Home Prosthetics and implantation Coordinates of a point symmetrical to a point relative to a straight line online. The simplest problems with a straight line on a plane

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Coordinates of a point symmetrical to a point relative to a straight line online. The simplest problems with a straight line on a plane

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

Solution plan.

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

Therefore the equation of the straight line will be

2. Find the point intersection of a straight line and planes (see problem 13).

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

Problem 14. Find a point symmetrical to the point relative to the plane.

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

Let's find the point of intersection of the line and the plane.

Where – the point of intersection of a line and a plane. is the middle of the segment, therefore

Those. .

Homogeneous plane coordinates. Affine transformations on the plane.

Let M X And at

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

Mae (X, at

Mae (X, at hu.

(hx, hy, h), h  0,

Comment

h(For example, h

In fact, considering h

Comment

Example 1.

b) to an angle (Fig. 9).

1st step.

2nd step. Rotate by angle 

matrix of the corresponding transformation.

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

matrix of the corresponding transformation.

Example 3

 along the x-axis and

1st step.

matrix of the corresponding transformation.

2nd step.

3rd step.

we will finally get it

Comment

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

Let M- arbitrary point of the plane with coordinates X And at, calculated relative 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, related to the given numbers x and y by the following relations:

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

Note that an arbitrary point on the line connecting the origin, point 0(0, 0, 0), with the point Mae (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 points 0 (0, 0, 0) and Mae (X, at, 1). This line intersects the z = 1 plane at the point (x, y, 1), which uniquely defines 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 that allows us to consider the numbers hx, hy, h as the new coordinates of this point.

Comment

Widely used in projective geometry, homogeneous coordinates make it possible to effectively describe the so-called improper elements (essentially those in which the projective plane differs from the familiar Euclidean plane). More details about the new possibilities 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, x1:x2:x3

(remember that here it is absolutely required that the numbers x 1, x 2, x 3 do not turn to zero at the same time).

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

Consider, for example, issues related to changes in scale. If the display device only works with integers (or if you need to work only with integers), then for an arbitrary value h(For example, h= 1) a point with homogeneous coordinates

impossible to imagine. 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. To prevent the transformation results from leading 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 when carrying out calculations. However, the main purpose of introducing homogeneous coordinates in computer graphics is their undoubted convenience in application to geometric transformations.

Using triples of homogeneous coordinates and third-order matrices, any affine transformation of a plane can be described.

In fact, considering h= 1, compare two entries: marked with the symbol * and the following matrix one:

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

Comment

Sometimes in the literature another notation is used - columnar notation:

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

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

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

Let us write down the corresponding third-order matrices.

A. Rotation matrix

B. Dilatation matrix

B. Reflection matrix

D. Transfer matrix (translation)

Let's consider examples of affine transformations of the plane.

Example 1.

Construct a rotation matrix around point A (a,b) to an angle (Fig. 9).

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

matrix of the corresponding transformation.

2nd step. Rotate by angle 

matrix of the corresponding transformation.

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

matrix of the corresponding transformation.

Let's multiply the matrices in the same order as they are written:

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

The elements of the resulting matrix (especially in the last row) are not so 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

Construct a stretch matrix with stretch coefficients along the x-axis and along the ordinate axis and with the center at point A(a, b).

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

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 vector A(a, b) to return the center of tension to its previous position; matrix of the corresponding transformation –

Multiplying matrices in the same order

we will finally get it

Comment

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

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

Scaling Transform (dilatation) relative to the origin has the form:

or in matrix form:

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

- scaling matrix.

When D > 1, expansion occurs, when 0<=D<1- сжатие

Rotation transformation 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. In fact, the squares of the lengths of 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 °.

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

Instructions

1. It is clear that both points - the given and the desired - must lie on the same line, and this line must be perpendicular to the given one. Thus, the first part of the problem is to discover the equation of a 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 specified in two ways. The canonical equation of a line looks like this: Ax + By + C = 0, where A, B, and C are constants. You can also determine a straight line using a linear function: y = kx + b, where k is the angular exponent, b is the displacement. These two methods are interchangeable, and you can move from each to the other. 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 displacement b = -C/B. For the task at hand, it is more comfortable to reason based on the canonical equation of the 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 look like Bx – Ay + D = 0, where D is a constant. In order to detect 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. After the perpendicular line has been discovered, 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), which 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 you can do it easier. If points (x0, y0) and (x, y) are at equal distances from 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. By 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. In addition, there is no point in considering the first equation any further, since it is known that the points (x0, y0) and (x1, y1) satisfy it, and the point (x, y) obviously lies on the same line.

The task is to find the coordinates of a point that is symmetrical to the point relative to the straight line . I suggest performing the steps yourself, but I will outline a solution algorithm with intermediate results:

1) Find a line that is perpendicular to the 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 midpoint of a segment we find .

It would be a good idea to check that the distance is also 2.2 units.

Difficulties may arise in calculations here, but a microcalculator is a great help in the tower, allowing you to calculate ordinary fractions. I have advised you many times and will recommend you again.

How to find the distance between two parallel lines?

Example 9

Find the distance between two parallel lines

This is another example for you to decide on your own. I’ll give you a little hint: there are infinitely many ways to solve this. Debriefing at the end of the lesson, but it’s better to try to guess for yourself, I think your ingenuity was well developed.

Angle between two straight lines

Every corner is a jamb:

In geometry, the angle between two straight lines is taken to be 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 the angle between intersecting lines. And his “green” neighbor or oppositely oriented"raspberry" 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. Firstly, the direction in which the angle is “scrolled” is fundamentally important. Secondly, a negatively oriented angle is written with a minus sign, for example if .

Why did I tell you this? It seems that we can get by with the usual concept of an angle. The fact is that the formulas by which we will find angles can easily result in a negative result, 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, be sure to indicate its orientation with an arrow (clockwise).

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

Example 10

Find the angle between lines

Solution And Method one

Let's consider two straight lines defined by equations in general form:

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

Let us pay close attention to the denominator - this is exactly scalar product directing vectors of straight lines:

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

Based on the above, it is convenient to formalize the solution in two steps:

1) Let's calculate the scalar product of the direction vectors of the lines:

2) Find the angle between straight lines using the formula:

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

Answer:

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

Well, minus, minus, no big deal. Here is a geometric illustration:

It is not surprising that the angle turned out to have a negative orientation, because in the problem statement the first number is a straight line and the “unscrewing” of the angle began precisely with it.

If you really want to get a positive angle, you need to swap the 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 .

I won’t hide it, I select the straight lines myself in the order so that the angle turns out to be positive. It's more beautiful, but nothing more.

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

Method two

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

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

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

Thus, the slopes are:

1) Let's check whether the lines are perpendicular:
, which means the lines are not perpendicular.

2) Use the formula:

Answer:

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

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

Here we are no longer talking about an oriented angle, but “just about an angle,” that is, the result will certainly be positive. The catch is that you may end up with an obtuse angle (not the one you need). In this case, you will have to make a reservation that the angle between straight 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 approach with an oriented angle, for the reason that it is widespread.

Example 11

Find the angle between the lines.

This is an example for you to solve on your own. 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 I’m not particularly steamed. To be honest, I thought the article would be much longer. But I’ll still take my recently acquired hat and 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:Solution : Let's find the direction vector of the line :

Let's compose the equation of the desired line using the point and direction vector . Since one of the coordinates of the direction vector is zero, Eq. let's rewrite it in the form:

Answer :

Example 5:Solution :
1) Equation of a line let's make up two points :

2) Equation of a line let's make up two points :

3) Corresponding coefficients for variables not proportional: , which means the lines intersect.
4) Find the 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 :

A straight line in space can always be defined as the 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 line is given as

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

Canonical equations of the line

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

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

. (9)

Parametric equations of a line

Let the canonical equations of the line be given

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

(10)

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

Equation of a line passing through two points
And
has the form:

Angle between straight lines

Angle between straight lines

And

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

Condition for parallel lines:

Condition for planes to be perpendicular:

Distance of a point from a line

P let's say the point is given
and straight

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

Condition for the intersection of lines

Two non-parallel lines

intersect if and only if

The relative position of a straight line and a plane.

Let the straight line be given
and plane. Corner between them can be found using the formula

Problem 73. Write the canonical equations of the line

(11)

Solution. 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 direction vector of the line.

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

,
, That

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

Since the point
any point on a 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 system (11):

From here,
.

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

or
.

Problem 74.

And
.

Solution. 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 coordinates of the direction vector
.

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

Let's find the coordinates of the vector
.

Let's calculate the vector product
:

Problem 75. Find a point symmetrical point
relatively straight

Solution. Let us write down the equation of the plane perpendicular to the given line and passing through the point . As its normal vector you can take the directing vector of a straight line. Then
. Hence,

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

Hence,
.

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

,
,
.

So,
.

Problem 76. Write the equation of a plane passing through a line
And

a) through a point
;

b) perpendicular to the plane.

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

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

a) Let's find
And from the condition that the plane passes through the point
, therefore, its coordinates must satisfy the equation of the plane. Let's substitute the coordinates of the point
into the equation of a bunch of planes:

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

b) Let's find
And from the condition that the desired plane is perpendicular to the plane. The normal vector of a given plane
, normal vector of the desired plane (see equation of a bunch of planes (12).