The measure of the angle between planes is the acute angle formed by two straight lines lying in these planes and drawn perpendicular to the line of their intersection.
Construction algorithm
- From an arbitrary point K, perpendiculars are drawn to each of the given planes.
- By rotating around the level line, the angle γ° with the vertex at point K is determined.
- Calculate the angle between the planes ϕ° = 180 – γ°, provided that γ° > 90°. If γ°< 90°, то ∠ϕ° = ∠γ°.
The figure shows the case when the planes α and β are given by traces. All necessary constructions were carried out according to the algorithm and are described below.
Solution
- In an arbitrary place in the drawing, mark point K. From it we lower perpendiculars m and n, respectively, to the planes α and β. The direction of the projections m and n is as follows: m""⊥f 0α , m"⊥h 0α , n""⊥f 0β , n"⊥h 0β .
- We determine the actual size ∠γ° between the lines m and n. To do this, around the frontal f we rotate the plane of the angle with vertex K to a position parallel to the frontal plane of projection. The radius of rotation R of point K is equal to the size of the hypotenuse of a right triangle O""K""K 0 , the side of which is K""K 0 = y K – y O .
- The desired angle is ϕ° = ∠γ°, since ∠γ° is acute.
The figure below shows the solution to a problem in which it is required to find the angle γ° between the planes α and β, given by parallel and intersecting lines, respectively.
Solution
- We determine the direction of the projections of the horizontals h 1, h 2 and the fronts f 1, f 2 belonging to the planes α and β, in the order indicated by the arrows. From an arbitrary point K on the square. α and β we omit the perpendiculars e and k. In this case, e""⊥f"" 1 , e"⊥h" 1 and k""⊥f"" 2 , k"⊥h" 2 .
- We define ∠γ° between the lines e and k. To do this, draw a horizontal line h 3 and around it we rotate point K to position K 1, at which △CKD will become parallel to the horizontal plane and will be reflected on it in natural size - △C"K" 1 D". The projection of the center of rotation O" is located on the drawn to h" 3 perpendicular to K"O". The radius R is determined from the right triangle O"K"K 0, whose side K"K 0 = Z O – Z K.
- The value of the desired value is ∠ϕ° = ∠γ°, since the angle γ° is acute.
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Consider two planes R 1 and R 2 with normal vectors n 1 and n 2. Angle φ between planes R 1 and R 2 is expressed through the angle ψ = \(\widehat((n_1; n_2))\) as follows: if ψ < 90°, then φ = ψ (Fig. 202, a); if ψ > 90°, then ψ = 180° - ψ (Fig. 202.6).
It is obvious that in any case the equality is true
cos φ = |cos ψ|
Since the cosine of the angle between non-zero vectors is equal to scalar product of these vectors divided by the product of their lengths, we have
$$ cos\psi=cos\widehat((n_1; n_2))=\frac(n_1\cdot n_2)(|n_1|\cdot |n_2|) $$
and, therefore, the cosine of the angle φ between the planes R 1 and R 2 can be calculated using the formula
$$ cos\phi=\frac(n_1\cdot n_2)(|n_1|\cdot |n_2|) (1)$$
If the planes are given by general equations
A 1 X+ B 1 y+ C 1 z+ D 1 = 0 and A 2 X+ B 2 y+ C 2 z+ D 2 = 0,
then for their normal vectors we can take the vectors n 1 = (A 1 ; B 1 ; C 1) and n 2 = (A 2; B 2; C 2).
Writing the right-hand side of formula (1) in terms of coordinates, we obtain
$$ cos\phi=\frac(|A_1 A_2 + B_1 B-2 + C_1 C_2|)(\sqrt((A_1)^2+(B_1)^2+(C_1)^2)\sqrt((A_2) ^2+(B_2)^2+(C_2)^2)) $$
Task 1. Calculate the angle between planes
X - √2 y + z- 2 = 0 and x+ √2 y - z + 13 = 0.
In this case, A 1 .=1, B 1 = - √2, C 1 = 1, A 2 =1, B 2 = √2, C 2 = - 1.
From formula (2) we get
$$ cos\phi=\frac(|1\cdot 1 - \sqrt2 \cdot \sqrt2 - 1 \cdot 1|)(\sqrt(1^2+(-\sqrt2)^2+1^2)\sqrt (1^2+(\sqrt2)^2+(-1)^2))=\frac(1)(2) $$
Therefore, the angle between these planes is 60°.
Planes with normal vectors n 1 and n 2:
a) are parallel if and only if the vectors n 1 and n 2 are collinear;
b) perpendicular if and only if the vectors n 1 and n 2 are perpendicular, i.e. when n 1 n 2 = 0.
From here we obtain the necessary and sufficient conditions for the parallelism and perpendicularity of two planes given by general equations.
To plane
A 1 X+ B 1 y+ C 1 z+ D 1 = 0 and A 2 X+ B 2 y+ C 2 z+ D 2 = 0
were parallel, it is necessary and sufficient for the equalities to hold
$$ \frac(A_1)(A_2)=\frac(B_1)(B_2)=\frac(C_1)(C_2) \;\; (3)$$
If any of the coefficients A 2 , B 2 , C 2 is equal to zero, it is assumed that the corresponding coefficient A 1 , B 1 , C 1 is also equal to zero
Failure to satisfy at least one of these two equalities means that the planes are not parallel, that is, they intersect.
For perpendicularity of planes
A 1 X+ B 1 y+ C 1 z+ D 1 = 0 and A 2 X+ B 2 y+ C 2 z+ D 2 = 0
it is necessary and sufficient for the equality to hold
A 1 A 2 + B 1 B 2 + C 1 C 2 = 0. (4)
Task 2. Among the following pairs of planes:
2X + 5at + 7z- 1 = 0 and 3 X - 4at + 2z = 0,
at - 3z+ 1 = 0 and 2 at - 6z + 5 = 0,
4X + 2at - 4z+ 1 = 0 and 2 X + at + 2z + 3 = 0
indicate parallel or perpendicular. For the first pair of planes
A 1 A 2 + B 1 B 2 + C 1 C 2 = 2 3 + 5 (- 4) + 7 2 = 0,
i.e., the perpendicularity condition is satisfied. The planes are perpendicular.
For the second pair of planes
\(\frac(B_1)(B_2)=\frac(C_1)(C_2)\), since \(\frac(1)(2)=\frac(-3)(-6)\)
and coefficients A 1 and A 2 are equal to zero. Therefore, the planes of the second pair are parallel. For the third pair
\(\frac(B_1)(B_2)\neq\frac(C_1)(C_2)\), since \(\frac(2)(1)\neq\frac(-4)(2)\)
and A 1 A 2 + B 1 B 2 + C 1 C 2 = 4 2 + 2 1 - 4 2 =/= 0, i.e. the planes of the third pair are neither parallel nor perpendicular.
This article is about the angle between planes and how to find it. First, the definition of the angle between two planes is given and a graphical illustration is given. After this, the principle of finding the angle between two intersecting planes using the coordinate method was analyzed, and a formula was obtained that allows you to calculate the angle between intersecting planes using the known coordinates of the normal vectors of these planes. In conclusion it is shown detailed solutions characteristic tasks.
Page navigation.
Angle between planes - definition.
Let us present arguments that will allow us to gradually approach the determination of the angle between two intersecting planes.
Let us be given two intersecting planes and . These planes intersect along a straight line, which we denote by the letter c. Let's construct a plane passing through point M of line c and perpendicular to line c. In this case, the plane will intersect the planes and. Let us denote the straight line along which the planes intersect as a, and the straight line along which the planes intersect as b. Obviously, lines a and b intersect at point M.
It is easy to show that the angle between intersecting lines a and b does not depend on the location of point M on the line c through which the plane passes.
Let's construct a plane perpendicular to the line c and different from the plane. The plane is intersected by planes and along straight lines, which we denote as a 1 and b 1, respectively.
From the method of constructing planes it follows that lines a and b are perpendicular to line c, and lines a 1 and b 1 are perpendicular to line c. Since lines a and a 1 lie in the same plane and are perpendicular to line c, then they are parallel. Similarly, lines b and b 1 lie in the same plane and are perpendicular to line c, therefore, they are parallel. Thus, it is possible to perform a parallel transfer of the plane to the plane, in which straight line a 1 coincides with straight line a, and straight line b with straight line b 1. Therefore, the angle between two intersecting lines a 1 and b 1 equal to angle between intersecting lines a and b.
This proves that the angle between intersecting lines a and b lying in intersecting planes and does not depend on the choice of point M through which the plane passes. Therefore, it is logical to take this angle as the angle between two intersecting planes.
Now you can voice the definition of the angle between two intersecting planes and.
Definition.
The angle between two planes intersecting in a straight line and- this is the angle between two intersecting lines a and b, along which the planes and intersect with the plane perpendicular to the line c.
The definition of the angle between two planes can be given a little differently. If on the straight line c along which the planes and intersect, mark a point M and draw straight lines a and b through it, perpendicular to the straight line c and lying in the planes and, respectively, then the angle between the straight lines a and b is the angle between the planes and. Usually in practice, just such constructions are performed in order to obtain the angle between the planes.
Since the angle between intersecting straight lines does not exceed , it follows from the stated definition that the degree measure of the angle between two intersecting planes is expressed real number from the interval . In this case, intersecting planes are called perpendicular, if the angle between them is ninety degrees. Angle between parallel planes either they do not determine it at all, or they consider it equal to zero.
Finding the angle between two intersecting planes.
Usually, when finding an angle between two intersecting planes, you first have to perform additional constructions to see the intersecting straight lines, the angle between which is equal to the desired angle, and then connect this angle with the original data using equality tests, similarity tests, the cosine theorem or definitions of sine, cosine and tangent of the angle. In the course of geometry high school similar problems occur.
As an example, let’s give the solution to Problem C2 from the Unified State Exam in Mathematics for 2012 (the condition was intentionally changed, but this does not affect the principle of the solution). In it, you just had to find the angle between two intersecting planes.
Example.
Solution.
First, let's make a drawing.
Let's perform additional constructions to “see” the angle between the planes.
First, let's define a straight line along which planes ABC and BED 1 intersect. Point B is one of their common points. Let's find the second common point of these planes. Lines DA and D 1 E lie in the same plane ADD 1, and they are not parallel, and therefore intersect. On the other hand, line DA lies in the plane ABC, and line D 1 E - in the plane BED 1, therefore, the intersection point of lines DA and D 1 E will be the common point of planes ABC and BED 1. So, let's continue the lines DA and D 1 E to their intersection, denoting the point of their intersection with the letter F. Then BF is the straight line along which planes ABC and BED 1 intersect.
It remains to construct two lines lying in the planes ABC and BED 1, respectively, passing through one point on the line BF and perpendicular to the line BF - the angle between these lines, by definition, will be equal to the desired angle between the planes ABC and BED 1. Let's do it.
Dot A is the projection of point E onto plane ABC. Let's draw a straight line intersecting line BF at right angles at point M. Then the straight line AM is the projection of the straight line EM onto the plane ABC, and by the theorem of three perpendiculars.
Thus, the required angle between planes ABC and BED 1 is equal to .
We can determine the sine, cosine or tangent of this angle (and therefore the angle itself) from the right triangle AEM if we know the lengths of its two sides. From the condition it is easy to find the length AE: since point E divides side AA 1 in the ratio of 4 to 3, counting from point A, and the length of side AA 1 is 7, then AE = 4. Let's find the length AM.
To do this, consider right triangle ABF with right angle A, where AM is the height. By condition AB = 2. We can find the length of side AF from the similarity of right triangles DD 1 F and AEF: 
Using the Pythagorean theorem, we find from triangle ABF. We find the length AM through the area of triangle ABF: on one side the area of triangle ABF is equal to 
, on the other side 
, where 
.
Thus, from the right triangle AEM we have 
.
Then the required angle between planes ABC and BED 1 is equal (note that 
).
Answer:
In some cases, to find the angle between two intersecting planes, it is convenient to set Oxyz and use the coordinate method. Let's stop there.
Let us set the task: find the angle between two intersecting planes and . Let us denote the desired angle as .
We will assume that in a given rectangular coordinate system Oxyz we know the coordinates of the normal vectors of intersecting planes and or have the opportunity to find them. Let 
is the normal vector of the plane, and 
is the normal vector of the plane. We will show how to find the angle between intersecting planes and through the coordinates of the normal vectors of these planes.
Let us denote the straight line along which the planes and intersect as c. Through point M on line c we draw a plane perpendicular to line c. The plane intersects the planes and along lines a and b, respectively, lines a and b intersect at point M. By definition, the angle between intersecting planes and is equal to the angle between intersecting lines a and b.
Let us plot the normal vectors and planes and from the point M in the plane. In this case, the vector lies on a line that is perpendicular to line a, and the vector lies on a line that is perpendicular to line b. Thus, in the plane the vector is the normal vector of the line a, is the normal vector of the line b.
In the article finding the angle between intersecting lines, we received a formula that allows us to calculate the cosine of the angle between intersecting lines using the coordinates of normal vectors. Thus, the cosine of the angle between lines a and b, and, consequently, cosine of the angle between intersecting planes and is found by the formula, where And are the normal vectors of the planes and, respectively. Then it is calculated as 
.
Let's solve the previous example using the coordinate method.
Example.
Dan cuboid ABCDA 1 B 1 C 1 D 1, in which AB=2, AD=3, AA 1 =7 and point E divides side AA 1 in the ratio of 4 to 3, counting from point A. Find the angle between planes ABC and BED 1.
Solution.
Since the sides of a rectangular parallelepiped at one vertex are pairwise perpendicular, it is convenient to introduce rectangular system coordinates Oxyz like this: align the beginning with the vertex C, and direct the coordinate axes Ox, Oy and Oz along the sides CD, CB and CC 1, respectively.
The angle between the ABC and BED 1 planes can be found through the coordinates of the normal vectors of these planes using the formula , where and are the normal vectors of the ABC and BED 1 planes, respectively. Let's determine the coordinates of normal vectors.
The article talks about finding the angle between planes. After giving the definition, let’s give a graphic illustration and consider detailed method finding by coordinate method. We obtain a formula for intersecting planes, which includes the coordinates of normal vectors.
Yandex.RTB R-A-339285-1
The material will use data and concepts that were previously studied in articles about the plane and line in space. First, it is necessary to move on to reasoning that allows us 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 associated with the intersection of these planes. The plane χ passes through the point M as a straight line c. The intersection of the planes γ 1 and γ 2 will be made using the plane χ. We take the designation of the line intersecting γ 1 and χ as line a, and the line intersecting γ 2 and χ as line b. We find that the intersection of lines a and b gives the point M.
The location of point M does not affect the angle between intersecting lines a and b, and point M is located on 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, lines a and b are perpendicular to line c, then a 1, b 1 are located perpendicular to line c. Finding straight lines a and a 1 in the plane γ 1 with perpendicularity to straight line c, then they can be considered parallel. In the same way, the location of b and b 1 in the γ 2 plane with perpendicularity to straight 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 straight lines a and a 1, b and b 1. We find that the angle between intersecting lines a and b 1 is equal to the angle of intersecting lines a and b.
Let's look at the figure below.
This proposition is proven 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 considered 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 called 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 line c.
Consider the figure below.
The determination may be submitted in another form. When the planes γ 1 and γ 2 intersect, where c is the line on which they intersected, mark a point M through which draw lines a and b perpendicular to line c and lying in the planes γ 1 and γ 2, then the angle between lines a and b will be the angle between the planes. In practice, this is applicable for constructing the angle between planes.
When intersecting, 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 signs of equality or similarity of a triangle, sines, and cosines of an angle.
Let's consider solving problems using an example from the Unified State Exam problems of block C 2.
Example 1
Given a rectangular parallelepiped A B C D A 1 B 1 C 1 D 1, where side A B = 2, A D = 3, A A 1 = 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.
Solution
For clarity, it is necessary to make a drawing. We get that
A visual representation is necessary to make it more convenient to work with the angle between planes.
We determine the straight line along which the intersection of planes A B C and B E D 1 occurs. Point B is a common point. Another common point of intersection should be found. Let's consider the straight lines D A and D 1 E, which are located in the same plane A D D 1. Their location does not indicate parallelism; it means that they have a common point of intersection.
However, straight line D A is located in the plane A B C, and D 1 E in B E D 1. From this we get that the straight lines D A And D 1 E have a common intersection point, which is 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 this we obtain that B F is the straight line along which planes A B C and B E D 1 intersect.
Let's look at the figure below.
To obtain the answer, it is necessary to construct straight lines located in planes A B C and B E D 1 passing through a point located on line B F and perpendicular to it. Then the resulting angle between these straight lines is considered the desired angle between planes A B C and B E D 1.
From this we can see that point A is the projection of point E onto the plane A B C. It is necessary to draw a straight line intersecting line B F at a right angle at point M. It can be seen that straight line A M is the projection of straight line E M onto the plane A B C, based on the theorem about those perpendiculars A M ⊥ B F . Consider the picture 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, and then the angle itself, only if its two sides are 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, which means the total length of the straight line is 7 parts, then A E = 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 = 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 side B F of 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 side A M is found through the area of 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 planes A B C and B E D 1 is equal to a r c t g 5, then upon simplification we obtain 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 specified using coordinate plane O x y z and the coordinate method. Let's take a closer look.
If a problem is given where it is necessary to find the angle between intersecting planes γ 1 and γ 2, we denote the desired angle as 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 the normal vector of the plane γ 1, and n 2 → = (n 2 x, n 2 y, n 2 z) - for the plane γ 2. Let us consider the detailed determination 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 c 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 point M. from the definition it follows 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 plot 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 → is located on a line perpendicular to line b. From here we get that given planeχ has a normal vector of line a equal to n 1 → and for 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 vectors. We found that the cosine of the angle between straight lines a and b is the same as the cosine between intersecting planes γ 1 and γ 2, which is derived from cos formulasα = 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
According to the condition, the parallelepiped A B C D A 1 B 1 C 1 D 1 is given , where A B = 2, A D = 3, A A 1 = 7, and point E divides side A A 1 4: 3. Find the angle between planes A B C and B E D 1.
Solution
From the condition it is clear that its sides are pairwise perpendicular. This means that it is necessary to introduce a coordinate system O x y z with the vertex at point C and coordinate axes O x, O y, O z. It is necessary to set the direction to the appropriate sides. Consider the figure below.
Intersecting planes A B C And B E D 1 form an angle that can be found by 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, in which 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 with the plane A B C, this means that the coordinates of the normal vector k → are equal to the value n 1 → = k → = (0, 0, 1).
The normal vector of the plane B E D 1 is taken to be 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 conditions 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 points A 2, 3, 0, A 1 2, 3, 7 we find E 2, 3, 4. We find 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 with the goal of finding the angle between intersecting planes with the existing 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 coordinate system O x y z 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 straight line equation 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. This means that n 1 → = 2, - 4, 1 and n 2 → = 0, 3, - 1 are normal vectors of the 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
From here we have that the cosine of the angle takes the form cos α = 13 210. Then the angle of intersecting lines is not obtuse. Substituting in trigonometric identity, we find that the value of the sine of the angle is equal to the expression. Let us calculate and find 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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