Find the angle phi between the lines. Finding the angle between lines
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 a 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), Besides X = (0, 0, …, 0).
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), Besides 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 are positive, and those of odd order are negative, i.e.: M 1 < 0, M 2 > 0, M 3 < 0, …, (–1)n
a. Let two lines be given. These lines, as it was indicated in Chapter 1, form various positive and negative angles, which, in this case, 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 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
with. 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 direction 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 win 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
I will be brief. The angle between two lines is 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:
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:
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:
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 down 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:
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 of coordinates. It remains to find the cosine of the angle:
It will be useful for every student who is preparing for the exam in mathematics to repeat the topic “Finding the angle between lines”. As statistics show, when passing an attestation test, tasks in this section of stereometry cause difficulties for a large number of students. At the same time, tasks requiring finding the angle between straight lines are found in the USE at both the basic and profile levels. This means that everyone should be able to solve them.
Basic moments
There are 4 types of mutual arrangement of lines in space. They can coincide, intersect, be parallel or intersecting. The angle between them can be acute or straight.
To find the angle between the lines in the Unified State Examination or, for example, in the solution, schoolchildren in Moscow and other cities can use several methods for solving problems in this section of stereometry. You can complete the task by classical constructions. To do this, it is worth learning the basic axioms and theorems of stereometry. The student needs to be able to logically build reasoning and create drawings in order to bring the task to a planimetric problem.
You can also use the vector-coordinate method, using simple formulas, rules and algorithms. The main thing in this case is to correctly perform all the calculations. The Shkolkovo educational project will help you hone your skills in solving problems in stereometry and other sections of the school course.