Finding the angle between planes (dihedral angle). The simplest problems with a straight line on a plane. Mutual arrangement of straight lines. Angle between straight lines Define an acute angle between straight lines

and. Let there be given two straight lines. These straight lines, as indicated in Chapter 1, form various positive and negative angles, which can be both acute and obtuse. Knowing one of these angles, we can easily find any other.

By the way, for all these angles the numerical value of the tangent is the same, the difference can only be in the sign

Equations of lines. The numbers are the projections of the direction vectors of the first and second straight lines. The angle between these vectors is equal to one of the angles formed by straight lines. Therefore, the task is reduced to determining the angle between the vectors, We get

For simplicity, we can agree on the angle between two straight lines to mean an acute positive angle (as, for example, in Fig. 53).

Then the tangent of this angle will always be positive. Thus, if a minus sign is obtained on the right side of formula (1), then we must discard it, that is, keep only the absolute value.

Example. Determine the angle between straight lines

By formula (1), we have

from. If it is indicated which of the sides of the angle is its beginning and which is the end, then, always counting the direction of the angle counterclockwise, we can extract something more from formula (1). As is easy to see from Fig. The 53rd sign obtained in the right-hand side of formula (1) will indicate which - acute or obtuse - angle forms the second straight line with the first.

(Indeed, from Fig, 53, we see that the angle between the first and second direction vectors is either equal to the desired angle between straight lines, or differs from it by ± 180 °.)

d. If the straight lines are parallel, then their direction vectors are also parallel. Applying the condition of parallelism of two vectors, we get!

This is a necessary and sufficient condition for the parallelism of two straight lines.

Example. Direct

are parallel because

e. If the lines are perpendicular, then their direction vectors are also perpendicular. Applying the condition of perpendicularity of two vectors, we obtain the condition of perpendicularity of two lines, namely

Example. Direct

are perpendicular due to the fact that

In connection with the conditions of parallelism and perpendicularity, we will solve the following two problems.

f. Draw a straight line through a point parallel to this straight line

The solution is carried out as follows. Since the desired straight line is parallel to the given one, then for its direction vector we can take the same one as for the given straight line, that is, a vector with projections A and B. And then the equation of the desired straight line will be written in the form (§ 1)

Example. Equation of a straight line passing through a point (1; 3) parallel to a straight line

will be next!

g. Draw a straight line through a point perpendicular to this straight line

Here, it is no longer suitable to take a vector with projections A and as a direction vector, but a vector that is perpendicular to it must be blown. The projections of this vector should be chosen, therefore, according to the condition of perpendicularity of both vectors, i.e., according to the condition

This condition can be fulfilled in countless ways, since there is one equation with two unknowns.But the easiest way is to take go Then the equation of the desired straight line will be written in the form

Example. Equation of a straight line passing through the point (-7; 2) in a perpendicular line

will be the following (according to the second formula)!

h. In the case when the straight lines are given by equations of the form

rewriting these equations differently, we have

Angle φ general equations A 1 x + B 1 y + C 1 \u003d 0 and A 2 x + B 2 y + C 2 \u003d 0, calculated by the formula:

Angle φ between two straight lines given canonical equations (x-x 1) / m 1 \u003d (y-y 1) / n 1 and (x-x 2) / m 2 \u003d (y-y 2) / n 2, calculated by the formula:

Distance from point to line

Each plane in space can be represented as a linear equation called general equation plane

Special cases.

o If in equation (8), then the plane passes through the origin.

o At (,) the plane is parallel to the axis (axis, axis), respectively.

o At (,) the plane is parallel to the plane (plane, plane).

Solution: use (7)

Answer: the general equation of the plane.

Example.

The plane in the rectangular coordinate system Oxyz is given by the general equation of the plane ... Write down the coordinates of all the normal vectors of this plane.

We know that the coefficients of the variables x, y and z in the general equation of the plane are the corresponding coordinates of the normal vector of this plane. Therefore, the normal vector of a given plane has coordinates. The set of all normal vectors can be specified as.

Write the equation of the plane, if in the rectangular coordinate system Oxyz in space it passes through the point , and is the normal vector of this plane.

Here are two solutions to this problem.

From the condition we have. We substitute this data into the general equation of the plane passing through the point:

Write the general equation of a plane parallel to the coordinate plane Oyz and passing through a point .

A plane that is parallel to the coordinate plane Oyz can be defined by the general incomplete equation of the view plane. Since the point belongs to the plane by condition, then the coordinates of this point must satisfy the equation of the plane, that is, equality must be true. From here we find. Thus, the required equation has the form.

Decision. The cross product, by Definition 10.26, is orthogonal to the vectors p and q. Therefore, it is orthogonal to the desired plane and the vector can be taken as its normal vector. Find the coordinates of the vector n:

i.e ... Using formula (11.1), we obtain

Opening the brackets in this equation, we come to the final answer.

Answer: .

Let's rewrite the normal vector in the form and find its length:

According to the above:

Answer:

Parallel planes have the same normal vector. 1) From the equation we find the normal vector of the plane :.

2) The equation of the plane is composed by the point and the normal vector:

Answer:

Vector equation of a plane in space

Parametric equation of a plane in space

Equation of a plane passing through a given point perpendicular to a given vector

Let a rectangular Cartesian coordinate system be given in three-dimensional space. Let us formulate the following problem:

Equate a plane passing through a given point M(x0, y0, z0) perpendicular to the given vector n \u003d ( A, B, C} .

Decision. Let be P(x, y, z) is an arbitrary point in space. Point P belongs to the plane if and only if the vector MP = {x − x0, y − y0, z − z0) is orthogonal to the vector n = {A, B, C) (Fig. 1).

Having written the orthogonality condition for these vectors (n, MP) \u003d 0 in coordinate form, we get:

Three-point plane equation

In vector form

In coordinates

Mutual arrangement of planes in space

- general equations of two planes. Then:

1) if , then the planes coincide;

2) if , then the planes are parallel;

3) if or, then the planes intersect and the system of equations

(6)

is the equations of the line of intersection of these planes.

Create parametric equations for the following straight lines:

Decision: Lines are given by canonical equations and at the first stage one should find some point belonging to the line and its direction vector.

a) From the equations remove point and direction vector:. You can choose another point (how to do it - described above), but it is better to take the most obvious one. By the way, to avoid mistakes, always substitute its coordinates in the equations.

Let's compose the parametric equations of this straight line:

The convenience of parametric equations is that it is very easy to find other points of a straight line with their help. For example, let's find a point whose coordinates, say, correspond to the parameter value:

Thus: b) Consider the canonical equations ... The choice of a point here is simple, but tricky: (be careful, do not mix up the coordinates !!!). How do I pull out the direction vector? You can speculate about what this line is parallel to, or you can use a simple formal trick: the "game" and "z" are in proportion, so we write down the direction vector, and put zero in the remaining space:.

Let's compose the parametric equations of the straight line:

c) Let's rewrite the equations in the form, that is, "z" can be anything. And if any, then let, for example,. Thus, the point belongs to this line. To find the direction vector, we use the following formal technique: in the original equations there are "x" and "game", and in the direction vector at these places we write zeros:. We put in the remaining space unit:. Instead of one, any number other than zero will do.

Let us write the parametric equations of the straight line:

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

The material will use data and concepts that were previously studied in articles about a plane and a straight line in space. First, you need to move on to reasoning that allows you 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 becomes c. The construction of the χ plane is associated with the intersection of these planes. The χ plane 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 take the notation of the line intersecting γ 1 and χ as line a, and intersecting γ 2 and χ as line b. We get that the intersection of lines a and b gives a point M.

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

It is necessary to construct the plane χ 1 with perpendicularity 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, straight lines a and b are perpendicular to line c, then a 1, b 1 are located perpendicular to line c. Finding the straight lines a and a 1 in the plane γ 1 with perpendicularity to the straight 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 straight line c indicates their parallelism. Hence, it is necessary to make a parallel transfer of the plane χ 1 to χ, where we get two coinciding straight lines a and a 1, b and b 1. We get that the angle between intersecting straight lines a and b 1 is equal to the angle of intersecting straight lines a and b.

Consider not the figure below.

This statement is proved by the fact that there is an angle between the intersecting straight lines a and b, which 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 us proceed 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 called the angle formed by the intersection of straight lines a and b, where the planes γ 1 and γ 2 intersect with the plane χ, perpendicular to the straight line c.

Consider the figure below.

The definition can be filed in another form. When the planes γ 1 and γ 2 intersect, where c is the line on which they intersect, mark the point M through which to draw lines a and b perpendicular to line c and lying in planes γ 1 and γ 2, then the angle between lines a and b will be the angle between the planes. This is practically applicable for constructing the angle between the 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 over an interval of this type (0, 90]. At the same time, these planes are called perpendicular if the intersection forms a right angle. The angle between parallel planes is considered equal to zero.

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

Let's consider the solution of problems using an example from the problems of the exam 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 divides side A A 1 in the ratio 4: 3. Find the angle between planes A B C and B E D 1.

Decision

For clarity, you need to complete the drawing. We get that

Visual representation is necessary in order to make it easier to work with the angle between the planes.

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

However, the line D A is located in the plane A B C, and D 1 E in the B E D 1. From this we obtain that the lines D A and D 1 Ehave a common point of intersection, which is common for the planes A B C and B E D 1. Indicates the point of intersection of lines D A and D 1 E the letter F. From this we obtain that B F is a line along which the planes A B C and B E D 1 intersect.

Consider in the figure below.

To get an answer, it is necessary to construct lines located in the planes A B C and B E D 1 with passing 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 point A is the projection of point E onto plane A B C. It is necessary to draw a straight line intersecting straight line BF at point M. It can be seen that straight line AM is the projection of straight line EM onto plane A B C, based on the theorem about those perpendiculars AM ⊥ BF. Consider the figure below.

∠ A M E is the desired angle formed by 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 sides known. By condition, we have that the length A E is found in this way: straight line A A 1 is divided by point E in the ratio 4: 3, that means the total length of the straight line is 7 parts, then A E \u003d 4 parts. Find A. M.

It is necessary to consider a right-angled 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 \u003d A F D F ⇔ A E D D 1 \u003d A F D A + A F ⇒ 4 7 \u003d A F 3 + A F ⇔ A F \u003d 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 \u003d A B 2 + A F 2 \u003d 2 2 + 4 2 \u003d 2 5. The length of the side A M is found through the area of \u200b\u200bthe triangle A B F. We have that the area can be equal to both S A B C \u003d 1 2 A B A F, and S A B C \u003d 1 2 B F A M.

We get that A M \u003d A B A F B F \u003d 2 4 2 5 \u003d 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 \u003d A E A M \u003d 4 4 5 5 \u003d 5

The sought 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, for simplification, we obtain a r c t g 5 \u003d a r c sin 30 6 \u003d a r c cos 6 6.

Answer: a r c t g 5 \u003d a r c sin 30 6 \u003d a r c cos 6 6.

Some cases of finding the angle between intersecting straight lines are specified using the coordinate plane O x y z and the method of coordinates. Let's take a closer look.

If a problem is given where it is necessary to find the angle between the intersecting planes γ 1 and γ 2, the sought angle will be denoted 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 → \u003d n 1 x, n 1 y, n 1 z is the normal vector of the plane γ 1, and n 2 → \u003d (n 2 x, n 2 y, n 2 z) is for the plane γ 2. Let's consider in detail finding 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 straight line c we have a point M, through which we draw the 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 intersecting straight lines a and b belonging to these planes, respectively.

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

From here we get a formula by which we can calculate the sine of the angle of intersecting straight lines using the coordinates of the vectors. We got 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 \u003d cos n 1 →, n 2 → ^ \u003d n 1 x n 2 x + n 1 y n 2 y + n 1 z n 2 zn 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 → \u003d (n 1 x, n 1 y, n 1 z) and n 2 → \u003d (n 2 x, n 2 y, n 2 z) are the coordinates of the vectors of the represented planes.

The angle between intersecting straight lines is calculated using the formula

α \u003d arc cos n 1 x n 2 x + n 1 y n 2 y + n 1 z n 2 zn 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, given a parallelepiped А В С D A 1 B 1 C 1 D 1 , where A B \u003d 2, A D \u003d 3, A A 1 \u003d 7, and point E separates side A A 1 4: 3. Find the angle between planes A B C and B E D 1.

Decision

It is 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 apex at point C and coordinate axes O x, O y, O z. It is necessary to put a direction on the corresponding sides. Consider the figure below.

Intersecting planes A B C and B E D 1form an angle that can be found by the formula α \u003d arc cos n 1 x n 2 x + n 1 y n 2 y + n 1 z n 2 zn 1 x 2 + n 1 y 2 + n 1 z 2 n 2 x 2 + n 2 y 2 + n 2 z 2, in which n 1 → \u003d (n 1 x, n 1 y, n 1 z) and n 2 → \u003d (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 → are equal to the value n 1 → \u003d k → \u003d (0, 0, 1).

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

We get that B (0, 3, 0), D 1 (2, 0, 7). Because A E E A 1 \u003d 4 3, from the coordinates of points A 2, 3, 0, A 1 2, 3, 7 we find E 2, 3, 4. We get that BE → \u003d (2, 0, 4), BD 1 → \u003d 2, - 3, 7 n 2 → \u003d BE → × BD 1 \u003d i → j → k → 2 0 4 2 - 3 7 \u003d 12 i → - 6 j → - 6 k → ⇔ n 2 → \u003d (12, - 6, - 6)

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

α \u003d arc cos 0 12 + 0 (- 6) + 1 (- 6) 0 2 + 0 2 + 1 2 12 2 + (- 6) 2 + (- 6) 2 \u003d arc cos 6 6 6 \u003d arc cos 6 6

The coordinate method gives a similar result.

Answer: a r c cos 6 6.

The final task 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 straight lines, which are defined in the O x y z coordinate system and given by the equations 2 x - 4 y + z + 1 \u003d 0 and 3 y - z - 1 \u003d 0.

Decision

When studying the topic of the general equation of a straight line of the form A x + B y + C z + D \u003d 0, it was revealed that A, B, C are coefficients equal to the coordinates of the normal vector. Hence, n 1 → \u003d 2, - 4, 1 and n 2 → \u003d 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 \u003d a r c cos 2 0 + - 4 3 + 1 (- 1) 2 2 + - 4 2 + 1 2 \u003d a r c cos 13 210

Hence we have that the cosine of the angle takes the form cos α \u003d 13 210. Then the angle of intersecting straight lines is not obtuse. Substituting into trigonometric identity, we find that the value of the sine of the angle is equal to the expression. We calculate and get that

sin α \u003d 1 - cos 2 α \u003d 1 - 13 210 \u003d 41 210

Answer: sin α \u003d 41 210, cos α \u003d 13 210, α \u003d a r c cos 13 210 \u003d a r c sin 41 210.

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Problem 1

Find the cosine of the angle between the straight lines $ \\ frac (x + 3) (5) \u003d \\ frac (y-2) (- 3) \u003d \\ frac (z-1) (4) $ and $ \\ left \\ (\\ begin (array ) (c) (x \u003d 2 \\ cdot t-3) \\\\ (y \u003d -t + 1) \\\\ (z \u003d 3 \\ cdot t + 5) \\ end (array) \\ right. $.

Let there be two lines in space: $ \\ frac (x-x_ (1)) (m_ (1)) \u003d \\ frac (y-y_ (1)) (n_ (1)) \u003d \\ frac (z-z_ (1 )) (p_ (1)) $ and $ \\ frac (x-x_ (2)) (m_ (2)) \u003d \\ frac (y-y_ (2)) (n_ (2)) \u003d \\ frac (z- z_ (2)) (p_ (2)) $. Choose an arbitrary point in space and draw through it two auxiliary lines parallel to the data. The angle between the given lines is any of two adjacent corners formed by construction lines. The cosine of one of the angles between the straight lines can be found by the well-known formula $ \\ cos \\ phi \u003d \\ 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\u003e 0 $, then an acute angle between the straight lines is obtained, if $ \\ cos \\ phi

Canonical equations of the first line: $ \\ frac (x + 3) (5) \u003d \\ frac (y-2) (- 3) \u003d \\ 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) \u003d \\ frac (y-1) (- 1) \u003d \\ frac (z-5) (3) $.

We calculate:

\\ [\\ cos \\ phi \u003d \\ frac (5 \\ cdot 2+ \\ left (-3 \\ right) \\ cdot \\ left (-1 \\ right) +4 \\ cdot 3) (\\ sqrt (5 ^ (2) + \\ \\ frac (25) (\\ sqrt (50) \\ cdot \\ sqrt (14)) \\ approx 0.9449. \\]

Problem 2

The first line goes through the given points $ A \\ left (2, -4, -1 \\ right) $ and $ B \\ left (-3,5,6 \\ right) $, the second line goes 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) \u003d \\ 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) \u003d - 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 vector $ \\ overline (AM) $:

\\ [\\ overline (AM) \u003d \\ left (x_ (M) -2 \\ right) \\ cdot \\ bar (i) + \\ left (y_ (M) - \\ left (-4 \\ right) \\ right) \\ cdot \\ 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 vectors $ \\ overline (a) \u003d x_ (1) \\ cdot \\ overline (i) + y_ (1) \\ cdot \\ overline (j) + z_ (1) \\ cdot \\ overline (k) $ and $ \\ overline (b) \u003d 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))) \u003d \\ frac (y _ ((\\ it 2))) ((\\ it y) _ ( (\\ it 1))) \u003d \\ frac (z _ ((\\ it 2))) ((\\ it z) _ ((\\ it 1))) $.

$ \\ frac (x_ (M) -2) (- 5) \u003d \\ frac (y_ (M) +4) (9) \u003d \\ frac (z_ (M) +1) (7) \u003d m $, where $ m $ is the result of division.

From here we get: $ x_ (M) -2 \u003d -5 \\ cdot m $; $ y_ (M) + 4 \u003d 9 \\ cdot m $; $ z_ (M) + 1 \u003d 7 \\ cdot m $.

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

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

\\ [\\ overline (CD) \u003d \\ left (6-1 \\ right) \\ cdot \\ bar (i) + \\ left (7- \\ left (-2 \\ right) \\ right) \\ cdot \\ bar (j) + \\ 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 build the vector $ \\ overline (CN) $:

\\ [\\ overline (CN) \u003d \\ left (x_ (N) -1 \\ right) \\ cdot \\ bar (i) + \\ left (y_ (N) - \\ left (-2 \\ right) \\ right) \\ cdot \\ \\ left (y_ (N) +2 \\ right) \\ cdot \\ bar (j) + \\ left (z_ (N) -8 \\ right) \\ cdot \\ bar (k). \\]

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

$ \\ frac (x_ (N) -1) (5) \u003d \\ frac (y_ (N) +2) (9) \u003d \\ frac (z_ (N) -8) (- 10) \u003d n $, where $ n $ is the result of division.

From here we get: $ x_ (N) -1 \u003d 5 \\ cdot n $; $ y_ (N) + 2 \u003d 9 \\ cdot n $; $ z_ (N) -8 \u003d -10 \\ cdot n $.

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

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

\\ [\\ overline (MN) \u003d \\ 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). \\]

Substitute the expressions for the coordinates of the points $ M $ and $ N $:

\\ [\\ overline (MN) \u003d \\ 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:

{!LANG-3b7e5da028282a1e49fcae383d02d5e9!}

\\ [\\ overline (MN) \u003d \\ 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 straight lines $ AB $ and $ MN $ are perpendicular, the scalar product of the corresponding vectors is zero, that is, $ \\ overline (AB) \\ cdot \\ overline (MN) \u003d 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 \\ After completing the steps, we get the first equation for determining $ m $ and $ n $: $ 155 \\ cdot m + 14 \\ cdot n \u003d 86 $.

Since the lines $ CD $ and $ MN $ are perpendicular, the scalar product of the corresponding vectors is equal to zero, that is, $ \\ overline (CD) \\ cdot \\ overline (MN) \u003d 0 $:

\\ \\ [- 5 + 25 \\ cdot n + 25 \\ cdot m + 18 + 81 \\ cdot n-81 \\ cdot m-90 + 100 \\ cdot n + 70 \\ cdot m \u003d 0. \\]

After completing the steps, we get the second equation for determining $ m $ and $ n $: $ 14 \\ cdot m + 206 \\ cdot n \u003d 77 $.

Find $ m $ and $ n $ by solving the system of equations $ \\ left \\ (\\ begin (array) (c) (155 \\ cdot m + 14 \\ cdot n \u003d 86) \\\\ (14 \\ cdot m + 206 \\ cdot n \u003d 77) \\ end (array) \\ right. $.

We apply Cramer's method:

\\ [\\ Delta \u003d \\ left | \\ begin (array) (cc) (155) & (14) \\\\ (14) & (206) \\ end (array) \\ right | \u003d 31734; \\] \\ [\\ Delta _ (m) \u003d \\ left | \\ begin (array) (cc) (86) & (14) \\\\ (77) & (206) \\ end (array) \\ right | \u003d 16638; \\] \\ [\\ Delta _ (n) \u003d \\ left | \\ begin (array) (cc) (155) & (86) \\\\ (14) & (77) \\ end (array) \\ right | \u003d 10731; \\ Find the coordinates of the points $ M $ and $ N $:

Finally:

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

\ \

$ \\ overline (MN) \u003d \\ 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) \u003d 3.3125 \\ cdot \\ bar (i) +0,3251 \\ cdot \\ bar ( j) +1.9479 \\ cdot \\ bar (k) $.

The distance between the straight lines $ AB $ and $ CD $ is the length of the vector $ \\ overline (MN) $: $ d \u003d \\ sqrt (3.3125 ^ (2) + 0.3251 ^ (2) + 1.9479 ^ ( 2)) \\ approx 3.8565 $ lin. units

Corner

between straight lines in space we will call any of the adjacent angles formed by two straight lines drawn through an arbitrary point parallel to the data.

Let two straight lines be given in space: Obviously, the angle between the straight lines can be taken as the angle between their direction vectors and. Since, then, according to the formula for the cosine of the angle between the vectors, we get

The conditions for parallelism and perpendicularity of two straight lines are equivalent to the conditions for parallelism and perpendicularity of their direction vectors and:

Two straight

parallel

if and only if their respective coefficients are proportional, i.e. l 1 parallel 2 if and only if parallel{!LANG-8de52ab0767cde8c1767c82a223677c3!} 2 if and only if parallel{!LANG-11bc5cdb612758420ba4d5ea569c2064!} .

if and only if their respective coefficients are proportional, i.e. perpendicular if and only if the sum of the products of the corresponding coefficients is zero:.

Have head between line and plane

Let it be straight d - not perpendicular to the plane θ;
d′ - projection of the straight line d on the plane θ;
The smallest of the angles between straight lines d and d′ We will call angle between line and plane.
We denote it as φ \u003d ( d,θ)
If a d⊥θ, then ( d, θ) \u003d π / 2

Oi→j→k→ - rectangular coordinate system.
Plane equation:

θ: Ax+By+Cz+D=0

We assume that the line is given by a point and a direction vector: d[M0,p→]
Vector n→(A,B,C)⊥θ
Then it remains to find out the angle between the vectors n→ and p→, we denote it as γ \u003d ( n→,p→).

If the angle γ<π/2 , то искомый угол φ=π/2−γ .

If the angle γ\u003e π / 2, then the sought angle φ \u003d γ − π / 2

sinφ \u003d sin (2π − γ) \u003d cosγ

sinφ \u003d sin (γ − 2π) \u003d - cosγ

Then, angle between line and planecan be calculated using the formula:

sinφ \u003d ∣cosγ∣ \u003d ∣ ∣ Ap1+Bp2+Cp3∣ ∣ √A2+B2+C2√p21+p22+p23

Question29. The concept of a quadratic form. Sign-definiteness of quadratic forms.

Quadratic form j (x 1, x 2, ..., x n) n real variables x 1, x 2, ..., x n called the sum of the form
, (1)

where a ij - some numbers called coefficients. Without loss of generality, we can assume that a ij = a ji.

The quadratic form is called valid, if a a ij Î GR. Matrix of quadratic form called a matrix composed of its coefficients. The quadratic form (1) corresponds to the only symmetric matrix
Ie. A T \u003d A... Therefore, the quadratic form (1) can be written in matrix form j ( x) = x T Axwhere x T = (x 1 x 2 … x n). (2)

And, conversely, every symmetric matrix (2) corresponds to a unique quadratic form up to the notation of variables.

By the rank of the quadratic form call the rank of its matrix. The quadratic form is called non-degenerate, if its matrix is \u200b\u200bnondegenerate AND... (recall that the matrix AND is called nondegenerate if its determinant is not zero). Otherwise, the quadratic form is degenerate.

positively defined (or strictly positive) if

j ( x) > 0 , for anyone x = (x 1 , x 2 , …, x n), Besides x = (0, 0, …, 0).

Matrix AND positive definite quadratic form j ( x) is also called positive definite. Therefore, a single positive definite matrix corresponds to a positive definite quadratic form and vice versa.

The quadratic form (1) is called negatively defined (or strictly negative) if

j ( x) < 0, для любого x = (x 1 , x 2 , …, x n), Besides x = (0, 0, …, 0).

Similarly as above, a matrix of negative definite quadratic form is also called negative definite.

Therefore, the positively (negatively) definite quadratic form j ( x) reaches the minimum (maximum) value j ( x *) \u003d 0 for x * = (0, 0, …, 0).

Note that most of the quadratic forms are not sign definite, that is, they are neither positive nor negative. Such quadratic forms vanish not only at the origin of the coordinate system, but also at other points.

When n \u003e 2, special criteria are required to check the definiteness of the quadratic form. Let's consider them.

Major Minors quadratic forms are called minors:

that is, these are minors of the order 1, 2, ..., n matrices ANDlocated in the upper left corner, the last of them coincides with the determinant of the matrix AND.

Positive definiteness criterion (Sylvester criterion)

x) = x T Ax was positive definite, it is necessary and sufficient that all the principal minors of the matrix AND were positive, that is: M 1 > 0, M 2 > 0, …, M n > 0. Negative certainty criterion In order for the quadratic form j ( x) = x T Ax was negative definite, it is necessary and sufficient that its principal minors of even order are positive, and that of odd order are negative, i.e.: M 1 < 0, M 2 > 0, M 3 < 0, …, (–1) n