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

but. Let two lines be given. These lines, as it was indicated in Chapter 1, form various positive and negative angles, which can be either acute or 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 directing vectors of the first and second lines. The angle between these vectors is equal to one of the angles formed by straight lines. Therefore, the problem is reduced to determining the angle between the vectors, We get

For simplicity, we can agree on an angle between two straight lines to understand 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, i.e., keep only the absolute value.

Example. Determine the angle between lines

By formula (1) we have

from. If it is indicated which of the sides of the angle is its beginning and which is its end, then, counting always the direction of the angle counterclockwise, we can extract something more from formulas (1). As is easy to see from Fig. 53 the sign obtained on the right side of the formula (1) will indicate which one - acute or obtuse - the angle forms the second 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 the lines, or differs from it by ±180°.)

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

This is a necessary and sufficient condition for two lines to be parallel.

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

perpendicular because

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

f. Draw a line parallel to a given line through a point

The decision is made like this. Since the desired line is parallel to the given one, then for its directing vector we can take the same one as that of the given line, i.e., a vector with projections A and B. And then the equation of the desired 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 line through a point perpendicular to the given line

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

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

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

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

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

rewriting these equations differently, we have

Injection φ general equations A 1 x + B 1 y + C 1 = 0 and A 2 x + B 2 y + C 2 = 0, is calculated by the formula:

Injection φ between two straight lines 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, is calculated by the formula:

Distance from point to line

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

Special cases.

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

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

o When (,) 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 normal vectors in 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 that plane. Therefore, the normal vector of the given plane has coordinates. The set of all normal vectors can be given as.

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

We present two solutions to this problem.

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

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

A plane that is parallel to the coordinate plane Oyz can be given by a general incomplete equation of the plane of the form . 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 desired equation has the form.

Solution. The vector 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 arrive at 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) We compose the equation of the plane according to 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's formulate the following problem:

Write an equation for a plane passing through a given point M(x 0, y 0, z 0) perpendicular to the given vector n = ( A, B, C} .

Solution. Let be P(x, y, z) is an arbitrary point in space. Dot P belongs to the plane if and only if the vector MP = {x − x 0, y − y 0, z − z 0) orthogonal to vector n = {A, B, C) (Fig. 1).

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

Equation of a plane by three points

In vector form

In coordinates

Mutual arrangement of planes in space

are 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)

are the equations of the line of intersection of the given planes.

Write parametric equations for the following lines:

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

Let us compose the parametric equations of this straight line:

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

Thus: b) Consider the canonical equations . The choice of a point here is simple, but insidious: (be careful not to mix up the coordinates!!!). How to pull out a guide vector? You can speculate what this line is parallel to, or you can use a simple formal trick: the proportion is “Y” and “Z”, so we write the direction vector , and put zero in the remaining space: .

We 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 initial equations there are "x" and "y", and in the direction vector at these places we write zeros: . In the remaining place we put unit: . Instead of one, any number, except zero, will do.

We write the parametric equations of the straight line:

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 of 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 USE tasks 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 BF 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 AM ⊥ BF . 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.

It is necessary to consider a 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 the O x y z coordinate plane 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 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 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 → = (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 = 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, 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 B E → and B D 1 → , where their coordinates are found by the coordinates of the 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 BE → = (2 , 0 , 4) , BD 1 → = 2 , - 3 , 7 n 2 → = BE → × BD 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 = arc cos 0 12 + 0 (- 6) + 1 (- 6) 0 2 + 0 2 + 1 2 12 2 + (- 6) 2 + (- 6) 2 = arc cos 6 6 6 = arc 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 the topic of the general equation of the straight line of the form A x + B y + C z + D = 0, it was 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 into the 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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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 angles formed by the auxiliary lines. The cosine of one of the angles between the lines can be found using the 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 passes through the given points $A\left(2,-4,-1\right)$ and $B\left(-3,5,6\right)$, the second line passes 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 vector $\overline(AM)$:

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

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

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 φ between the 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 of parallelism and perpendicularity of two lines are equivalent to the conditions of parallelism and perpendicularity of their direction vectors and:

Two straight are parallel if and only if their respective coefficients are proportional, i.e. l 1 parallel l 2 if and only if parallel .

Two straight perpendicular if and only if the sum of the products of the corresponding coefficients is equal to zero: .

At goal between line and plane

Let the line d- not perpendicular to the plane θ;
d′− projection of a straight line d to the plane θ;
The smallest of the angles between straight lines d And d′ we will call angle between line and plane.
Let's denote it as φ=( d,θ)
If d⊥θ , then ( d,θ)=π/2

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

θ: Ax+By+cz+D=0

We consider that the line is given by a point and a direction vector: d[M 0,p→]
Vector n→(A,B,C)⊥θ
Then it remains to find out the angle between the vectors n→ and p→, denote it as γ=( n→,p→).

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

If the angle γ>π/2 , then the required angle φ=γ−π/2

sinφ=sin(2π−γ)=cosγ

sinφ=sin(γ−2π)=−cosγ

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

sinφ=∣cosγ∣=∣ ∣ Ap 1+bp 2+cp 3∣ ∣ √A 2+B 2+C 2√p 21+p 22+p 23

Question 29. The concept of a quadratic form. The sign-definiteness of quadratic forms.

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

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

The quadratic form is called valid, if aij О GR. Matrix of quadratic form is called the matrix composed of its coefficients. Quadratic form (1) corresponds to a unique symmetric matrix
i.e. A T = A. Therefore, quadratic form (1) can be written in matrix form j ( X) = x T Ah, where x T = (X 1 X 2 … x n). (2)

And vice versa, any symmetric matrix (2) corresponds to a unique quadratic form up to the notation of variables.

The rank of the quadratic form is called the rank of its matrix. The quadratic form is called non-degenerate, if its matrix is ​​nonsingular BUT. (recall that the matrix BUT is called non-degenerate if its determinant is non-zero). Otherwise, the quadratic form is degenerate.

positive definite(or strictly positive) if

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

The matrix BUT positive definite quadratic form j ( X) is also called positive definite. Therefore, a positive definite quadratic form corresponds to a unique positive definite matrix and vice versa.

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

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

Similarly as above, a negative-definite quadratic matrix is ​​also called negative-definite.

Therefore, a positively (negatively) definite quadratic form j ( X) reaches the minimum (maximum) value j ( X*) = 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> 2, special criteria are required to check the sign-definiteness of a quadratic form. Let's consider them.

Major Minors quadratic form are called minors:

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

Criterion for positive definiteness (Sylvester criterion)

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