Let the plane `alpha` be parallel to the plane `beta`, line `b` lie in the plane `beta`, point `B` lie on line `b`. Obviously, the distance from point `B` to the `alpha` plane is equal to the distance from the line `b` to the `alpha` plane and is equal to the distance between the `alpha` and `beta` planes.
Consider two crossing lines `a` and `b` . Let us draw a plane through line `a` parallel to line `b`. Let's draw a plane through straight line `b`, perpendicular to the plane`alpha`, let the line of intersection of these planes be `b_1` (this line is the projection of the line `b` onto the plane `alpha`). Let us denote the intersection point of lines `a` and `b_1` as `A`. Point `A` is a projection of some point `B` straight `b`. From the fact that `AB_|_alpha` it follows that `AB_|_a` and `AB_|_b_1`; in addition `b``||``b_1`, means `AB_|_b` - . Line `AB` intersects skew lines `a` and `b` and is perpendicular to both. The segment `AB` is called common perpendicular two intersecting lines.
The length of the common perpendicular of intersecting lines is equal to the distance from any point on the line`b` to plane`alpha`.
* Distance between crossing lines equal to the length of their common perpendicular. Let a straight line `l_1` be given in space with a known direction vector `veca_1` ( guide vector a straight line is a non-zero vector parallel to this straight line), a straight line `l_2` with a known direction vector `veca_2`, points `A_1` and `A_2` lying on `l_1` and `l_2` respectively, in addition, the vector `vec( A_1A_2)=vecr`. Let the segment `P_1P_2` be a common perpendicular to `l_1` and `l_2` (see Fig. 9). The task is to find the length of this segment. Let's represent the vector `vec(P_1P_2)` as the sum `vec(P_1A_1)+vec(A_1A_2)+vec(A_2P_2)`. Then, using the collinearity of the vectors `vec(P_1A_1)` and `veca_1`, `vec(A_2P_2)` and `veca_2`, we obtain for the vector `vec(P_1P_2)` the representation `vec(P_1P_2)=xveca_1+yveca_2+vecr`, where `x` and `y` are currently unknown numbers. These numbers can be found from the condition that the vector `vec(P_1P_2)` is perpendicular to the vectors `veca_1` and `veca_2`, i.e. from the following system of linear equations:
x a → 1 + y a → 2 + r → · a → 1 = 0, x a → 1 + y a → 2 + r → · a → 2 = 0. \left\(\begin(array)(l)\left(x(\overrightarrow a)_1+y(\overrightarrow a)_2+\overrightarrow r\right)\cdot(\overrightarrow a)_1=0,\\\ left(x(\overrightarrow a)_1+y(\overrightarrow a)_2+\overrightarrow r\right)\cdot(\overrightarrow a)_2=0.\end(array)\right.
After this, we find the length of the vector `vec(P_1P_2):`
`P_1P_2=sqrt((xveca_1+yveca_2+vecr)^2)`.
Calculate the distance between the crossing diagonals of two adjacent faces of a cube with edge `a`.
Let a cube `A...D_1` with edge `a` be given. Let's find the distance between the lines `AD_1` and `DC_1` (Fig. 10). Let's introduce the basis `veca=vec(DA)`, `vecb=vec(DC)`, `vecc=vec(DD_1)`. For the direction vectors of the lines `AD_1` and `DC_1` we can take `vec(AD_1)=vecc-veca` and `vec(DC_1)=vecb+vecc`. If `P_1P_2` is a common perpendicular to the lines under consideration, then `vec(P_1P_2)=x(vecc-veca)+y(vecb+vecc)+veca`.
Let's create a system of equations to find the unknown numbers `x` and `y`:
x c → - a → + y b → + c → + a → · c → - a → = 0 , x c → - a → + y b → + c → + a → · b → + c → = 0 . \left\(\begin(array)(l)\left(x\left(\overrightarrow c-\overrightarrow a\right)+y\left(\overrightarrow b+\overrightarrow c\right)+\overrightarrow a\right) \cdot\left(\overrightarrow c-\overrightarrow a\right)=0,\\\left(x\left(\overrightarrow c-\overrightarrow a\right)+y\left(\overrightarrow b+\overrightarrow c\right )+\overrightarrow a\right)\cdot\left(\overrightarrow b+\overrightarrow c\right)=0.\end(array)\right.
Let us reduce this system to an equivalent one:
2 x + y - 1 = 0, x + 2 y = 0. \left\(\begin(array)(l)2x+y-1=0,\\x+2y=0.\end(array)\right.
From here we find `x=2/3`, `y=-1/3`. Then
`vec(P_1P_2)=2/3(vecc-veca)-1/3(vecb+vecc)+veca=1/3veca-1/3vecb+1/3vecc`,
Not even a minute had passed before I created a new Verdov file and continued like this fascinating topic. You need to capture moments of a working mood, so there will be no lyrical introduction. There will be a prosaic spanking =)
Two straight spaces can:
1) interbreed;
2) intersect at the 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 arm up and extend the other arm forward - here is an example of crossing lines. In points No. 2-4 the straight lines must lie in one plane.
How to find out the relative positions of lines in space?
Consider two direct spaces:
– a straight line defined by a point and a direction vector;
– a straight line defined by a point and a direction vector.
For a better understanding, let's make a schematic drawing: 
The drawing shows intersecting 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 the vectors not coplanar(see lesson Linear (non) dependence of vectors. Basis of vectors), and, therefore, the determinant composed of their coordinates is non-zero. Or, which is actually the same thing, it will be non-zero: 
.
In cases No. 2-4, our structure “falls” into one plane, while the vectors coplanar, and the mixed product is linear dependent vectors equals zero: 
.
Let's expand the algorithm further. Let's pretend that 
Therefore, the lines either intersect, are parallel, or coincide.
If the direction vectors collinear, then the lines are either parallel or coincident. For the final nail, I propose the following technique: take any point on one line and substitute its coordinates into the equation of the second line; if the coordinates “fit,” then the lines coincide; if they “don’t fit,” then the lines are parallel.
The algorithm is simple, but practical examples still won't hurt:
Example 11
Find out the relative position of two lines
Solution: as in many geometry problems, it is convenient to formulate the solution point by point:
1) We take out 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) Let's check the direction vectors for collinearity.
Let's create a system from the corresponding coordinates of these vectors:
From everyone equations it follows that, therefore, the system is consistent, the corresponding coordinates of the vectors are proportional, and the vectors are collinear.
Conclusion: the lines are parallel or coincide.
5) Let's find out whether the straight lines have common points. Let's 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:
Interesting example For independent decision:
Example 12
Find out the relative positions of the lines
This is an example for you to solve on your own. Please note that the second line has the letter as a parameter. Logical. IN general case– these are two different lines, so each line has its own parameter.
And again I urge you not to skip the examples, the tasks I propose are far from random ;-)
Problems with a line in space
In the final part of the lesson I will try to consider maximum amount various tasks with spatial lines. In this case, the original order of the story will be observed: first we will consider problems with crossing lines, then with intersecting 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 for several cases of the location of lines at once, 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 I hope everyone will find what they need.
Crossing lines
Let me remind you that straight lines intersect if there is no plane in which they both lie. When I was thinking through the practice, a monster problem came to mind, and now I’m glad to present to your attention a dragon with four heads:
Example 13
Given straight lines. Required:
a) prove that lines intersect;
b) find the equations of a line passing through a point perpendicular to the given lines;
c) compose equations of a straight line that contains common perpendicular crossing lines;
d) find the distance between the lines.
Solution: The one who walks will master the road:
a) Let us prove that lines intersect. Let's find the points and direction vectors of these lines: 
Let's find the vector:
Let's calculate mixed product of vectors:
Thus, the vectors not coplanar, which means that the lines intersect, which is what needed to be proven.
Probably everyone has long noticed that for crossing lines the verification algorithm is the shortest.
b) Find the equations of the line that passes through the point and is perpendicular to the lines. Let's make a schematic drawing: 
For a change I posted a direct BEHIND straight, look how it is a little erased at the crossing points. Crossbreeding? Yes, in general, the straight line “de” will be crossed with the original straight lines. Although this moment we are not interested in it yet, 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. There is not enough guide vector.
According to the 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, let’s find the vector product:
Let’s compose the equations of the straight line “de” using a point and a direction vector:
Ready. 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, you need to substitute the coordinates of the point into the resulting straight line equations, then use scalar 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 will be more difficult. I recommend that dummies skip this point, I don’t want to dampen your sincere sympathy for analytical geometry=) By the way, it may be better for more prepared readers to hold off too; the fact is that, due to its complexity, the example should be placed last in the article, but according to the logic of presentation, it should be located here.
So, you need to find the equations of a line that contains the common perpendicular of skew lines.
- this is a segment connecting these lines and perpendicular to these lines: 
Here is our handsome guy: - common perpendicular of intersecting lines. He's the only one. There is no other like it. We need to create equations for the line that contains this segment.
What is known about the direct “um”? Its direction vector is known, found in previous paragraph. But, unfortunately, we do not know a single point belonging to the straight line “em”, nor do we know the ends of the perpendicular – the 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 at all clear how to solve the problem... But there is tricky move, associated with the use of parametric straight line equations.
We will formulate the decision point by point:
1) Let's rewrite the equations of the first line in parametric form: 
Let's consider the point. We don't know the coordinates. BUT. If a point belongs to a given line, then its coordinates correspond to , let us denote it by . Then the coordinates of the point will be written in the form:
Life is getting better, one unknown is still not three unknowns.
2) The same outrage must be carried out on the second point. Let us rewrite the equations of the second 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) Vector, like the previously found vector, will be the directing vector of the straight line. How to construct a vector from two points was discussed in time immemorial at the lesson Vectors for dummies. Now the difference is that the coordinates of the vectors are written with unknown values parameters. 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 the vector:
4) Since the direction vectors are collinear, one vector is linearly expressed through the other with a certain proportionality coefficient “lambda”:
Or coordinate-by-coordinate: 
It turned out to be the most ordinary system of linear equations with three unknowns, which is standardly solvable, for example, Cramer's method. But here it is possible to get off with little loss; from the third equation we will express “lambda” and substitute it into the first and second equations:
Thus: 
, and we don’t need “lambda”. The fact that the parameter values turned out to be the same is purely an accident.
5) The sky is completely clearing, let’s substitute the found values 
to our points:
The direction vector is not particularly needed, since its counterpart has already been found.
It's always interesting to check after a long journey.
:
The correct equalities are obtained.
Let's substitute the coordinates of the point into the equations 
:
The correct equalities are obtained.
6) Final chord: let’s create the equations of a straight line using a point (you can take it) and a direction vector:
In principle, you can select a “good” point with intact coordinates, but this is cosmetic.
How to find the distance between intersecting lines?
d) We cut off the fourth head of the dragon.
Method one. Not even a way, but a small one special case. The distance between crossing lines is equal to the length of their common perpendicular: 
.
Extreme points 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 intersecting straight lines, and the distance between these planes is equal to the distance between these straight lines. In particular, a common perpendicular sticks out between these planes.
In the course of analytical geometry, from the above considerations, a formula is derived for finding the distance between intersecting straight lines: 
(instead of our points “um one, two” you can take arbitrary points straight lines).
Mixed product of vectors already found in point "a": 
.
Vector product of vectors found in paragraph "be": 
, let's calculate its length:
Thus:
Let's proudly display the trophies in one row:
Answer:
A) 
, which means that straight lines intersect, which is what was required to be proved;
b) 
;
V) 
;
G) 
What else can you tell about crossing lines? There is a defined angle between them. But universal formula We will consider the angle in the next paragraph:
Intersecting straight spaces necessarily lie in the same plane: 
The first thought is to lean on the intersection point with all your might. And I immediately thought, why deny yourself the right desires?! Let's get on top of her 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 discussed in detail in Example No. 7 of this lesson (see. Equations of a line in space). And by the way, I took the straight lines themselves from Example No. 12. I won’t lie, I’m too lazy to come up with new ones.
The solution is standard and has already been encountered when we were trying to figure out the equations for the common perpendicular of intersecting lines.
The point of intersection of the lines belongs to the line, therefore its coordinates satisfy the parametric equations of this line, and corresponds to them quite specific meaning parameter:
But this same point also belongs to the second line, therefore: 
We equate the corresponding equations and carry out simplifications: 
Received system of three linear equations with two unknowns. If the lines intersect (which is proven in Example No. 12), then the system is necessarily consistent and has a unique solution. It can be solved Gaussian method, but we won’t sin with such kindergarten fetishism, we’ll do it simpler: 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 needed to be checked. Meticulous readers can substitute the coordinates of the point into the original canonical equations of 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 superstition says: where the intersection of 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) Write down 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” – significant. Through the point
you can draw an infinite number of perpendicular lines that will intersect with the straight line “el”. Only decision occurs in the case when, through this point a straight line is drawn perpendicular two given by a straight line (see Example No. 13, point “b”).
A) Solution: We denote the unknown line by . Let's make a schematic drawing:
What is known about the straight line? According to the condition, a point is given. In order to compose the equations of a straight line, it is necessary to find the direction vector. The vector is quite suitable as such a vector, so we will deal with it. More precisely, let's take the unknown end of the vector by the scruff of the neck.
1) Let’s take out its direction vector from the equations of the straight line “el”, and rewrite the equations themselves in parametric form: 
Many guessed that now for the third time during the lesson the magician will get white swan from a hat. Consider a point with unknown coordinates. Since the point is , its coordinates satisfy the parametric equations of the straight line “el” and they correspond to a specific parameter value:
Or in one line:
2) According to the 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 a straight line using 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 to the solution is formalized as follows: let’s compose the equations of a straight line using a point and a direction vector 
. Indeed, if a vector is the guiding vector of a straight line, then the collinear vector , naturally, will also be the guiding 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 line, they should “fit” both there and there.
There was a lot of talk about typical actions, so I checked on a draft.
By the way, I forgot another point - build a “zyu” point symmetrical point"en" is relatively straight "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 the only difference will be in the additional “Z” coordinate.
How to find the distance from a point to a line in space?
b) Solution: Let's find the distance from a point to a line.
Method one. This distance 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 sealed secret, 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 directing vector of the straight line “el”, and – free a point belonging to a given line.
1) From the equations of the line 
we take out 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 guide vector:
5) Thus, the distance from a point to a line:
Geometry. Grade 11
Lesson topic: Distance between crossing lines
Ter-Ovanesyan G.L., teacher highest category, laureate of the Soros Foundation Prize
Moscow
Let's consider the problem of finding the distance between crossing lines. The distance between crossing lines is the length of the common perpendicular to these lines.
Let us be given a cube ABCDA 1 B 1 C 1 D 1, the edge of which is equal to the unit AB = 1. You need to find the distance between straight lines AB and DC 1: ρ(AB;DC 1) - ?
These two lines lie in parallel planes: AB lies in the plane AA 1 B 1 B, DC 1 lies in the plane D 1 DC 1 C. Let us first find the perpendicular to these two planes. There are many such perpendiculars in the figure. This is the segment BC, B 1 C 1, A 1 D 1 and AD. Of these, it makes sense to choose the segment that is not only perpendicular to these planes, and therefore perpendicular to our straight lines AB and DC 1, but also passes through these straight lines. Such a segment is AD. It is simultaneously perpendicular to the straight line AB, because it is perpendicular to the plane AA 1 B 1 B and to the straight line DC 1, because it is perpendicular to the plane D 1 DC 1 C. This means that AD is the common perpendicular to the crossing straight lines AB and DC 1. The distance between these straight lines is the length of this perpendicular, that is, the length of the segment AD. But AD is an edge of the cube. Therefore the distance is 1:
ρ(AB;DC 1)=AD=1
Let's consider another problem, a little more complex, about finding the distance between intersecting lines.
Let us again be given a cube whose edge is equal to one. You need to find the distance between the diagonals of opposite faces. That is, given a cube ABCDA 1 B 1 C 1 D 1. Edge AB=1. You need to find the distance between straight lines BA 1 and DC 1: ρ(A 1 B; DC 1) - ?
These two lines intersect, which means the distance is the length of the common perpendicular. Instead of drawing a general perpendicular, you can formulate in the following way: this is the length of the perpendicular between parallel planes, in which these lines lie. Straight line BA 1 lies in the plane АВВ 1 А 1 , and straight line DC 1 lies in the plane D 1 DCC 1 . They are parallel, which means the distance between them is the distance between these straight lines. And the distance between the faces of the cube is the length of the edge. For example, the length of the rib BC. Because BC is perpendicular to both the plane АВВ 1 А 1 and the plane DСС 1 D 1. This means that the distance between the straight lines given in the condition is equal to the distance between parallel planes and is equal to 1:
ρ(A 1 B;DC 1)=BC=1
Let's consider another problem about finding the distance between crossing lines.
Let us be given the correct one triangular prism, for which all edges are known. You need to find the distance between the edges of the upper and lower bases. That is, we are given a prism ABCA 1 B 1 C 1. Moreover, AB=3=AA 1. You need to find the distance between straight lines BC and A 1 C 1: ρ(BC;A 1 C 1) - ?
Since these lines intersect, the distance between them is the length of the common perpendicular, or the length of the perpendicular to the parallel planes in which they lie. Let's find these parallel planes.
Direct sun lies in ABC plane, and straight line A 1 C 1 lies in the plane A 1 B 1 C 1. These two planes are parallel because they are the top and bottom bases of the prism. This means that the distance between our straight lines is the distance between these parallel planes. And the distance between them is exactly equal to the length lateral rib AA 1, that is, equal to 3:
ρ(BC;A 1 C 1)=AA 1 =3
In this specific task you can find not only the length of the common perpendicular, but also construct it. To do this, from all the side edges we select one that has common points with straight line BC and A 1 C 1. In our figure this is edge CC 1. It will be perpendicular to straight line A 1 C 1, since it is perpendicular to the plane of the upper base, and to straight line BC, since it is perpendicular to the plane of the lower base. Thus, we can find not only the distance, but also construct this general perpendicular.
Today in the lesson we remembered how to find the length of the common perpendicular between intersecting lines.
The article is aimed at finding the distance between crossing lines using the coordinate method. Determination of the distance between these lines will be considered, we will obtain an algorithm with the help of which we will transform the determination of the distance between crossing lines. Let's consolidate the topic by solving similar examples.
Yandex.RTB R-A-339285-1
It is first necessary to prove a theorem that defines the connection between given crossing lines.
Chapter relative position straight lines in space says that if two straight lines are called intersecting if their location is not in the same plane.
Theorem
Through each pair of intersecting straight lines there can pass a plane parallel to the given one, and only one.
Proof
By condition, we are given skew lines a and b. It is necessary to prove the permeability of a single plane through a line b parallel to a given line a. A similar proof must be applied for a line a, through which passes a plane parallel to a given line b.
First you need to mark point Q on line b. If we follow from the definition of parallelism of lines, we find that through a point in space it is possible to draw a line parallel to a given line, and only one. This means that only one line passes through point Q, parallel to line a. Let's take the notation a a 1 .
In the section on methods for specifying a plane, it was said that the passage of a single plane is possible through two intersecting lines. This means that we find that lines b and a 1 are intersecting lines through which the plane denoted χ passes.
Based on the sign that a line is parallel to a plane, we can conclude that a given straight line a is parallel to the plane χ, because straight line a is parallel to straight line a 1 located in the plane χ.
The χ plane is unique, since a line passing through a given line located in space is parallel to the given line. Let's look at the picture provided below.
When moving from determining the distance between intersecting straight lines, we determine the distance through the distance between the straight line and the plane parallel to it.
Definition 1
The distance between one of the intersecting lines and a plane parallel to it passing through the other line is called.
That is, the distance between a straight line and a plane is the distance from a given point to the plane. Then the formulation for determining the distance between crossing lines is applicable.
Definition 2
Distance between crossing lines call the distance from a certain point of intersecting lines to a plane passing through another line parallel to the first line.
We will produce detailed consideration straight lines a and b. Point M 1 is located on line a, through line b a plane χ is drawn parallel to line a. From point M 1 we draw a perpendicular M 1 H 1 to the plane χ. The length of this perpendicular is the distance between the crossing lines a and b. Let's look at the figure below.
Finding the distance between crossing lines - theory, examples, solutions
The distances between intersecting lines are found when constructing a segment. The required distance is equal to the length of this segment. According to the conditions of the problem, its length is determined by the Pythagorean theorem, by the signs of equality or similarity of triangles, or others.
When we have a three-dimensional space with a coordinate system O x y z with straight lines a and b given in it, then calculations should be carried out starting from the distance between the given crossing ones using the coordinate method. Let's take a detailed look.
Let, by condition, χ be a plane passing through the line b, which is parallel to the line a. The required distance between the crossing lines a and b is equal to the distance from the point M 1 located on the line a to the plane _ χ. In order to obtain the normal equation of the χ plane, it is necessary to determine the coordinates of the point M 1 (x 1, y 1, z 1) located on the straight line a. Then we obtain cos α · x + cos β · y + cos γ · z - p = 0, which is necessary to determine the distance M 1 H 1 from the point M 1 x 1, y 1, z 1 to the χ plane. Calculations are made using the formula M 1 H 1 = cos α · x 1 + cos β · y 1 + cos γ · z 1 - p. The required distance is equal to the required distance between the crossing lines.
This task involves obtaining the coordinates of point M 1, which is located on straight line a, finding normal equation plane χ.
Determining the coordinates of point M 1 is necessary and possible if you know the basic types of equations of a straight line in space. To obtain the equation of the χ plane, it is necessary to take a closer look at the calculation algorithm.
If the coordinates x 2 , y 2 , z 2 are determined using the point M 2 through which the plane χ is drawn, we obtain the normal vector of the plane χ in the form of a vector n → = (A, B, C). Following from this, we can write general equation plane χ in the form A · x - x 2 + B · (y - y 2) + C · (z - z 2) = 0.
Instead of point M 2, any other point belonging to line b can be taken, because the plane χ passes through it. This means that the coordinates of point M 2 have been found. It is necessary to proceed to finding the normal vector of the plane χ.
We have that the plane χ passes through the line b, and is parallel to the line a. This means that the normal vector of the plane χ is perpendicular to the direction vector of the line a, denoted a →, and to the direction vector of the line b, denoted b →. Vector n → will be equal to vector product a → and b →, which means n → = a → × b →. After determining the coordinates a x , a y , a z and b x , b y , b z of the direction vectors of the given lines a and b , we calculate
n → = a → × b → = i → j → k → a x a y a z b x b y b z
From here we find the value of the coordinates A, B, C of the normal vector to the χ plane.
We know that the general equation of the χ plane has the form A · (x - x 2) + B · (y - y 2) + C · (z - z 2) = 0.
It is necessary to bring the equation to the normal form cos α · x + cos β · y + cos γ · z - p = 0 . Then you need to calculate the required distance between the crossing lines a and b, based on the formula M 1 H 1 = cos α · x 1 + cos β · y 1 + cos γ · z 1 - p.
To find the distance between crossing lines a and b, you must follow the algorithm:
- determination of coordinates (x 1, y 1, z 1) and x 2, y 2, z 2 of points M 1 and M 2 located on straight lines a and b, respectively;
- obtaining coordinates a x , a y , a z and b x , b y , b z belonging to the direction vectors of lines a and b ;
- finding the coordinates A, B, C belonging to the vector n → on the plane χ passing through the line b located parallel to a, by the equality n → = a → × b → = i → j → k → a x a y a z b x b y b z;
- writing the general equation of the χ plane in the form A · x - x 2 + B · (y - y 2) + C · (z - z 2) = 0;
- bringing the resulting equation of the χ plane to the normal equation type cosα · x + cos β · y + cos γ · z - p = 0 ;
- calculating the distance M 1 H 1 from M 1 x 1, y 1, z 1 to the χ plane, based on the formula M 1 H 1 = cos α · x 1 + cos β · y 1 + cos γ · z 1 - p.
There are two intersecting lines at rectangular system coordinates O x y z three-dimensional space. Line a is defined parametric equation line in space x = - 2 y = 1 + 2 λ z = 4 - 3 λ, line b using canonical equation straight line in space x 1 = y - 1 - 2 = z + 4 6. Find the distance between intersecting lines.
Solution
It is clear that straight line a intersects point M 1 (- 2, 1, 4) with the direction vector a → = (0, 2, - 3), and straight line b intersects point M 2 (0, 1, - 4) with the direction vector b → = (1 , - 2 , 6) .
First, you should calculate the direction vectors a → = (0, 2, - 3) and b → = (1, - 2, 6) using the formula. Then we get that
a → × b → = i → j → k → 0 2 - 3 1 - 2 6 = 6 i → - 3 j → - 2 k →
From here we get that n → = a → × b → is a vector of the plane χ, which passes through the line b parallel to a with coordinates 6, - 3, - 2. We get:
6 (x - 0) - 3 (y - 1) - 2 (z - (- 4)) = 0 ⇔ 6 x - 3 y - 2 z - 5 = 0
We find the normalizing factor for the general equation of the plane 6 x - 3 y - 2 z - 5 = 0. Let's calculate using the formula 1 6 2 + - 3 2 + - 2 2 = 1 7. This means that the normal equation will take the form 6 7 x - 3 7 y - 2 7 z - 5 7 = 0.
It is necessary to use the formula to find the distance from the point M 1 - 2, 1, 4 to the plane, given by the equation 6 7 x - 3 7 y - 2 7 z - 5 7 = 0 . We get that
M 1 H 1 = 6 7 (- 2) - 3 7 1 - 2 7 4 - 5 7 = - 28 7 = 4
It follows that the required distance is the distance between the given crossing lines, the value is 4.
Answer: 4 .
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This article focuses on finding the distance between crossing lines using the coordinate method. First, the definition of the distance between intersecting lines is given. Next, an algorithm is obtained that allows one to find the distance between crossing lines. In conclusion, the solution to the example is analyzed in detail.
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Distance between crossing lines - definition.
Before giving the definition of the distance between skew lines, let us recall the definition of skew lines and prove a theorem related to skew lines.
Definition.
- this is the distance between one of the intersecting lines and a plane parallel to it passing through the other line.
In turn, the distance between a straight line and a plane parallel to it is the distance from some point of the straight line to the plane. Then the following formulation of the definition of the distance between crossing lines is valid.
Definition.
Distance between crossing lines is the distance from a certain point of one of the intersecting lines to a plane passing through another line parallel to the first line.
Consider the crossing lines a and b. Let's mark a certain point M 1 on line a, draw a plane parallel to line a through line b, and from point M 1 lower a perpendicular M 1 H 1 to the plane. The length of the perpendicular M 1 H 1 is the distance between the crossing lines a and b.
Finding the distance between crossing lines - theory, examples, solutions.
When finding the distance between crossing lines, the main difficulty is often to see or construct a segment whose length is equal to the desired distance. If such a segment is constructed, then, depending on the conditions of the problem, its length can be found using the Pythagorean theorem, signs of equality or similarity of triangles, etc. This is what we do when finding the distance between intersecting lines in geometry lessons in grades 10-11.
If in three-dimensional space Oxyz is introduced and intersecting lines a and b are given in it, then the coordinate method allows us to cope with the task of calculating the distance between given intersecting lines. Let's look at it in detail.
Let be a plane passing through line b, parallel to line a. Then the required distance between the crossing lines a and b is, by definition, equal to the distance from some point M 1 lying on the line a to the plane. Thus, if we determine the coordinates of a certain point M 1 lying on a line a, and obtain the normal equation of the plane in the form, then we can calculate the distance from the point 
to the plane using the formula (this formula was obtained in the article finding the distance from a point to a plane). And this distance is equal to the required distance between the crossing lines.
Now in detail.
The problem comes down to obtaining the coordinates of the point M 1 lying on the line a and finding the normal equation of the plane.
There are no difficulties in determining the coordinates of point M 1 if you know well the basic types of equations of a straight line in space. But it is worth dwelling in more detail on obtaining the equation of the plane.
If we determine the coordinates of a certain point M 2 through which the plane passes, and also obtain the normal vector of the plane in the form 
, then we can write the general equation of the plane as .
As a point M 2, you can take any point lying on the line b, since the plane passes through the line b. Thus, the coordinates of point M 2 can be considered found.
It remains to obtain the coordinates of the normal vector of the plane. Let's do it.
The plane passes through line b and is parallel to line a. Consequently, the normal vector of the plane is perpendicular to both the direction vector of the line a (let's denote it) and the direction vector of the line b (let's denote it). Then we can take and as a vector, that is, . Having determined the coordinates and direction vectors of straight lines a and b and calculated 
, we will find the coordinates of the normal vector of the plane.
So, we have the general equation of the plane: .
All that remains is to bring the general equation of the plane to normal form and calculate the required distance between the crossing lines a and b using the formula.
Thus, to find the distance between crossing lines a and b you need:
Let's look at the solution to the example.
Example.
In three-dimensional space in the rectangular coordinate system Oxyz, two intersecting straight lines a and b are given. The straight line a is determined