TEXT CODE OF THE LESSON:

You already know two cases of mutual arrangement of straight lines in space:

1. intersecting straight lines;

2. Parallel lines.

Let us recall their definitions.

Definition. Lines in space are called intersecting if they lie in the same plane and have one common point

Definition. Lines in space are called parallel if they lie in the same plane and have no common points.

Common to these definitions is that the lines lie in the same plane.

This is not always the case in space. We can deal with several planes, and not every two straight lines will lie in the same plane.

For example, the edges of the cube ABCDA1B1C1D1

AB and A1D1 lie in different planes.

Definition. Two lines are called intersecting if there is no plane that would pass through these lines. From the definition it is clear that these lines do not intersect and are not parallel.

Let us prove a theorem that expresses the criterion for intersecting lines.

Theorem (a sign of intersecting lines).

If one of the lines lies in a certain plane, and the other line intersects this plane at a point not belonging to this line, then these lines are intersecting.

Line AB lies in the plane α. Line CD intersects the plane α at point C, which does not belong to line AB.

Prove that lines AB and DC are crossed.

Evidence

The proof will be carried out by contradiction.

Suppose AB and CD lie in the same plane, we denote it β.

Then the plane β passes through line AB and point C.

By the corollary of the axioms, a plane can be drawn through line AB and a point C not lying on it, and, moreover, only one.

But we already have such a plane - the plane α.

Consequently, the planes β and α coincide.

But this is impossible, since line CD intersects α, but does not lie in it.

We have come to a contradiction, therefore, our assumption is wrong. AB and CD lie in

different planes and are crossed.

The theorem is proved.

So, there are three possible ways of mutual arrangement of straight lines in space:

A) Lines intersect, that is, they have only one common point.

B) Lines are parallel, i.e. lie in the same plane and have no common points.

C) Straight lines are crossed, i.e. do not lie in the same plane.

Consider another intersecting line theorem

Theorem. Through each of the two crossing lines there is a plane parallel to the other line, and moreover, only one.

AB and CD - crossing straight lines

Prove that there is a plane α such that line AB lies in the plane α, and line CD is parallel to the plane α.

Evidence

Let us prove the existence of such a plane.

1) Through point A, draw a line AE parallel to CD.

2) Since the straight lines AE and AB intersect, a plane can be drawn through them. Let us denote it by α.

3) Since the line CD is parallel to AE, and AE lies in the plane α, then the line CD ∥ of the plane α (by the theorem on the perpendicularity of the line and the plane).

Plane α is the desired plane.

Let us prove that the plane α is the only one that satisfies the condition.

Any other plane passing through line AB will intersect AE, and hence line CD parallel to it. That is, any other plane passing through AB intersects with the line CD, therefore it is not parallel to it.

Consequently, the plane α is unique. The theorem is proved.

In this article, we will first give the definition of the angle between crossing lines and give a graphic illustration. Next, we will answer the question: "How to find the angle between crossing straight lines, if the coordinates of the direction vectors of these straight lines in a rectangular coordinate system are known?" In conclusion, we will practice finding the angle between crossing lines when solving examples and problems.

Page navigation.

Angle between crossed lines - definition.

We will approach the definition of the angle between crossed lines gradually.

First, recall the definition of intersecting lines: two lines in three-dimensional space are called interbreedingif they do not lie in the same plane. It follows from this definition that crossing lines do not intersect, are not parallel, and, moreover, do not coincide, otherwise they would both lie in a certain plane.

Here are some additional arguments.

Let two intersecting straight lines a and b be given in three-dimensional space. Let's construct lines a 1 and b 1 so that they are parallel to intersecting lines a and b, respectively, and pass through some point of the space M 1. Thus, we get two intersecting lines a 1 and b 1. Let the angle between the intersecting straight lines a 1 and b 1 be equal to the angle. Now we will construct lines a 2 and b 2, parallel to intersecting lines a and b, respectively, passing through point М 2, different from point М 1. The angle between the intersecting straight lines a 2 and b 2 will also be equal to the angle. This statement is true, since the straight lines a 1 and b 1 coincide with the straight lines a 2 and b 2, respectively, if you perform a parallel translation, in which point M 1 goes to point M 2. Thus, the measure of the angle between two intersecting straight lines at the point M, respectively parallel to the given intersecting straight lines, does not depend on the choice of the point M.

Now we are ready to define the angle between crossing lines.

Definition.

The angle between crossing lines Is the angle between two intersecting straight lines, which are respectively parallel to the given intersecting straight lines.

It follows from the definition that the angle between crossing lines will also not depend on the choice of point M. Therefore, as a point M, you can take any point belonging to one of the intersecting lines.

Let us give an illustration of the definition of the angle between crossing lines.

Finding the angle between crossed lines.

Since the angle between intersecting straight lines is determined through the angle between intersecting straight lines, then finding the angle between intersecting straight lines is reduced to finding the angle between the corresponding intersecting straight lines in three-dimensional space.

Undoubtedly, the methods taught in geometry lessons in high school are suitable for finding the angle between crossed lines. That is, having completed the necessary constructions, you can associate the desired angle with any angle known from the condition, based on the equality or similarity of the figures, in some cases it will help cosine theorem, and sometimes the result is definition of sine, cosine and tangent of an angle right triangle.

However, it is very convenient to solve the problem of finding the angle between crossing straight lines by the coordinate method. This is what we will consider.

Let Oxyz be introduced in three-dimensional space (however, in many problems it has to be entered independently).

Let us set ourselves the task: find the angle between the crossing straight lines a and b, which correspond to some equations of a straight line in space in the rectangular coordinate system Oxyz.

Let's solve it.

Take an arbitrary point of the three-dimensional space M and assume that straight lines a 1 and b 1 pass through it, parallel to the intersecting lines a and b, respectively. Then the required angle between the intersecting straight lines a and b is equal to the angle between the intersecting straight lines a 1 and b 1 by definition.

Thus, it remains for us to find the angle between the intersecting straight lines a 1 and b 1. To apply the formula for finding the angle between two intersecting straight lines in space, we need to know the coordinates of the direction vectors of the straight lines a 1 and b 1.

How can we get them? It's very simple. The definition of the direction vector of a straight line allows us to assert that the sets of direction vectors of parallel straight lines coincide. Therefore, as the direction vectors of lines a 1 and b 1, we can take the direction vectors and lines a and b, respectively.

So, the angle between two crossed straight lines a and b is calculated by the formula
where and - direction vectors of straight lines a and b, respectively.

Formula for finding the cosine of the angle between crossed straight lines a and b has the form .

Allows you to find the sine of the angle between crossed lines, if the cosine is known: .

It remains to analyze the solutions of examples.

Example.

Find the angle between crossing straight lines a and b, which are defined in the rectangular coordinate system Oxyz by the equations and .

Decision.

The canonical equations of a straight line in space allow you to immediately determine the coordinates of the directing vector of this straight line - they are given by the numbers in the denominators of fractions, that is, ... Parametric equations of a straight line in space also make it possible to immediately write down the coordinates of the direction vector - they are equal to the coefficients in front of the parameter, that is, - directing vector of a straight line ... Thus, we have all the necessary data to apply the formula by which the angle between crossing lines is calculated:

Answer:

The angle between the given crossing lines is.

Example.

Find the sine and cosine of the angle between crossed straight lines on which the edges AD and BC of the pyramid ABCD lie, if the coordinates of its vertices are known:.

Decision.

Directing vectors of crossing lines AD and BC are vectors and. Let's calculate their coordinates as the difference of the corresponding coordinates of the points of the end and the beginning of the vector:

According to the formula we can calculate the cosine of the angle between the specified crossing lines:

Now let's calculate the sine of the angle between crossing lines:

Answer:

In conclusion, let us consider the solution to the problem in which it is required to find the angle between crossing straight lines, and the rectangular coordinate system has to be entered independently.

Example.

Given a rectangular parallelepiped ABCDA 1 B 1 C 1 D 1, in which AB \u003d 3, AD \u003d 2 and AA 1 \u003d 7 units. Point E lies on edge AA 1 and divides it in a ratio of 5 to 2 counting from point A. Find the angle between the crossed lines BE and A 1 C.

Decision.

Since the edges of a rectangular parallelepiped at one vertex are mutually perpendicular, it is convenient to enter a rectangular coordinate system, and determine the angle between the indicated crossing lines using the coordinate method through the angle between the direction vectors of these lines.

Let us introduce a rectangular coordinate system Oxyz as follows: let the origin of coordinates coincide with the vertex A, the Ox axis coincides with the AD line, the Oy axis with the AB line, and the Oz axis with the AA 1 line.

Then point B has coordinates, point E - (if necessary, see the article), point A1 -, and point C -. From the coordinates of these points, we can calculate the coordinates of the vectors and. We have , .

It remains to apply the formula to find the angle between the crossing lines along the coordinates of the direction vectors:

Answer:

List of references.

• Atanasyan L.S., Butuzov V.F., Kadomtsev S.B., Kiseleva L.S., Poznyak E.G. Geometry. Textbook for grades 10-11 of secondary school.
• Pogorelov A.V., Geometry. Textbook for grades 7-11 of educational institutions.
• Bugrov Y.S., Nikolsky S.M. Higher mathematics. Volume One: Elements of Linear Algebra and Analytic Geometry.
• Ilyin V.A., Poznyak E.G. Analytic geometry.

Crossed straight lines are easy to recognize by these features. Sign 1. If there are four points on two lines that do not lie in the same plane, then these lines intersect (Fig. 1.21).

Indeed, if these straight lines would intersect or be parallel, then they would lie in one plane, and then these points would lie in one plane, which contradicts the condition.

Sign 2. If the line O lies in the plane, and the line b intersects the plane a at some point

M, not lying on the straight line a, then the straight lines a and b intersect (Fig. 1.22).

Indeed, taking any two points on the line a and any two points on the line b, we arrive at criterion 1, i.e. a and b are crossed.

Real examples of intersecting straight lines are given by transport interchanges (Fig. 1.23).

In space, there are more pairs of intersecting straight lines than there are pairs of parallel or intersecting straight lines. This can be explained as follows.

Take in space some point A and some straight line a that does not pass through point A. To draw a straight line through point A parallel to line a, it is necessary to draw plane a through point A and straight line a (Proposition 2 in clause 1.1), and then in the plane and draw a straight line b parallel to a straight line a (Fig. 1.24).

There is only one such straight line b. All lines passing through point A and intersecting line O also lie in the plane a and fill it all with the exception of line b. All the other straight lines going through A and filling all the space except the plane a will intersect with the straight line a. We can say that intersecting lines in space are a general case, and intersecting and parallel lines are special cases. "Small perturbations" of crossing lines leave them crossing. But the properties of being parallel or intersecting with "small perturbations" in space are not preserved.

Mutual arrangement two straight lines in space.

The relative position of two lines and space is characterized by the following three possibilities.

Lines lie in the same plane and have no common points - parallel lines.

Lines lie on the same plane and have one common point - the lines intersect.

In space, two straight lines can also be located so that they do not lie in any plane. Such straight lines are called crossing (do not intersect and are not parallel).

EXAMPLE:

PROBLEM 434 In the plane lies a triangle ABC, a

Triangle ABC lies in the plane, and point D is not in this plane. Points M, N and K, respectively, are the midpoints of segments DA, DB and DC

Theorem. If one of two straight lines lies in a certain plane, and the other intersects this plane and to a point that does not lie on the first straight line, then these straight lines intersect.

In fig. 26 line a lies in the plane, and line c intersects at point N. Lines a and c are intersecting.

Theorem.Only one plane passes through each of the two intersecting lines, parallel to the other line.

In fig. 26 straight lines a and b cross. Black straight line and drawn plane a (alpha) || b (line a1 || b is indicated in plane B (beta)).

Theorem 3.2.

Two straight lines parallel to the third are parallel.

This property is called transitivityparallelism of straight lines.

Evidence

Let lines a and b be simultaneously parallel to line c. Suppose that a is not parallel to b, then line a intersects with line b at some point A that does not lie on line c by hypothesis. Therefore, we have two lines a and b passing through point A, not lying on the given line c, and simultaneously parallel to it. This contradicts Axiom 3.1. The theorem is proved.

Theorem 3.3.

Through a point that does not lie on a given straight line, one and only one straight line parallel to the given one can be drawn.

Evidence

Let (AB) be a given line, C be a point not lying on it. Line AC splits the plane into two half-planes. Point B lies in one of them. In accordance with axiom 3.2, it is possible to postpone the angle (ACD) equal to the angle (CAB) from the ray C A to another half-plane. ACD and CAB are equal inner criss-cross lines under the lines AB and CD and the secant (AC) Then by Theorem 3.1 (AB) || (CD). Taking into account axiom 3.1. The theorem is proved.

The property of parallel lines is given by the following theorem, which is the converse to Theorem 3.1.

Theorem 3.4.

If two parallel lines are intersected by a third line, then the inner angles lying crosswise are equal.

Evidence

Let (AB) || (CD). Suppose ACD ≠ BAC. Draw a line AE through point A so that EAC \u003d ACD. But then, by Theorem 3.1 (AE) || (CD), and by hypothesis - (AB) || (CD). According to Theorem 3.2 (AE) || (AB). This contradicts Theorem 3.3, according to which a single straight line parallel to it can be drawn through a point A that does not lie on CD. The theorem is proved.

Figure 3.3.1.

Based on this theorem, the following properties are easily justified.

If two parallel lines are crossed by a third line, then the corresponding angles are equal.

If two parallel lines are intersected by a third line, then the sum of the inner one-sided angles is 180 °.

Corollary 3.2.

If a line is perpendicular to one of the parallel lines, then it is perpendicular to the other.

The concept of parallelism allows us to introduce the following new concept, which will be needed later in Chapter 11.

The two beams are called equally directedif there is a straight line such that, firstly, they are perpendicular to this straight line, and secondly, the rays lie in the same half-plane relative to this straight line.

The two beams are called oppositely directedif each of them is equally directed with a ray complementary to the other.

Equally directed rays AB and CD will be denoted: and oppositely directed rays AB and CD -

Figure 3.3.2.

A sign of crossing lines.

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

Cases of mutual arrangement of straight lines in space.

1. There are four different cases of two straight lines in space:

   
   

   - straight crossing, i.e. do not lie in the same plane;

   - straight lines intersect, i.e. lie in the same plane and have one common point;

   - straight lines parallel, i.e. lie in the same plane and do not intersect;

   - straight lines coincide.

   
   

   Let us obtain signs of these cases of mutual arrangement of straight lines given by the canonical equations

   
   
   where - points belonging to straight lines and respectively, a - direction vectors (Figure 4.34). Let us denote by a vector connecting the given points.
   
   

   The above cases of the mutual arrangement of straight lines and correspond to the following signs:

   
   

   - direct and crossed vectors are not coplanar;

   
   

   - straight lines and intersecting vectors are coplanar, but vectors are not collinear;

   
   

   - direct and parallel vectors are collinear, but vectors are not collinear;

   
   

   - straight and coincide vectors are collinear.

   
   

   These conditions can be written using the properties of mixed and vector products. Recall that the mixed product of vectors in a right-handed rectangular coordinate system is found by the formula:

   
   
   and the determinant intersects is zero, and its second and third lines are not proportional, i.e.
   

   - straight and parallel second and third lines of the determinant are proportional, i.e. and the first two lines are not proportional, i.e.

   
   

   - all lines of the determinant are straight and coincide proportional, i.e.

Proof of the sign of crossed lines.

If one of the two lines lies in a plane, and the other intersects this plane at a point that does not belong to the first line, then these two lines intersect.

Evidence

Let a belong to α, b intersect α \u003d A, A does not belong to a (drawing 2.1.2). Suppose that lines a and b are not intersecting, that is, they intersect. Then there is a plane β to which the lines a and b belong. Line a and point A lie in this plane β. Since line a and point A outside it define a unique plane, then β \u003d α. But b leads β and b does not belong to α, hence the equality β \u003d α is impossible.

In less than a minute, I created a new Vord 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 a prosaic whipping \u003d)

Two straight spaces can:

1) interbreed;

2) intersect at a point;

3) be parallel;

4) match.

Case No. 1 is fundamentally different from other cases. Two straight lines intersect if they do not lie in the same plane... Raise one hand up, and extend the other hand forward - here's an example of crossing straight lines. In points 2-4, the straight lines must lie in one plane.

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

Consider two straight spaces:

- straight, given point and direction vector;
- a straight line given by a point and a direction vector.

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

The drawing shows crossed straight lines as an example.

How to deal with these straight lines?

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

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

In cases No. 2-4, our construction "falls" into one plane, while the vectors coplanar, and the mixed product of linearly dependent vectors equals zero: .

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

If the direction vectors collinear, then the lines are either parallel or coincide. As a final nail, I suggest 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 "fit", then the straight lines coincide; if they "did not fit," then the straight lines are parallel.

The flow of the algorithm is simple, but practical examples still do not hurt:

Example 11

Find out the relative position of two lines

Decision: as in many problems of geometry, it is convenient to draw up the solution according to the points:

1) We take out the points and direction vectors from the equations:

2) Find the vector:

Thus, 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 of the corresponding coordinates of these vectors:

Of each the equation implies that, therefore, the system is consistent, the corresponding coordinates of the vectors are proportional, and the vectors are collinear.

Conclusion: straight lines are parallel or coincide.

5) Let us find out if the lines have common points. Take a point belonging to the first line and substitute its coordinates into the equations of the line:

Thus, the lines have no common points, and they have no choice but to be parallel.

Answer:

An interesting example for an independent solution:

Example 12

Find out the relative position of the straight lines

This is an example for a do-it-yourself solution. Note that the second line has a letter as a parameter. It is logical. In general, these are two different straight lines, so each straight line has its own parameter.

And again I urge you not to skip examples, I will flog the problems I proposed 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 starting order of the narrative will be respected: first we will consider problems with intersecting straight lines, then with intersecting straight lines, and at the end we will talk about parallel lines in space. However, I must say that some tasks of this lesson can be formulated at once for several cases of the arrangement of straight lines, and in this regard, the division of the section into paragraphs is somewhat arbitrary. There are more simple examples, there are more complex examples, and hopefully everyone will find what they need.

Crossed straight lines

I remind you that straight lines intersect if there is no plane in which they both lie. When I was thinking over the practice, a monster problem came to my 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 straight lines intersect;

b) find the equations of a straight line passing through a point perpendicular to these straight lines;

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

d) find the distance between the lines.

Decision: The road will be mastered by the walking:

a) Let us prove that the lines intersect. Find the points and direction vectors of these lines:

Find the vector:

We calculate mixed product of vectors:

Thus the vectors not coplanar, which means that the lines intersect, which is what was required to be proved.

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

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

For a change, I have placed a straight line BEHIND straight, see how it is slightly erased at the crossing points. Crossbreeds? Yes, in the general case the straight line "de" will intersect with the original straight 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 her is known. Direction vector is missing.

By condition, the straight line must be perpendicular to the straight lines, which means that its direction vector will be orthogonal to the direction vectors. Already familiar from Example No. 9 motive, find the cross product:

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

Done. In principle, you 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 vectorsmake sure that the vector is really orthogonal to the direction vectors "pe one" and "pe two".

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

c) This task will be more difficult. I recommend dummies to skip this point, I don't want to cool your sincere sympathy for analytic geometry \u003d) By the way, it may be better for more prepared readers to postpone 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 intersecting straight lines.

Is a line segment connecting the given lines and perpendicular to the given lines:

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

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

We will issue the decision according to the points:

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

Consider a point. We do not know the coordinates. BUT... If a point belongs to a given straight line, then it corresponds to its coordinates, we denote it by. Then the coordinates of the point will be written in the form:

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

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

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

Or:

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

There are two points: .

Find the vector:

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

Or coordinatewise:

It turned out the most, that neither is the usual system of linear equations with three unknowns, which is solvable in the standard, for example, cramer's method... But here there is an opportunity to get rid of a little blood, from the third equation we express the "lambda" and substitute it into the first and second equations:

Thus: , and we do not need a lambda. The fact that the values \u200b\u200bof the parameters turned out to be the same is pure coincidence.

5) The sky is completely clear, substitute the found values to our points:

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

After a long journey, it's always fun to check.

:

The correct equalities are obtained.

Substitute the coordinates of the point into the equations :

The correct equalities are obtained.

6) Final chord: compose the equations of a straight line along a point (you can take it) and a direction vector:

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

How to find the distance between crossed lines?

d) Cut off the fourth dragon head.

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

The 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. Parallel planes can be drawn through two crossing lines, and the distance between these planes is equal to the distance between these lines. In particular, a common perpendicular sticks out between these planes.

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

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

Vector product of vectors found in item "bae": , let's calculate its length:

Thus:

Let's proudly lay out the trophies in one row:

Answer:
and) , which means that the lines intersect, which was required to prove;
b) ;
in) ;
d)

What else can you tell us about crossing straight lines? An angle is defined between them. But consider the universal angle formula in the next paragraph:

Intersecting straight lines of space necessarily lie in the same plane:

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

How to find the point of intersection of spatial lines?

Example 14

Find the point of intersection of lines

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

This task was discussed 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 will not lie, I am too lazy to invent new ones.

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

The intersection point of the straight lines belongs to the straight line, therefore its coordinates satisfy the parametric equations of the given straight line, and they correspond to quite specific parameter value:

But the same point belongs to the second straight 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 proved in Example 12), then the system is necessarily consistent and has a unique solution. It can be solved gaussian method, but we will not sin with such kindergarten fetishism, we will 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, in fact, the same, and it follows from them that. Then:

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 verified. Meticulous readers can substitute the coordinates of a point in the original canonical equations of straight lines.

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

A well-known mathematical sign says: where they discuss the intersection of straight lines, it always smells like perpendiculars.

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

(lines intersect)

Example 15

a) Make up the equations of a straight line passing through a point perpendicular to a straight line (lines intersect).

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

Note : clause "lines intersect" - essential... Through point
you can draw infinitely many perpendicular straight lines that will intersect with the straight "ale". The only solution takes place when a straight line is drawn through this point, perpendicular to two given by a straight line (see Example No. 13, point "b").

and) Decision: The unknown line is denoted by. Let's execute a schematic drawing:

What is known about the straight 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. A vector is quite suitable as such a vector, and we will deal with it. More precisely, let's take the unknown end of the vector by the scruff.

1) Let us take out its direction vector from the equations of the straight line "el", and rewrite the equations themselves in parametric form:

Many have 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 value of the parameter:

Or in one line:

2) By condition, the straight 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, find a point:

And the direction vector:
.

4) Let us compose the equations of a straight line by a point and a 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 formed as follows: we compose the equations of a straight line along a point and a direction vector. Indeed, if a vector is a directing vector of a straight line, then a vector collinear to it will naturally also be a directing vector of a given straight line.

The check consists of two stages:

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

2) we substitute the coordinates of the point into the equations of each straight line, they must "fit" both there and there.

A lot was said about typical actions, so I checked on a draft.

By the way, I also forgot the point - to build a point "siu" symmetrical to the point "en" relative to the straight line "el". However, there is a good "flat analog", 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 "zeta" coordinate.

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

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

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

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

The distance from a point to a straight line is expressed by the formula:
, where is the directing vector of the straight line "el", and - arbitrarypoint belonging to this 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) Find cross product and calculate its length:

4) Calculate the length of the direction vector:

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