find the distance between intersecting lines?

d) We cut down the fourth head of the dragon.

Method one. Not even a way, but a small special case. The distance between intersecting lines is equal to the length of their common perpendicular: .

Extreme points of the common perpendicular found in the previous paragraph, and the task is elementary:

Method two. In practice, most often the ends of the common perpendicular are unknown, so a different approach is used. It is possible to draw parallel planes through two intersecting lines, and the distance between the given planes is equal to the distance between the given lines. In particular, a common perpendicular sticks out between these planes.

In the course of analytic geometry, from the above considerations, a formula was derived for finding the distance between skew lines:
(instead of our points "em one, two" we can take arbitrary points of straight lines).

Mixed product of vectors already found in paragraph "a": .

Cross product of vectors found in paragraph "be": , calculate its length:

Thus:

Proudly lay out the trophies in one row:

Answer:
A) , hence, the lines intersect, which was required to be proved;
b) ;
V) ;
G)

What else can be said about intersecting lines? An angle is defined between them. But consider the universal angle formula in the next paragraph:

Intersecting straight lines necessarily lie in the same plane:

The first thought is to lean on the intersection point with all your might. And immediately I thought, why deny yourself the right desires ?! Let's jump on it right now!

How to find the point of intersection of spatial lines?

Example 14

Find the point of intersection of lines

Solution: Let's rewrite the equations of lines in parametric form:

This task was considered in detail in Example No. 7 of this lesson (see. Equations of a straight line in space). And the straight lines themselves, by the way, I took from Example No. 12. I won’t lie, I’m too lazy to invent new ones.

The solution is standard and has already been encountered when we worked out the equations of the common perpendicular of skew lines.

The point of intersection of the lines belongs to the line, therefore its coordinates satisfy the parametric equations of this line, and they correspond to a very specific parameter value:

But the same point belongs to the second line, therefore:

We equate the corresponding equations and make simplifications:

A system of three linear equations with two unknowns is obtained. If the lines intersect (as proven in Example 12), then the system is necessarily consistent and has a unique solution. It can be solved Gauss method, but we won’t sin with such kindergarten fetishism, let’s do it easier: from the first equation we express “te zero” and substitute it into the second and third equations:

The last two equations turned out to be essentially the same, and it follows from them that . Then:

Let's substitute the found value of the parameter into the equations:

Answer:

To check, we substitute the found value of the parameter into the equations:
The same coordinates were obtained as required to be checked. Meticulous readers can substitute the coordinates of the point in the original canonical equations of the lines.

By the way, it was possible to do the opposite: find the point through “es zero”, and check it through “te zero”.

A well-known mathematical sign says: where the intersection of straight lines is discussed, there is always a smell of perpendiculars.

How to construct a line of space perpendicular to a given one?

(lines intersect)

Example 15

a) Compose the equations of a line passing through a point perpendicular to the line (lines intersect).

b) Find the distance from the point to the line.

Note : clause "lines intersect" - essential. Through the dot
it is possible to draw an infinite number of perpendicular lines that will intersect with the line "el". The only solution occurs when a line is drawn through a given point perpendicular to two given straight lines (see Example No. 13, paragraph "b").

A) Solution: Denote the unknown line by . Let's make a schematic drawing:

What is known about the line? By condition, a point is given. In order to compose the equations of a straight line, it is necessary to find the direction vector. As such a vector, the vector is quite suitable, and we will deal with it. More precisely, let's take the unknown end of the vector by the scruff.

1) We will extract its directing vector from the equations of the straight line "el", and we will rewrite the equations themselves in parametric form:

Many guessed that now for the third time in a lesson the magician will get a white swan out of his hat. Consider a point with unknown coordinates. Since the point , then its coordinates satisfy the parametric equations of the straight line "el" and they correspond to a specific parameter value:

Or in one line:

2) By condition, the lines must be perpendicular, therefore, their direction vectors are orthogonal. And if the vectors are orthogonal, then their scalar product equals zero:

What happened? The simplest linear equation with one unknown:

3) The value of the parameter is known, let's find the point:

And the direction vector:
.

4) We will compose the equations of the straight line by the point and the direction vector:

The denominators of the proportion turned out to be fractional, and this is exactly the case when it is appropriate to get rid of fractions. I'll just multiply them by -2:

Answer:

Note : a more rigorous ending of the solution is drawn up as follows: we compose the equations of a straight line by a point and a directing vector . Indeed, if a vector is a directing vector of a straight line, then the vector collinear to it will naturally also be a directing vector of this straight line.

The verification consists of two stages:

1) check the direction vectors of the lines for orthogonality;

2) we substitute the coordinates of the point into the equations of each straight line, they should “fit” both here and there.

There was a lot of talk about typical actions, so I did a check on a draft.

By the way, I forgot another fad - to build a point "sue" symmetrical to the point "en" with respect to the straight line "el". However, there is a good “flat analogue”, which can be found in the article The simplest problems with a straight line on a plane. Here, all the difference will be in the additional "Z" coordinate.

How to find the distance from a point to a line in space?

b) Solution: Find the distance from a point to a line.

Method one. This distance is exactly equal to the length of the perpendicular: . The solution is obvious: if the points are known , That:

Method two. In practical problems, the base of the perpendicular is often a mystery, so it is more rational to use a ready-made formula.

The distance from a point to a line is expressed by the formula:
, where is the direction vector of the straight line "el", and - arbitrary a point on a given line.

1) From the equations of the straight line we get the direction vector and the most accessible point .

2) The point is known from the condition, sharpen the vector:

3) Let's find vector product and calculate its length:

4) Calculate the length of the direction vector:

5) Thus, the distance from a point to a line:

lines l1 and l2 are called intersecting if they do not lie in the same plane. Let a and b be the direction vectors of these lines, and the points M1 and M2 belong respectively to the lines and l1 and l2

Then the vectors a, b, M1M2> are not coplanar, and therefore their mixed product is not equal to zero, i.e. (a, b, M1M2>) =/= 0. The converse is also true: if (a, b, M1M2> ) =/= 0, then the vectors a, b, M1M2> are not coplanar, and, consequently, the lines l1 and l2 do not lie in the same plane, i.e., they intersect. Thus, two lines intersect if and only if condition(a, b, M1M2>) =/= 0, where a and b are the direction vectors of the lines, and M1 and M2 are the points belonging respectively to the given lines. The condition (a, b, M1M2>) = 0 is a necessary and sufficient condition for the lines to lie in the same plane. If the lines are given by their canonical equations

then a = (a1; a2; a3), b = (b1; b2; b3), M1 (x1; y1; z1), M2(x2; y2; z2) and condition (2) is written as follows:

Distance between intersecting lines

this is the distance between one of the skew lines and a plane parallel to it passing through the other line. The distance between the skew lines is the distance from some point of one of the skew lines to a plane passing through the other line parallel to the first line.

26. Definition of an ellipse, canonical equation. Derivation of the canonical equation. Properties.

An ellipse is the locus of points in a plane for which the sum of the distances to two focused points F1 and F2 of this plane, called foci, is a constant. This does not exclude the coincidence of the foci of the ellipse. If the foci coincide, then the ellipsis is a circle. coordinate system such that the ellipse will be described by the equation (the canonical equation of the ellipse):

It describes an ellipse centered at the origin, whose axes coincide with the coordinate axes.

If on the right side there is a unit with a minus sign, then the resulting equation:

describes an imaginary ellipse. It is impossible to draw such an ellipse in the real plane. Let's denote the foci as F1 and F2, and the distance between them as 2c, and the sum of the distances from an arbitrary point of the ellipse to the foci as 2a

To derive the equation of the ellipse, we choose the Oxy coordinate system so that the foci F1 and F2 lie on the Ox axis, and the origin of coordinates coincides with the middle of the segment F1F2. Then the foci will have the following coordinates: u Let M(x; y) be an arbitrary point of the ellipse. Then, according to the definition of an ellipse, i.e.

This, in fact, is the equation of an ellipse.

27. Definition of a hyperbola, canonical equation. Derivation of the canonical equation. Properties

A hyperbola is a locus of points in a plane for which the absolute value of the difference between the distances to two fixed points F1 and F2 of this plane, called foci, is a constant. Let M(x;y) be an arbitrary point of the hyperbola. Then according to the definition of a hyperbola |MF 1 – MF 2 |=2a or MF 1 – MF 2 =±2a,

28. Definition of a parabola, canonical equation. Derivation of the canonical equation. Properties. A parabola is a GMT of a plane for which the distance to some fixed point F of this plane is equal to the distance to some fixed straight line, also located in the plane under consideration. F is the focus of the parabola; the fixed straight line is the directrix of the parabola. r=d,

r=; d=x+p/2; (x-p/2) 2 +y 2 =(x+p/2) 2 ; x 2 -xp + p 2 / 4 + y 2 \u003d x 2 + px + p 2 / 4; y 2 =2px;

Properties: 1. The parabola has an axis of symmetry (the axis of the parabola); 2.All

the parabola is located in the right half-plane of the Oxy plane at p>0, and in the left

if p<0. 3.Директриса параболы, определяемая каноническим уравнением, имеет уравнение x= -p/2.