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

Oh-oh-oh-oh-oh ... well, it's tinny, as if you read the sentence to yourself =) However, then relaxation will help, especially since today I bought suitable accessories. Therefore, let's proceed to the first section, I hope, by the end of the article I will keep a cheerful mood.

Mutual arrangement of two straight lines

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

1) match;

2) be parallel: ;

3) or intersect at a single point: .

Help for dummies : please remember the mathematical sign of the intersection , it will occur very often. The entry means that the line intersects with the line at the point.

How to determine the relative position of two lines?

Let's start with the first case:

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

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

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

The second case when the lines are parallel:

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

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

However, it is clear that .

And the third case, when the lines intersect:

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

So, for straight lines we will compose a system:

From the first equation it follows that , and from the second equation: , hence, the system is inconsistent(no solutions). Thus, the coefficients at the variables are not proportional.

Conclusion: lines intersect

AT practical tasks the solution scheme just discussed can be used. By the way, it is very similar to the algorithm for checking vectors for collinearity, which we considered in the lesson. The concept of linear (non) dependence of vectors. Vector basis. But there is a more civilized package:

Example 1

Find out the relative position of the lines:

Decision based on the study of directing vectors of straight lines:

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

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

Just in case, I will put a stone with pointers at the crossroads:

The rest jump over the stone and follow on, straight to Kashchei the Deathless =)

b) Find the direction vectors of the lines:

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

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

Let's find out if the equality is true:

Thus,

c) Find the direction vectors of the lines:

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

The proportionality factor "lambda" is easy to see directly from the ratio of collinear direction vectors. However, it can also be found through the coefficients of the equations themselves: .

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

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

Thus, the lines coincide.

Answer:

Very soon you will learn (or even have already learned) to solve the considered problem verbally literally in a matter of seconds. In this regard, I see no reason to offer anything for independent decision, it is better to lay another important brick in the geometric foundation:

How to draw a line parallel to a given one?

For ignorance of this the simplest task severely punishes the Nightingale the Robber.

Example 2

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

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

We take out the direction vector from the equation:

Answer:

The geometry of the example looks simple:

Analytical verification consists of the following steps:

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

2) Check if the point satisfies the resulting equation.

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

Examples for self-solving today will be creative. Because you still have to compete with Baba Yaga, and she, you know, is a lover of all kinds of riddles.

Example 3

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

There is a rational and not very rational way to solve. The shortest way is at the end of the lesson.

We did a little work with parallel lines and will return to them later. The case of coinciding lines is of little interest, so consider a problem that is well known to you from school curriculum:

How to find the point of intersection of two lines?

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

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

Here's to you geometric meaning two linear equations with two unknowns are two intersecting (most often) straight lines on a plane.

Example 4

Find the point of intersection of lines

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

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

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

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

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

To solve the system, the method of termwise addition of equations was used. To develop the relevant skills, visit the lesson How to solve a system of equations?

Answer:

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

Example 5

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

This is a do-it-yourself example. It is convenient to divide the problem into several stages. Analysis of the condition suggests that it is necessary:
1) Write the equation of a straight line.
2) Write the equation of a straight line.
3) Find out the relative position of the lines.
4) If the lines intersect, then find the point of intersection.

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

Complete Solution and the answer at the end of the lesson:

A pair of shoes has not yet been worn out, as we got to the second section of the lesson:

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

Let's start with a typical and very important task. In the first part, we learned how to build a straight line parallel to the given one, and now the hut on chicken legs will turn 90 degrees:

How to draw a line perpendicular to a given one?

Example 6

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

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

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

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

Answer:

Let's unfold the geometric sketch:

Hmmm... Orange sky, orange sea, orange camel.

Analytical verification of the solution:

1) Extract the direction vectors from the equations and with the help dot product of vectors we conclude that the lines are indeed perpendicular: .

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

2) Check if the point satisfies the resulting equation .

Verification, again, is easy to perform verbally.

Example 7

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

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

Our exciting journey continues:

Distance from point to line

Before us is a straight strip of the river and our task is to reach it in the shortest way. There are no obstacles, and the most optimal route will be movement along the perpendicular. That is, the distance from a point to a line is the length of the perpendicular segment.

The distance in geometry is traditionally denoted by the Greek letter "ro", for example: - the distance from the point "em" to the straight line "de".

Distance from point to line is expressed by the formula

Example 8

Find the distance from a point to a line

Decision: all you need is to carefully substitute the numbers into the formula and do the calculations:

Answer:

Let's execute the drawing:

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

Consider another task according to the same drawing:

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

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

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

Both actions are discussed in detail in this lesson.

3) The point is the midpoint of the segment. We know the coordinates of the middle and one of the ends. By formulas for the coordinates of the middle of the segment find .

It will not be superfluous to check that the distance is also equal to 2.2 units.

Difficulties here may arise in calculations, but in the tower a microcalculator helps out a lot, allowing you to count common fractions. Have advised many times and will recommend again.

How to find the distance between two parallel lines?

Example 9

Find the distance between two parallel lines

This is another example for an independent solution. A little hint: there are infinitely many ways to solve. Debriefing at the end of the lesson, but better try to guess for yourself, I think you managed to disperse your ingenuity well.

Angle between two lines

Whatever the corner, then the jamb:

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

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

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

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

How to find the angle between two lines? There are two working formulas:

Example 10

Find the angle between lines

Decision and Method one

Consider two straight lines given by equations in general view:

If straight not perpendicular, then oriented the angle between them can be calculated using the formula:

Most close attention turn to the denominator - this is exactly scalar product direction vectors of straight lines:

If , then the denominator of the formula vanishes, and the vectors will be orthogonal and the lines will be perpendicular. That is why a reservation was made about the non-perpendicularity of the lines in the formulation.

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

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

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

Via inverse function easy to find the corner itself. In this case, we use the oddness of the arc tangent (see Fig. Graphs and properties of elementary functions):

Answer:

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

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

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

If you really want to get a positive angle, you need to swap the straight lines, that is, take the coefficients from the second equation , and take the coefficients from the first equation . In short, you need to start with a direct .

Formulation of the problem. Find the coordinates of a point symmetrical to a point relative to the plane.

Solution plan.

1. We find the equation of a straight line that is perpendicular to a given plane and passes through a point . Since the line is perpendicular to the given plane, then the vector of the normal of the plane can be taken as its direction vector, i.e.

Therefore, the equation of a straight line will be

2. Find a point line intersection and planes (see Problem 13).

3. Point is the midpoint of the segment, where the point is a point symmetric to a point , That's why

Task 14. Find a point symmetrical to a point with respect to the plane.

The equation of a straight line that passes through a point perpendicular to a given plane will be:

Find the point of intersection of the line and the plane.

Where - the point of intersection of the line and the plane. is the midpoint of the segment, therefore

Those. .

Homogeneous plane coordinates. Affine transformations on the plane.

Let be M X and at

M(X, atMe (X, at, 1) in space (Fig. 8).

Me (X, at

Me (X, at hu.

(hx, hy, h), h  0,

Comment

h(For example, h

Indeed, considering h

Comment

Example 1

b) at the corner (Fig. 9).

1st step.

2nd step. Angle rotation 

the matrix of the corresponding transformation.

3rd step. Transfer to the vector A(a, b)

the matrix of the corresponding transformation.

Example 3

 along the x-axis and

1st step.

the matrix of the corresponding transformation.

2nd step.

3rd step.

finally get

Comment

[R],[D],[M],[T],

Let be M- arbitrary point of the plane with coordinates X and at calculated with respect to a given rectilinear coordinate system. The homogeneous coordinates of this point are any triple of simultaneously non-zero numbers x 1, x 2, x 3, associated with the given numbers x and y by the following relations:

When solving computer graphics problems, homogeneous coordinates are usually introduced as follows: an arbitrary point M(X, at) the plane is assigned a point Me (X, at, 1) in space (Fig. 8).

Note that an arbitrary point on the line connecting the origin, the point 0(0, 0, 0), with the point Me (X, at, 1) can be given by a triple of numbers of the form (hx, hy, h).

The vector with coordinates hx, hy is the direction vector of the straight line connecting the points 0 (0, 0, 0) and Me (X, at, one). This line intersects the plane z = 1 at the point (x, y, 1), which uniquely determines the point (x, y) of the coordinate plane hu.

Thus, between an arbitrary point with coordinates (x, y) and a set of triples of numbers of the form

(hx, hy, h), h  0,

a (one-to-one) correspondence is established, which allows us to consider the numbers hx, hy, h as new coordinates of this point.

Comment

Homogeneous coordinates widely used in projective geometry make it possible to effectively describe the so-called improper elements (essentially those in which the projective plane differs from the Euclidean plane familiar to us). More details about the new features provided by the introduced homogeneous coordinates are discussed in the fourth section of this chapter.

In projective geometry, for homogeneous coordinates, the following notation is accepted:

x: y: 1, or, more generally, x 1: x 2: x 3

(recall that here it is absolutely required that the numbers x 1, x 2, x 3 at the same time do not vanish).

The use of homogeneous coordinates turns out to be convenient even when solving the simplest problems.

Consider, for example, issues related to scaling. If the display device works only with integers (or if it is necessary to work only with integers), then for an arbitrary value h(For example, h= 1) a point with homogeneous coordinates

cannot be imagined. However, with a reasonable choice of h, it is possible to ensure that the coordinates of this point are integers. In particular, for h = 10, for the example under consideration, we have

Let's consider another case. So that the results of the transformation do not lead to arithmetic overflow, for a point with coordinates (80000 40000 1000) you can take, for example, h=0.001. As a result, we get (80 40 1).

The examples given show the usefulness of using homogeneous coordinates in calculations. However, the main purpose of introducing homogeneous coordinates in computer graphics is their undoubted convenience in applying to geometric transformations.

With the help of triples of homogeneous coordinates and matrices of the third order, any affine transformation of the plane can be described.

Indeed, considering h= 1, compare two entries: marked with * and the following, matrix:

It is easy to see that after multiplying the expressions on the right side of the last relation, we get both formulas (*) and the correct numerical equality 1=1.

Comment

Sometimes in the literature another notation is used - a notation by columns:

This notation is equivalent to the line notation above (and is obtained from it by transposition).

The elements of an arbitrary matrix of an affine transformation do not carry an explicit geometric meaning. Therefore, in order to implement a particular mapping, that is, to find the elements of the corresponding matrix according to a given geometric description, special techniques are needed. Usually, the construction of this matrix, in accordance with the complexity of the problem under consideration and with the particular cases described above, is divided into several stages.

At each stage, a matrix is searched for that corresponds to one or another of the above cases A, B, C, or D, which have well-defined geometric properties.

Let us write out the corresponding matrices of the third order.

A. Rotation matrix, (rotation)

B. Dilatation Matrix

B. Reflection Matrix

D. Transfer Matrix (translation)

Consider examples of affine transformations of the plane.

Example 1

Build a rotation matrix around point A (a,b) at the corner (Fig. 9).

1st step. Transfer to the vector - A (-a, -b) to align the center of rotation with the origin;

the matrix of the corresponding transformation.

2nd step. Angle rotation 

the matrix of the corresponding transformation.

3rd step. Transfer to the vector A(a, b) to return the center of rotation to its previous position;

the matrix of the corresponding transformation.

We multiply the matrices in the same order as they are written out:

As a result, we get that the desired transformation (in matrix notation) will look like this:

The elements of the resulting matrix (especially in the last row) are not easy to remember. At the same time, each of the three multiplied matrices can be easily constructed from the geometric description of the corresponding mapping.

Example 3

Build Stretch Matrix with Stretch Factors along the x-axis and along the y-axis and centered at point A(a, b).

1st step. Transfer to the vector -А(-а, -b) to match the stretching center with the origin;

the matrix of the corresponding transformation.

2nd step. Stretching along the coordinate axes with coefficients  and , respectively; the transformation matrix has the form

3rd step. Transfer to the vector A(a, b) to return the stretching center to its previous position; the matrix of the corresponding transformation is

Multiply matrices in the same order

finally get

Comment

Arguing in a similar way, that is, breaking the proposed transformation into stages supported by matrices[R],[D],[M],[T], one can construct the matrix of any affine transformation from its geometric description.

Shift is implemented by addition, and scaling and rotation by multiplication.

Scale transformation (dilation) relative to the origin has the form:

or in matrix form:

where Dx,Dy are the scaling factors along the axes, and

- scaling matrix.

For D > 1, expansion occurs, for 0<=D<1- сжатие

Rotate Transform relative to the origin has the form:

or in matrix form:

where φ is the angle of rotation, and

- rotation matrix.

Comment: The columns and rows of the rotation matrix are mutually orthogonal unit vectors. Indeed, the squares of the lengths of the row vectors are equal to one:

cosφ cosφ+sinφ sinφ = 1 and (-sinφ) (-sinφ)+cosφ cosφ = 1,

and the scalar product of row vectors is

cosφ (-sinφ) + sinφ cosφ= 0.

Since the scalar product of vectors A · B = |A| ·| B| ·cosψ, where | A| - vector length A, |B| - vector length B, and ψ is the smallest positive angle between them, then from the equality 0 of the scalar product of two row vectors of length 1 it follows that the angle between them is 90 ° .

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

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

Both actions are discussed in detail in this lesson.

3) The point is the midpoint of the segment. We know the coordinates of the middle and one of the ends. By formulas for the coordinates of the middle of the segment find .

It will not be superfluous to check that the distance is also equal to 2.2 units.

Difficulties here may arise in calculations, but in the tower a microcalculator helps out a lot, allowing you to count ordinary fractions. Have advised many times and will recommend again.

How to find the distance between two parallel lines?

Example 9

Find the distance between two parallel lines

Angle between two lines

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

How to find the angle between two lines? There are two working formulas:

Example 10

Find the angle between lines

Decision and Method one

Consider two straight lines given by equations in general form:

If straight not perpendicular, then oriented the angle between them can be calculated using the formula:

Let's pay close attention to the denominator - this is exactly scalar product direction vectors of straight lines:

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

1) Calculate the scalar product of directing vectors of straight lines:

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

Using the inverse function, it is easy to find the angle itself. In this case, we use the oddness of the arc tangent (see Fig. Graphs and Properties elementary functions ):

Answer:

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

I will not hide, I myself select the straight lines in the order that the angle is positive. It's more beautiful, but nothing more.

To check the solution, you can take a protractor and measure the angle.

Method two

If the lines are given by equations with slope and not perpendicular, then oriented the angle between them can be found using the formula:

The condition of perpendicularity of straight lines is expressed by the equality, from which, by the way, a very useful relationship of slope coefficients of perpendicular straight lines follows: , which is used in some problems.

The solution algorithm is similar to the previous paragraph. But first, let's rewrite our lines in the required form:

Thus, the slope coefficients:

1) Check if the lines are perpendicular:
so the lines are not perpendicular.

2) We use the formula:

Answer:

The second method is appropriate to use when the equations of lines are initially set with a slope. It should be noted that if at least one line is parallel to the y-axis, then the formula is not applicable at all, since the slope is not defined for such lines (see article Equation of a straight line on a plane).

There is also a third solution. The idea is to calculate the angle between the direction vectors of the lines using the formula discussed in the lesson Dot product of vectors:

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

Those who wish can solve the problem in a third way. But I still recommend sticking to the first angle-oriented approach, because it's widely used.

Example 11

Find the angle between the lines.

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

Somehow the fairy tale died out along the way .... Because there is no Kashchei the Immortal. There is me, and not particularly steamed. To be honest, I thought the article would be much longer. But all the same, I will take a recently acquired hat with glasses and go for a swim in the September lake water. Perfectly relieves fatigue and negative energy.

See you soon!

And remember, Baba Yaga has not been canceled =)

Solutions and answers:

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

We will compose the equation of the desired straight line using the point and direction vector . Since one of the direction vector coordinates is zero, the equation rewrite in the form:

Answer :

Example 5:Decision :
1) Straight line equation make two points :

2) Straight line equation make two points :

3) Corresponding coefficients for variables out of proportion: , so the lines intersect.
4) Find a point :

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

Let some straight line given by a linear equation and a point given by its coordinates (x0, y0) and not lying on this straight line be given. It is required to find a point that would be symmetrical to a given point with respect to a given straight line, that is, would coincide with it, if the plane is mentally bent in half along this straight line.

Instruction

1. It is clear that both points - given and desired - must lie on the same straight line, and this line must be perpendicular to the given one. Thus, the first part of the problem is to find the equation of a straight line that would be perpendicular to some given line and at the same time pass through a given point.

2. A straight line can be defined in two ways. The canonical equation of a straight line looks like this: Ax + By + C = 0, where A, B, and C are constants. Also, a straight line can be determined using a linear function: y \u003d kx + b, where k is the angular exponent, b is the displacement. These two methods are interchangeable, and it is allowed to move from each to another. If Ax + By + C = 0, then y = – (Ax + C)/B. In other words, in a linear function y = kx + b, the angular exponent k = -A/B, and the offset b = -C/B. For the task at hand, it is more comfortable to reason on the basis of the canonical equation of a straight line.

3. If two lines are perpendicular to each other, and the equation of the first line is Ax + By + C = 0, then the equation of the 2nd line should be Bx - Ay + D = 0, where D is a constant. In order to find a certain value of D, it is necessary to additionally know through which point the perpendicular line passes. In this case, this is the point (x0, y0). Consequently, D must satisfy the equality: Bx0 – Ay0 + D = 0, that is, D = Ay0 – Bx0.

4. Later, after the perpendicular line is found, it is necessary to calculate the coordinates of the point of its intersection with the given one. To do this, you need to solve a system of linear equations: Ax + By + C = 0, Bx - Ay + Ay0 - Bx0 = 0. Its solution will give the numbers (x1, y1) that serve as the coordinates of the point of intersection of the lines.

5. The desired point must lie on the detected line, and its distance to the intersection point must be equal to the distance from the intersection point to the point (x0, y0). The coordinates of a point symmetrical to the point (x0, y0) can thus be found by solving the system of equations: Bx - Ay + Ay0 - Bx0 = 0,?((x1 - x0)^2 + (y1 - y0)^2 = ?((x – x1)^2 + (y – y1)^2).

6. But let's make it easier. If the points (x0, y0) and (x, y) are at equal distances from the point (x1, y1), and all three points lie on the same straight line, then: x - x1 = x1 - x0,y - y1 = y1 - y0. Consequently, x = 2×1 – x0, y = 2y1 – y0. Substituting these values into the second equation of the first system and simplifying the expressions, it is easy to make sure that its right side becomes the same as the left side. In addition, it makes no sense to consider the first equation more closely, because it is known that the points (x0, y0) and (x1, y1) satisfy it, and the point (x, y) certainly lies on the same line.

A straight line in space can always be defined as a line of intersection of two non-parallel planes. If the equation of one plane is the equation of the second plane, then the equation of the straight line is given as

here non-collinear
. These equations are called general equations straight line in space.

Canonical equations of the straight line

Any non-zero vector lying on a given line or parallel to it is called a directing vector of this line.

If the point is known
line and its direction vector
, then the canonical equations of the line have the form:

. (9)

Parametric equations of a straight line

Let the canonical equations of the line be given

From here, we obtain the parametric equations of the straight line:

(10)

These equations are useful for finding the point of intersection of a line and a plane.

Equation of a line passing through two points
and
looks like:

Angle between lines

Angle between lines

and

is equal to the angle between their direction vectors. Therefore, it can be calculated by formula (4):

Condition of parallel lines:

Condition of perpendicularity of planes:

Distance of a point from a straight line

P given point
and direct

From the canonical equations of the line, the point is known
, belonging to the line, and its direction vector
. Then the point distance
from a straight line is equal to the height of a parallelogram built on vectors and
. Hence,

Line intersection condition

Two non-parallel lines

intersect if and only if

Mutual arrangement of a straight line and a plane.

Let the straight line
and flat. Injection between them can be found by the formula

Problem 73. Write the canonical equations of the line

(11)

Decision. In order to write down the canonical equations of the line (9), it is necessary to know any point belonging to the line and the directing vector of the line.

Let's find the vector parallel to the given line. Since it must be perpendicular to the normal vectors of these planes, i.e.

,
, then

From the general equations of the straight line, we have that
,
. Then

Since the point
any point of the line, then its coordinates must satisfy the equations of the line, and one of them can be specified, for example,
, we find the other two coordinates from the system (11):

From here,
.

Thus, the canonical equations of the desired line have the form:

or
.

Problem 74.

and
.

Decision. From the canonical equations of the first line, the coordinates of the point are known
belonging to the line, and the coordinates of the direction vector
. From the canonical equations of the second line, the coordinates of the point are also known
and direction vector coordinates
.

The distance between parallel lines is equal to the distance of a point
from the second line. This distance is calculated by the formula

Let's find the coordinates of the vector
.

Compute the vector product
:

Problem 75. Find a point symmetrical point
relatively straight

Decision. We write the equation of the plane perpendicular to the given line and passing through the point . As its normal vector we can take the directing vector as a straight line. Then
. Hence,

Let's find a point
the point of intersection of the given line and the plane P. To do this, we write the parametric equations of the line using equations (10), we obtain

Hence,
.

Let be
point symmetrical to point
about this line. Then the point
midpoint
. To find the coordinates of a point we use the formulas for the coordinates of the middle of the segment:

,
,
.

So,
.

Problem 76. Write the equation for a plane passing through a straight line
and

a) through a dot
;

b) perpendicular to the plane.

Decision. Let us write down the general equations of this straight line. To do this, consider two equalities:

This means that the desired plane belongs to a pencil of planes with generators and its equation can be written in the form (8):

a) find
and from the condition that the plane passes through the point
, therefore, its coordinates must satisfy the equation of the plane. Substitute the coordinates of the point
into the equation of a beam of planes:

Found value
we substitute into equation (12). we obtain the equation of the desired plane:

b) find
and from the condition that the desired plane is perpendicular to the plane. The normal vector of a given plane
, the normal vector of the desired plane (see the equation for a bundle of planes (12).