Definition. If two lines are given y = k 1 x + b 1 , y = k 2 x + b 2 , then sharp corner between these lines will be defined as

Two lines are parallel if k 1 = k 2 . Two lines are perpendicular if k 1 = -1/ k 2 .

Theorem. The straight lines Ax + Vy + C \u003d 0 and A 1 x + B 1 y + C 1 \u003d 0 are parallel when the coefficients A 1 \u003d λA, B 1 \u003d λB are proportional. If also С 1 = λС, then the lines coincide. The coordinates of the point of intersection of two lines are found as a solution to the system of equations of these lines.

Equation of a straight line passing through given point

Perpendicular to this line

Definition. The line passing through the point M 1 (x 1, y 1) and perpendicular to the line y \u003d kx + b is represented by the equation:

Distance from point to line

Theorem. If a point M(x 0, y 0) is given, then the distance to the line Ax + Vy + C \u003d 0 is defined as

.

Proof. Let the point M 1 (x 1, y 1) be the base of the perpendicular dropped from the point M to the given line. Then the distance between points M and M 1:

(1)

The x 1 and y 1 coordinates can be found as a solution to the system of equations:

The second equation of the system is the equation of a straight line passing through a given point M 0 perpendicular to a given straight line. If we transform the first equation of the system to the form:

A(x - x 0) + B(y - y 0) + Ax 0 + By 0 + C = 0,

then, solving, we get:

Substituting these expressions into equation (1), we find:

The theorem has been proven.

Example. Determine the angle between the lines: y = -3 x + 7; y = 2 x + 1.

k 1 \u003d -3; k2 = 2; tgφ = ; φ= p /4.

Example. Show that the lines 3x - 5y + 7 = 0 and 10x + 6y - 3 = 0 are perpendicular.

Solution. We find: k 1 \u003d 3/5, k 2 \u003d -5/3, k 1 * k 2 \u003d -1, therefore, the lines are perpendicular.

Example. The vertices of the triangle A(0; 1), B (6; 5), C (12; -1) are given. Find the equation for the height drawn from vertex C.

Solution. We find the equation of the side AB: ; 4 x = 6 y - 6;

2x – 3y + 3 = 0;

The desired height equation is: Ax + By + C = 0 or y = kx + b. k = . Then y = . Because the height passes through point C, then its coordinates satisfy this equation: whence b = 17. Total: .

Answer: 3x + 2y - 34 = 0.

Equation of a line passing through a given point in a given direction. Equation of a straight line passing through two given points. Angle between two lines. Condition of parallelism and perpendicularity of two lines. Determining the point of intersection of two lines

1. Equation of a line passing through a given point A(x 1 , y 1) in a given direction, determined by the slope k,

y - y 1 = k(x - x 1). (1)

This equation defines a pencil of lines passing through a point A(x 1 , y 1), which is called the center of the beam.

2. Equation of a straight line passing through two points: A(x 1 , y 1) and B(x 2 , y 2) is written like this:

The slope of a straight line passing through two given points is determined by the formula

3. Angle between straight lines A and B is the angle by which the first straight line must be rotated A around the point of intersection of these lines counterclockwise until it coincides with the second line B. If two lines are given by slope equations

y = k 1 x + B 1 ,

y = k 2 x + B 2 , (4)

then the angle between them is determined by the formula

It should be noted that in the numerator of the fraction, the slope of the first straight line is subtracted from the slope of the second straight line.

If the equations of a straight line are given in general view

A 1 x + B 1 y + C 1 = 0,

A 2 x + B 2 y + C 2 = 0, (6)

the angle between them is determined by the formula

4. Conditions for parallelism of two lines:

a) If the lines are given by equations (4) with a slope, then the necessary and sufficient condition for their parallelism is the equality of their slopes:

k 1 = k 2 . (8)

b) For the case when the lines are given by equations in general form (6), the necessary and sufficient condition for their parallelism is that the coefficients at the corresponding current coordinates in their equations are proportional, i.e.

5. Conditions for perpendicularity of two lines:

a) In the case when the lines are given by equations (4) with a slope, the necessary and sufficient condition for their perpendicularity is that their slopes are reciprocal in magnitude and opposite in sign, i.e.

This condition can also be written in the form

k 1 k 2 = -1. (11)

b) If the equations of straight lines are given in general form (6), then the condition for their perpendicularity (necessary and sufficient) is to fulfill the equality

A 1 A 2 + B 1 B 2 = 0. (12)

6. The coordinates of the point of intersection of two lines are found by solving the system of equations (6). Lines (6) intersect if and only if

1. Write the equations of the lines passing through the point M, one of which is parallel and the other is perpendicular to the given line l.

Task 1

Find the cosine of the angle between the lines $\frac(x+3)(5) =\frac(y-2)(-3) =\frac(z-1)(4) $ and $\left\(\begin(array )(c) (x=2\cdot t-3) \\ (y=-t+1) \\ (z=3\cdot t+5) \end(array)\right.$.

Let two lines be given in space: $\frac(x-x_(1) )(m_(1) ) =\frac(y-y_(1) )(n_(1) ) =\frac(z-z_(1 ) )(p_(1) ) $ and $\frac(x-x_(2) )(m_(2) ) =\frac(y-y_(2) )(n_(2) ) =\frac(z- z_(2) )(p_(2) ) $. We choose an arbitrary point in space and draw two auxiliary lines through it, parallel to the data. The angle between the given lines is any of the two adjacent corners formed by auxiliary lines. The cosine of one of the angles between the lines can be found from well-known formula$\cos \phi =\frac(m_(1) \cdot m_(2) +n_(1) \cdot n_(2) +p_(1) \cdot p_(2) )(\sqrt(m_(1) ^(2) +n_(1)^(2) +p_(1)^(2) ) \cdot \sqrt(m_(2)^(2) +n_(2)^(2) +p_(2) ^(2) ) ) $. If the value $\cos \phi >0$, then an acute angle between the lines is obtained, if $\cos \phi

Canonical equations of the first line: $\frac(x+3)(5) =\frac(y-2)(-3) =\frac(z-1)(4) $.

The canonical equations of the second straight line can be obtained from the parametric ones:

\ \ \

Thus, the canonical equations of this line are: $\frac(x+3)(2) =\frac(y-1)(-1) =\frac(z-5)(3) $.

We calculate:

\[\cos \phi =\frac(5\cdot 2+\left(-3\right)\cdot \left(-1\right)+4\cdot 3)(\sqrt(5^(2) +\ left(-3\right)^(2) +4^(2) ) \cdot \sqrt(2^(2) +\left(-1\right)^(2) +3^(2) ) ) = \frac(25)(\sqrt(50) \cdot \sqrt(14) ) \approx 0.9449.\]

Task 2

The first line goes through given points$A\left(2,-4,-1\right)$ and $B\left(-3,5,6\right)$, the second line is through the given points $C\left(1,-2, 8\right)$ and $D\left(6,7,-2\right)$. Find the distance between these lines.

Let some line be perpendicular to lines $AB$ and $CD$ and intersect them at points $M$ and $N$, respectively. Under these conditions, the length of the segment $MN$ is equal to the distance between the lines $AB$ and $CD$.

We build the vector $\overline(AB)$:

\[\overline(AB)=\left(-3-2\right)\cdot \bar(i)+\left(5-\left(-4\right)\right)\cdot \bar(j)+ \left(6-\left(-1\right)\right)\cdot \bar(k)=-5\cdot \bar(i)+9\cdot \bar(j)+7\cdot \bar(k ).\]

Let the segment representing the distance between the lines pass through the point $M\left(x_(M) ,y_(M) ,z_(M) \right)$ on the line $AB$.

We build the $\overline(AM)$ vector:

\[\overline(AM)=\left(x_(M) -2\right)\cdot \bar(i)+\left(y_(M) -\left(-4\right)\right)\cdot \ bar(j)+\left(z_(M) -\left(-1\right)\right)\cdot \bar(k)=\] \[=\left(x_(M) -2\right)\ cdot \bar(i)+\left(y_(M) +4\right)\cdot \bar(j)+\left(z_(M) +1\right)\cdot \bar(k).\]

The vectors $\overline(AB)$ and $\overline(AM)$ are the same, hence they are collinear.

It is known that if the vectors $\overline(a)=x_(1) \cdot \overline(i)+y_(1) \cdot \overline(j)+z_(1) \cdot \overline(k)$ and $ \overline(b)=x_(2) \cdot \overline(i)+y_(2) \cdot \overline(j)+z_(2) \cdot \overline(k)$ are collinear, then their coordinates are proportional, then is $\frac(x_((\it 2)) )((\it x)_((\it 1)) ) =\frac(y_((\it 2)) )((\it y)_( (\it 1)) ) =\frac(z_((\it 2)) )((\it z)_((\it 1)) ) $.

$\frac(x_(M) -2)(-5) =\frac(y_(M) +4)(9) =\frac(z_(M) +1)(7) =m$, where $m $ is the result of the division.

From here we get: $x_(M) -2=-5\cdot m$; $y_(M) +4=9\cdot m$; $z_(M) +1=7\cdot m$.

Finally, we obtain expressions for the coordinates of the point $M$:

We build the $\overline(CD)$ vector:

\[\overline(CD)=\left(6-1\right)\cdot \bar(i)+\left(7-\left(-2\right)\right)\cdot \bar(j)+\ left(-2-8\right)\cdot \bar(k)=5\cdot \bar(i)+9\cdot \bar(j)-10\cdot \bar(k).\]

Let the segment representing the distance between the lines pass through the point $N\left(x_(N) ,y_(N) ,z_(N) \right)$ on the line $CD$.

We construct the vector $\overline(CN)$:

\[\overline(CN)=\left(x_(N) -1\right)\cdot \bar(i)+\left(y_(N) -\left(-2\right)\right)\cdot \ bar(j)+\left(z_(N) -8\right)\cdot \bar(k)=\] \[=\left(x_(N) -1\right)\cdot \bar(i)+ \left(y_(N) +2\right)\cdot \bar(j)+\left(z_(N) -8\right)\cdot \bar(k).\]

The vectors $\overline(CD)$ and $\overline(CN)$ are the same, hence they are collinear. We apply the condition of collinear vectors:

$\frac(x_(N) -1)(5) =\frac(y_(N) +2)(9) =\frac(z_(N) -8)(-10) =n$ where $n $ is the result of the division.

From here we get: $x_(N) -1=5\cdot n$; $y_(N) +2=9\cdot n$; $z_(N) -8=-10\cdot n$.

Finally, we obtain expressions for the coordinates of point $N$:

We build the $\overline(MN)$ vector:

\[\overline(MN)=\left(x_(N) -x_(M) \right)\cdot \bar(i)+\left(y_(N) -y_(M) \right)\cdot \bar (j)+\left(z_(N) -z_(M) \right)\cdot \bar(k).\]

We substitute the expressions for the coordinates of the points $M$ and $N$:

\[\overline(MN)=\left(1+5\cdot n-\left(2-5\cdot m\right)\right)\cdot \bar(i)+\] \[+\left(- 2+9\cdot n-\left(-4+9\cdot m\right)\right)\cdot \bar(j)+\left(8-10\cdot n-\left(-1+7\cdot m\right)\right)\cdot \bar(k).\]

After completing the steps, we get:

\[\overline(MN)=\left(-1+5\cdot n+5\cdot m\right)\cdot \bar(i)+\left(2+9\cdot n-9\cdot m\right )\cdot \bar(j)+\left(9-10\cdot n-7\cdot m\right)\cdot \bar(k).\]

Since the lines $AB$ and $MN$ are perpendicular, the scalar product of the corresponding vectors is equal to zero, i.e. $\overline(AB)\cdot \overline(MN)=0$:

\[-5\cdot \left(-1+5\cdot n+5\cdot m\right)+9\cdot \left(2+9\cdot n-9\cdot m\right)+7\cdot \ left(9-10\cdot n-7\cdot m\right)=0;\] \

After completing the steps, we get the first equation for determining $m$ and $n$: $155\cdot m+14\cdot n=86$.

Since the lines $CD$ and $MN$ are perpendicular, the scalar product of the corresponding vectors is equal to zero, i.e. $\overline(CD)\cdot \overline(MN)=0$:

\ \[-5+25\cdot n+25\cdot m+18+81\cdot n-81\cdot m-90+100\cdot n+70\cdot m=0.\]

After completing the steps, we obtain the second equation for determining $m$ and $n$: $14\cdot m+206\cdot n=77$.

Find $m$ and $n$ by solving the system of equations $\left\(\begin(array)(c) (155\cdot m+14\cdot n=86) \\ (14\cdot m+206\cdot n =77) \end(array)\right.$.

We apply the Cramer method:

\[\Delta =\left|\begin(array)(cc) (155) & (14) \\ (14) & (206) \end(array)\right|=31734; \] \[\Delta _(m) =\left|\begin(array)(cc) (86) & (14) \\ (77) & (206) \end(array)\right|=16638; \] \[\Delta _(n) =\left|\begin(array)(cc) (155) & (86) \\ (14) & (77) \end(array)\right|=10731;\ ]\

Find the coordinates of points $M$ and $N$:

\ \

Finally:

Finally, we write the vector $\overline(MN)$:

$\overline(MN)=\left(2.691-\left(-0.6215\right)\right)\cdot \bar(i)+\left(1.0438-0.7187\right)\cdot \bar (j)+\left(4,618-2,6701\right)\cdot \bar(k)$ or $\overline(MN)=3,3125\cdot \bar(i)+0,3251\cdot \bar( j)+1.9479\cdot\bar(k)$.

The distance between lines $AB$ and $CD$ is the length of the vector $\overline(MN)$:$d=\sqrt(3.3125^(2) +0.3251^(2) +1.9479^( 2) ) \approx 3.8565$ lin. units

Oh-oh-oh-oh-oh ... well, it's tinny, as if you read the sentence to yourself =) However, then relaxation will help, especially since I bought suitable accessories today. 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 reduce all the coefficients of the equation 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

In practical problems, the solution scheme just considered 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:

Solution 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:

In this way,

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 something for an independent solution, it is better to lay one more important brick in the geometric foundation:

How to draw a line parallel to a given one?

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

Example 2

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

Solution: 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 orally. 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 sense systems of 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

Solution: 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 focus on this repeatedly.

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.

Solution: 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

Solution: 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 your ingenuity was well dispersed.

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

Solution and Method one

Consider two lines given by equations in general:

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:

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 the scalar product of directing vectors of straight lines:
so the lines are not perpendicular.

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

By using 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, we indicate the exact value, as well as the approximate value (preferably both in degrees and in 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 .

I will be brief. Angle between two lines equal to the angle between their direction vectors. Thus, if you manage to find the coordinates of the direction vectors a \u003d (x 1; y 1; z 1) and b \u003d (x 2; y 2; z 2), you can find the angle. More precisely, the cosine of the angle according to the formula:

Let's see how this formula works on specific examples:

A task. Points E and F are marked in the cube ABCDA 1 B 1 C 1 D 1 - the midpoints of the edges A 1 B 1 and B 1 C 1, respectively. Find the angle between lines AE and BF.

Since the edge of the cube is not specified, we set AB = 1. We introduce a standard coordinate system: the origin is at point A, and the x, y, z axes are directed along AB, AD, and AA 1, respectively. The unit segment is equal to AB = 1. Now let's find the coordinates of the direction vectors for our lines.

Find the coordinates of the vector AE. To do this, we need points A = (0; 0; 0) and E = (0.5; 0; 1). Since the point E is the middle of the segment A 1 B 1 , its coordinates are equal to the arithmetic mean of the coordinates of the ends. Note that the origin of the vector AE coincides with the origin, so AE = (0.5; 0; 1).

Now let's deal with the BF vector. Similarly, we analyze the points B = (1; 0; 0) and F = (1; 0.5; 1), because F - the middle of the segment B 1 C 1 . We have:
BF = (1 - 1; 0.5 - 0; 1 - 0) = (0; 0.5; 1).

So, the direction vectors are ready. The cosine of the angle between the lines is the cosine of the angle between the direction vectors, so we have:

A task. In a regular trihedral prism ABCA 1 B 1 C 1 , all edges of which are equal to 1, points D and E are marked - the midpoints of the edges A 1 B 1 and B 1 C 1, respectively. Find the angle between lines AD and BE.

We introduce a standard coordinate system: the origin is at point A, the x-axis is directed along AB, z - along AA 1 . We direct the y axis so that the OXY plane coincides with the ABC plane. The unit segment is equal to AB = 1. Find the coordinates of the direction vectors for the desired lines.

First, let's find the coordinates of the AD vector. Consider the points: A = (0; 0; 0) and D = (0.5; 0; 1), because D - the middle of the segment A 1 B 1 . Since the beginning of the vector AD coincides with the origin, we get AD = (0.5; 0; 1).

Now let's find the coordinates of the vector BE. Point B = (1; 0; 0) is easy to calculate. With point E - the middle of the segment C 1 B 1 - a little more difficult. We have:

It remains to find the cosine of the angle:

A task. In a regular hexagonal prism ABCDEFA 1 B 1 C 1 D 1 E 1 F 1 , all edges of which are equal to 1, the points K and L are marked - the midpoints of the edges A 1 B 1 and B 1 C 1, respectively. Find the angle between lines AK and BL.

We introduce a standard coordinate system for a prism: we place the origin of coordinates at the center of the lower base, direct the x-axis along FC, the y-axis through the midpoints of segments AB and DE, and the z-axis vertically upwards. The unit segment is again equal to AB = 1. Let us write out the coordinates of the points of interest to us:

Points K and L are the midpoints of the segments A 1 B 1 and B 1 C 1, respectively, so their coordinates are found through the arithmetic mean. Knowing the points, we find the coordinates of the direction vectors AK and BL:

Now let's find the cosine of the angle:

A task. In a regular quadrangular pyramid SABCD, all edges of which are equal to 1, points E and F are marked - the midpoints of the sides SB and SC, respectively. Find the angle between lines AE and BF.

We introduce a standard coordinate system: the origin is at point A, the x and y axes are directed along AB and AD, respectively, and the z axis is directed vertically upwards. The unit segment is equal to AB = 1.

Points E and F are the midpoints of the segments SB and SC, respectively, so their coordinates are found as the arithmetic mean of the ends. We write down the coordinates of the points of interest to us:
A = (0; 0; 0); B = (1; 0; 0)

Knowing the points, we find the coordinates of the direction vectors AE and BF:

The coordinates of the vector AE coincide with the coordinates of point E, since point A is the origin. It remains to find the cosine of the angle:

The article talks about finding the angle between planes. After bringing the definition, we will set a graphic illustration, consider a detailed method for finding coordinates by the method. We obtain a formula for intersecting planes, which includes the coordinates normal vectors.

The material will use data and concepts that were previously studied in articles about the plane and the line in space. To begin with, it is necessary to move on to reasoning that allows one to have a certain approach to determining the angle between two intersecting planes.

Two intersecting planes γ 1 and γ 2 are given. Their intersection will take the designation c . The construction of the χ plane is connected with the intersection of these planes. The plane χ passes through the point M as a straight line c. The planes γ 1 and γ 2 will be intersected using the χ plane. We accept the designations of the line intersecting γ 1 and χ for the line a, and intersecting γ 2 and χ for the line b. We get that the intersection of lines a and b gives the point M .

The location of the point M does not affect the angle between the intersecting lines a and b, and the point M is located on the line c through which the plane χ passes.

It is necessary to construct a plane χ 1 perpendicular to the line c and different from the plane χ . The intersection of the planes γ 1 and γ 2 with the help of χ 1 will take the designation of lines a 1 and b 1 .

It can be seen that when constructing χ and χ 1, the lines a and b are perpendicular to the line c, then a 1, b 1 are perpendicular to the line c. Finding lines a and a 1 in the plane γ 1 with perpendicularity to the line c, then they can be considered parallel. In the same way, the location of b and b 1 in the plane γ 2 with the perpendicularity of the line c indicates their parallelism. This means that it is necessary to make a parallel transfer of the plane χ 1 to χ, where we get two coinciding lines a and a 1 , b and b 1 . We get that the angle between the intersecting lines a and b 1 is equal to the angle of the intersecting lines a and b.

Consider the figure below.

This judgment is proved by the fact that between the intersecting lines a and b there is an angle that does not depend on the location of the point M, that is, the point of intersection. These lines are located in the planes γ 1 and γ 2 . In fact, the resulting angle can be thought of as the angle between two intersecting planes.

Let's move on to determining the angle between the existing intersecting planes γ 1 and γ 2 .

Definition 1

The angle between two intersecting planes γ 1 and γ 2 call the angle formed by the intersection of lines a and b, where the planes γ 1 and γ 2 intersect with the plane χ perpendicular to the line c.

Consider the figure below.

The definition may be submitted in another form. At the intersection of the planes γ 1 and γ 2, where c is the line on which they intersect, mark the point M, through which draw lines a and b, perpendicular to the line c and lying in the planes γ 1 and γ 2, then the angle between the lines a and b will be the angle between the planes. In practice, this is applicable to constructing an angle between planes.

At the intersection, an angle is formed that is less than 90 degrees in value, that is, the degree measure of the angle is valid on an interval of this type (0, 90] . At the same time, these planes are called perpendicular if a right angle is formed at the intersection. The angle between parallel planes is considered equal to zero.

The usual way to find the angle between intersecting planes is to perform additional constructions. This helps to determine it with accuracy, and this can be done using the signs of equality or similarity of the triangle, sines, cosines of the angle.

Consider solving problems using an example from the problems of the Unified State Examination of block C 2.

Example 1

A rectangular parallelepiped A B C D A 1 B 1 C 1 D 1 is given, where side A B \u003d 2, A D \u003d 3, A A 1 \u003d 7, point E separates side A A 1 in a ratio of 4: 3. Find the angle between planes A B C and B E D 1 .

Solution

For clarity, you need to make a drawing. We get that

A visual representation is necessary in order to make it more convenient to work with the angle between the planes.

We make the definition of a straight line along which the planes A B C and B E D 1 intersect. Point B is a common point. One more common point of intersection should be found. Consider the lines D A and D 1 E , which are located in the same plane A D D 1 . Their location does not indicate parallelism, which means they have a common intersection point.

However, the line D A is located in the plane A B C, and D 1 E in B E D 1 . Hence we get that the lines D A and D 1 E have a common point of intersection, which is also common for planes A B C and B E D 1 . Indicates the point of intersection of lines D A and D 1 E letter F. From here we get that B F is a straight line along which the planes A B C and B E D 1 intersect.

Consider the figure below.

To obtain an answer, it is necessary to construct straight lines located in the planes A B C and B E D 1 with the passage through a point located on the line B F and perpendicular to it. Then the resulting angle between these lines is considered the desired angle between the planes A B C and B E D 1.

From this it can be seen that the point A is the projection of the point E onto the plane A B C. It is necessary to draw a line intersecting the line B F at a right angle at the point M. It can be seen that the line A M is the projection of the line E M onto the plane A B C, based on the theorem about those perpendiculars A M ⊥ B F . Consider the figure below.

∠ A M E is the desired angle formed by the planes A B C and B E D 1 . From the resulting triangle A E M we can find the sine, cosine or tangent of the angle, after which the angle itself, only with its two known sides. By condition, we have that the length of A E is found in this way: the line A A 1 is divided by the point E in a ratio of 4: 3, which means the total length of the line is 7 parts, then A E \u003d 4 parts. We find A.M.

Need to consider right triangle A B F . We have a right angle A with height A M. From the condition A B \u003d 2, then we can find the length A F by the similarity of triangles D D 1 F and A E F. We get that A E D D 1 = A F D F ⇔ A E D D 1 = A F D A + A F ⇒ 4 7 = A F 3 + A F ⇔ A F = 4

It is necessary to find the length of the side B F from the triangle A B F using the Pythagorean theorem. We get that B F   = A B 2 + A F 2 = 2 2 + 4 2 = 2 5 . The length of the side A M is found through the area of ​​the triangle A B F. We have that the area can be equal to both S A B C = 1 2 · A B · A F , and S A B C = 1 2 · B F · A M .

We get that A M = A B A F B F = 2 4 2 5 = 4 5 5

Then we can find the value of the tangent of the angle of the triangle A E M. We get:

t g ∠ A M E = A E A M = 4 4 5 5 = 5

The desired angle obtained by the intersection of the planes A B C and B E D 1 is equal to a r c t g 5, then, when simplified, we get a r c t g 5 = a r c sin 30 6 = a r c cos 6 6 .

Answer: a r c t g 5 = a r c sin 30 6 = a r c cos 6 6 .

Some cases of finding the angle between intersecting lines are given using coordinate plane About x y z and the coordinate method. Let's consider in more detail.

If a problem is given where it is necessary to find the angle between the intersecting planes γ 1 and γ 2, we denote the desired angle by α.

Then the given coordinate system shows that we have the coordinates of the normal vectors of the intersecting planes γ 1 and γ 2 . Then we denote that n 1 → = n 1 x , n 1 y , n 1 z is a normal vector of the plane γ 1 , and n 2 → = (n 2 x , n 2 y , n 2 z) - for the plane γ 2 . Consider a detailed finding of the angle located between these planes according to the coordinates of the vectors.

It is necessary to designate the straight line along which the planes γ 1 and γ 2 intersect with the letter c. On the line with we have a point M, through which we draw a plane χ, perpendicular to c. The plane χ along the lines a and b intersects the planes γ 1 and γ 2 at the point M . it follows from the definition that the angle between the intersecting planes γ 1 and γ 2 is equal to the angle of the intersecting lines a and b belonging to these planes, respectively.

In the χ plane, we set aside the normal vectors from the point M and denote them n 1 → and n 2 →. Vector n 1 → is located on a line perpendicular to line a, and vector n 2 → on a line perpendicular to line b. From here we get that the given plane χ has a normal vector of the straight line a equal to n 1 → and for the straight line b equal to n 2 → . Consider the figure below.

From here we obtain a formula by which we can calculate the sine of the angle of intersecting lines using the coordinates of the vectors. We found that the cosine of the angle between the straight lines a and b is the same as the cosine between the intersecting planes γ 1 and γ 2 is derived from the formula cos α = cos n 1 → , n 2 → ^ = n 1 x n 2 x + n 1 y n 2 y + n 1 z n 2 z n 1 x 2 + n 1 y 2 + n 1 z 2 n 2 x 2 + n 2 y 2 + n 2 z 2 , where we have that n 1 → = (n 1 x , n 1 y , n 1 z) and n 2 → = (n 2 x , n 2 y , n 2 z) are the coordinates of the vectors of the represented planes.

The angle between intersecting lines is calculated using the formula

α = a r c cos n 1 x n 2 x + n 1 y n 2 y + n 1 z n 2 z n 1 x 2 + n 1 y 2 + n 1 z 2 n 2 x 2 + n 2 y 2 + n 2 z 2

Example 2

By condition, a parallelepiped А В С D A 1 B 1 C 1 D 1 is given , where A B \u003d 2, A D \u003d 3, A A 1 \u003d 7, and point E separates the side A A 1 4: 3. Find the angle between planes A B C and B E D 1 .

Solution

It can be seen from the condition that its sides are pairwise perpendicular. This means that it is necessary to introduce a coordinate system O x y z with a vertex at point C and coordinate axes O x, O y, O z. It is necessary to put the direction on the appropriate sides. Consider the figure below.

Intersecting planes A B C and B E D 1 form an angle, which can be found by the formula 2 x 2 + n 2 y 2 + n 2 z 2 , where n 1 → = (n 1 x , n 1 y , n 1 z) and n 2 → = (n 2 x , n 2 y , n 2 z ) are normal vectors of these planes. It is necessary to determine the coordinates. From the figure, we see that the coordinate axis O x y coincides in the plane A B C, which means that the coordinates of the normal vector k → equal to the value n 1 → = k → = (0, 0, 1) .

The normal vector of the plane B E D 1 is the vector product of B E → and B D 1 → , where their coordinates are found by the coordinates extreme points B, E, D 1 , which are determined based on the condition of the problem.

We get that B (0 , 3 , 0) , D 1 (2 , 0 , 7) . Because A E E A 1 = 4 3 , from the coordinates of the points A 2 , 3 , 0 , A 1 2 , 3 , 7 we find E 2 , 3 , 4 . We get that B E → = (2 , 0 , 4) , B D 1 → = 2 , - 3 , 7 n 2 → = B E → × B D 1 = i → j → k → 2 0 4 2 - 3 7 = 12 i → - 6 j → - 6 k → ⇔ n 2 → = (12, - 6, - 6)

It is necessary to substitute the found coordinates into the formula for calculating the angle through the arc cosine. We get

α = a r c cos 0 12 + 0 (- 6) + 1 (- 6) 0 2 + 0 2 + 1 2 12 2 + (- 6) 2 + (- 6) 2 = a r c cos 6 6 6 = a r c cos 6 6

The coordinate method gives a similar result.

Answer: a r c cos 6 6 .

The final problem is considered in order to find the angle between the intersecting planes with the available known equations of the planes.

Example 3

Calculate the sine, cosine of the angle, and the value of the angle formed by two intersecting lines, which are defined in the O x y z coordinate system and given by the equations 2 x - 4 y + z + 1 = 0 and 3 y - z - 1 = 0 .

Solution

When studying a topic general equation line of the form A x + B y + C z + D = 0 revealed that A, B, C are coefficients equal to the coordinates of the normal vector. Hence, n 1 → = 2 , - 4 , 1 and n 2 → = 0 , 3 , - 1 are normal vectors of given lines.

It is necessary to substitute the coordinates of the normal vectors of the planes into the formula for calculating the desired angle of intersecting planes. Then we get that

α = a r c cos 2 0 + - 4 3 + 1 (- 1) 2 2 + - 4 2 + 1 2 = a r c cos 13 210

Hence we have that the cosine of the angle takes the form cos α = 13 210 . Then the angle of the intersecting lines is not obtuse. Substituting in trigonometric identity, we get that the value of the sine of the angle is equal to the expression. We calculate and get that

sin α = 1 - cos 2 α = 1 - 13 210 = 41 210

Answer: sin α = 41 210 , cos α = 13 210 , α = a r c cos 13 210 = a r c sin 41 210 .

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