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When deciding geometric problems in space there are often those where it is necessary to calculate the angles between different spatial objects. In this article we will consider the issue of finding angles between planes and between them and a straight line.
Straight line in space
It is known that absolutely any straight line in the plane can be defined by the following equality:
Here a and b are some numbers. If we imagine a straight line in space using the same expression, we will get a plane parallel to the z axis. For mathematical definition spatial straight line, a different method of solution is used than in the two-dimensional case. It consists in using the concept of “direction vector”.
Examples of solving problems on determining the angle of intersection of planes
Knowing how to find the angle between planes, we will solve the following problem. Given two planes, the equations of which have the form:
3 * x + 4 * y - z + 3 = 0;
X - 2 * y + 5 * z +1 = 0
What is the angle between the planes?
To answer the question of the problem, remember that the coefficients associated with the variables in the general plane equation are the coordinates of the guide vector. For these planes we have the following coordinates of their normals:
n 1 ¯(3; 4; -1);
n 2 ¯(-1; -2; 5)
Now we find the scalar product of these vectors and their modules, we have:
(n 1 ¯ * n 2 ¯) = -3 -8 -5 = -16;
|n 1 ¯| = √(9 + 16 + 1) = √26;
|n 2 ¯| = √(1 + 4 + 25) = √30
Now you can substitute the found numbers into the given in previous paragraph formula. We get:
α = arccos(|-16 | / (√26 * √30) ≈ 55.05 o
The resulting value corresponds to the acute angle of intersection of the planes specified in the problem statement.
Now let's look at another example. Two planes are given:
Do they intersect? Let's write down the values of the coordinates of their direction vectors and calculate scalar product them and modules:
n 1 ¯(1; 1; 0);
n 2 ¯(3; 3; 0);
(n 1 ¯ * n 2 ¯) = 3 + 3 + 0 = 6;
|n 1 ¯| = √2;
|n 2 ¯| = √18
Then the angle of intersection is:
α = arccos(|6| / (√2 * √18) =0 o .
This angle indicates that the planes do not intersect, but are parallel. The fact that they do not coincide with each other is easy to check. To do this, take an arbitrary point belonging to the first of them, for example, P(0; 3; 2). Substituting its coordinates into the second equation, we get:
3 * 0 +3 * 3 + 8 = 17 ≠ 0
That is, point P belongs only to the first plane.
Thus, two planes are parallel when their normals are so.
Flat and straight
In case of consideration relative position There are slightly more options between a plane and a straight line than with two planes. This fact is due to the fact that a straight line is a one-dimensional object. A straight line and a plane can be:
- mutually parallel, in this case the plane does not intersect the line;
- the latter may belong to the plane, while it will also be parallel to it;
- both objects can intersect at some angle.
Let's consider first last case, since it requires the introduction of the concept of intersection angle.
Straight line and plane, the value of the angle between them
If a plane intersects a straight line, then it is called inclined with respect to it. The point of intersection is usually called the base of the inclined line. To determine the angle between these geometric objects, it is necessary to lower a straight perpendicular from any point onto the plane. Then the point of intersection of the perpendicular with the plane and the intersection of the inclined line with it form a straight line. The latter is called the projection of the original line onto the plane under consideration. Sharp and its projection is the desired one.
The somewhat confusing definition of the angle between a plane and an inclined one will be clarified by the figure below.
Here angle ABO is the angle between straight line AB and plane a.
To write down the formula for it, consider an example. Let there be a straight line and a plane, which are described by the equations:
(x; y; z) = (x 0; y 0; z 0) + λ * (a; b; c);
A * x + B * x + C * x + D = 0
You can easily calculate the desired angle for these objects if you find the scalar product between the direction vectors of the straight line and the plane. Received sharp corner should be subtracted from 90 o, then it is obtained between a straight line and a plane.
The figure above demonstrates the described algorithm for finding the angle in question. Here β is the angle between the normal and the line, and α is between the line and its projection onto the plane. It can be seen that their sum is 90 o.
Above was presented a formula that answers the question of how to find an angle between planes. Now we give the corresponding expression for the case of a straight line and a plane:
α = arcsin(|a * A + b * B + c * C| / (√(a 2 + b 2 + c 2) * √(A 2 + B 2 + C 2)))
The module in the formula allows you to calculate only acute angles. The arcsine function appeared instead of the arccosine thanks to the use of the corresponding reduction formula between trigonometric functions (cos(β) = sin(90 o-β) = sin(α)).
Problem: a plane intersects a line
Now we will show how to work with the given formula. Let's solve the problem: we need to calculate the angle between the y-axis and the plane, given by the equation:
This plane is shown in the figure.
It can be seen that it intersects the y and z axes at points (0; -12; 0) and (0; 0; 12), respectively, and is parallel to the x axis.
The direction vector of the straight line y has coordinates (0; 1; 0). Vector perpendicular given plane, characterized by coordinates (0; 1; -1). We apply the formula for the angle of intersection of a straight line and a plane, we get:
α = arcsin(|1| / (√1 * √2)) = arcsin(1 / √2) = 45 o
Problem: a line parallel to a plane
Now let's solve a similar one previous task, whose question is posed differently. The equations of a plane and a line are known:
x + y - z - 3 = 0;
(x; y; z) = (1; 0; 0) + λ * (0; 2; 2)
It is necessary to find out whether these geometric objects are parallel each other to a friend.
We have two vectors: the directing line is equal to (0; 2; 2) and the directing plane is equal to (1; 1; -1). We find their scalar product:
0 * 1 + 1 * 2 - 1 * 2 = 0
The resulting zero indicates that the angle between these vectors is 90 o, which proves the parallelism of the straight line and the plane.
Now let’s check whether this line is only parallel or also lies in the plane. To do this, select an arbitrary point on a line and check whether it belongs to the plane. For example, let's take λ = 0, then the point P(1; 0; 0) belongs to the line. We substitute plane P into the equation:
Point P does not belong to the plane, and therefore the entire line does not lie in it.
Where is it important to know the angles between the considered geometric objects?
The above formulas and examples of problem solving are not only of theoretical interest. They are often used to determine important physical quantities real volumetric figures, such as prisms or pyramids. It is important to be able to determine the angle between planes when calculating the volumes of figures and the areas of their surfaces. Moreover, if in the case of a straight prism it is possible not to use these formulas to determine the indicated quantities, then for any type of pyramid their use turns out to be inevitable.
Below we will consider an example of using the stated theory to determine the corners of a pyramid with a square base.
Pyramid and its corners
The figure below shows a pyramid, at the base of which lies a square with side a. The height of the figure is h. You need to find two angles:
- between the side surface and the base;
- between the side rib and the base.
To solve the problem, you must first introduce a coordinate system and determine the parameters of the corresponding vertices. The figure shows that the origin coincides with the point in the center square base. In this case, the base plane is described by the equation:
That is, for any x and y, the value of the third coordinate is always zero. The lateral plane ABC intersects the z axis at point B(0; 0; h), and the y axis at the point with coordinates (0; a/2; 0). It does not intersect the x axis. This means that the equation of the ABC plane can be written as:
y/(a/2) + z/h = 1 or
2 * h * y + a * z - a * h = 0
Vector AB¯ is a side edge. The coordinates of its beginning and end are equal: A(a/2; a/2; 0) and B(0; 0; h). Then the coordinates of the vector itself:
We have found all the necessary equations and vectors. Now it remains to use the considered formulas.
Let us first calculate the angle in the pyramid between the planes of the base and the side. The corresponding normal vectors are equal: n 1 ¯(0; 0; 1) and n 2 ¯(0; 2*h; a). Then the angle will be:
α = arccos(a / √(4 * h 2 + a 2))
The angle between the plane and edge AB will be equal to:
β = arcsin(h / √(a 2 / 2 + h 2))
All that remains is to substitute specific values sides of the base a and height h to obtain the required angles.
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. So you can do parallel transfer plane to 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 lines does not exceed , it follows from the stated definition that degree measure 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 point of intersection of lines DA and D 1 E will be common point ABC planes 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 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 a 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.
Given a rectangular parallelepiped 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 rectangular parallelepiped when one vertex is pairwise perpendicular, it is convenient to introduce rectangular system coordinates Oxyz like this: the beginning is aligned with vertex C, and coordinate axes Ox, Oy and Oz are directed to 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.
\(\blacktriangleright\) Dihedral angle is an angle formed by two half-planes and a straight line \(a\), which is their common boundary.
\(\blacktriangleright\) To find the angle between the planes \(\xi\) and \(\pi\) , you need to find linear angle(and spicy or straight) dihedral angle, formed by the planes \(\xi\) and \(\pi\) :
Step 1: let \(\xi\cap\pi=a\) (the line of intersection of the planes). In the plane \(\xi\) we mark an arbitrary point \(F\) and draw \(FA\perp a\) ;
Step 2: carry out \(FG\perp \pi\) ;
Step 3: according to TTP (\(FG\) – perpendicular, \(FA\) – oblique, \(AG\) – projection) we have: \(AG\perp a\) ;
Step 4: The angle \(\angle FAG\) is called the linear angle of the dihedral angle formed by the planes \(\xi\) and \(\pi\) .
Note that the triangle \(AG\) is right-angled.
Note also that the plane \(AFG\) constructed in this way is perpendicular to both planes \(\xi\) and \(\pi\) . Therefore, we can say it differently: angle between planes\(\xi\) and \(\pi\) is the angle between two intersecting lines \(c\in \xi\) and \(b\in\pi\) forming a plane perpendicular to and \(\xi\ ) , and \(\pi\) .
Task 1 #2875
Task level: More difficult than the Unified State Exam
Dana quadrangular pyramid, all edges of which are equal, and the base is a square. Find \(6\cos \alpha\) , where \(\alpha\) is the angle between its adjacent side faces.
Let \(SABCD\) – this pyramid(\(S\) is a vertex) whose edges are equal to \(a\) . Therefore, everything side faces are equal equilateral triangles. Let's find the angle between the faces \(SAD\) and \(SCD\) .
Let's do \(CH\perp SD\) . Because \(\triangle SAD=\triangle SCD\), then \(AH\) will also be the height of \(\triangle SAD\) . Therefore, by definition, \(\angle AHC=\alpha\) is the linear angle of the dihedral angle between the faces \(SAD\) and \(SCD\) .
Since the base is a square, then \(AC=a\sqrt2\) . Note also that \(CH=AH\) is the height equilateral triangle with side \(a\) , therefore, \(CH=AH=\frac(\sqrt3)2a\) .
Then, by the cosine theorem from \(\triangle AHC\) : \[\cos \alpha=\dfrac(CH^2+AH^2-AC^2)(2CH\cdot AH)=-\dfrac13 \quad\Rightarrow\quad 6\cos\alpha=-2.\]
Answer: -2
Task 2 #2876
Task level: More difficult than the Unified State Exam
The planes \(\pi_1\) and \(\pi_2\) intersect at an angle whose cosine is equal to \(0.2\). The planes \(\pi_2\) and \(\pi_3\) intersect at right angles, and the line of intersection of the planes \(\pi_1\) and \(\pi_2\) is parallel to the line of intersection of the planes \(\pi_2\) and \(\ pi_3\) . Find the sine of the angle between the planes \(\pi_1\) and \(\pi_3\) .
Let the line of intersection of \(\pi_1\) and \(\pi_2\) be a straight line \(a\), the line of intersection of \(\pi_2\) and \(\pi_3\) be a straight line \(b\), and the line of intersection \(\pi_3\) and \(\pi_1\) – straight line \(c\) . Since \(a\parallel b\) , then \(c\parallel a\parallel b\) (according to the theorem from the section of the theoretical reference “Geometry in space” \(\rightarrow\) “Introduction to stereometry, parallelism”).
Let's mark the points \(A\in a, B\in b\) so that \(AB\perp a, AB\perp b\) (this is possible since \(a\parallel b\) ). Let us mark \(C\in c\) so that \(BC\perp c\) , therefore, \(BC\perp b\) . Then \(AC\perp c\) and \(AC\perp a\) .
Indeed, since \(AB\perp b, BC\perp b\) , then \(b\) is perpendicular to the plane \(ABC\) . Since \(c\parallel a\parallel b\), then the lines \(a\) and \(c\) are also perpendicular to the plane \(ABC\), and therefore to any line from this plane, in particular, the line \ (AC\) .
It follows that \(\angle BAC=\angle (\pi_1, \pi_2)\), \(\angle ABC=\angle (\pi_2, \pi_3)=90^\circ\), \(\angle BCA=\angle (\pi_3, \pi_1)\). It turns out that \(\triangle ABC\) is rectangular, which means \[\sin \angle BCA=\cos \angle BAC=0.2.\]
Answer: 0.2
Task 3 #2877
Task level: More difficult than the Unified State Exam
Given straight lines \(a, b, c\) intersecting at one point, and the angle between any two of them is equal to \(60^\circ\) . Find \(\cos^(-1)\alpha\) , where \(\alpha\) is the angle between the plane formed by lines \(a\) and \(c\) and the plane formed by lines \(b\ ) and \(c\) . Give your answer in degrees.
Let the lines intersect at the point \(O\) . Since the angle between any two of them is equal to \(60^\circ\), then all three straight lines cannot lie in the same plane. Let us mark the point \(A\) on the line \(a\) and draw \(AB\perp b\) and \(AC\perp c\) . Then \(\triangle AOB=\triangle AOC\) as rectangular along the hypotenuse and acute angle. Therefore, \(OB=OC\) and \(AB=AC\) .
Let's do \(AH\perp (BOC)\) . Then by the theorem about three perpendiculars \(HC\perp c\) , \(HB\perp b\) . Since \(AB=AC\) , then \(\triangle AHB=\triangle AHC\) as rectangular along the hypotenuse and leg. Therefore, \(HB=HC\) . This means that \(OH\) is the bisector of the angle \(BOC\) (since the point \(H\) is equidistant from the sides of the angle).
Note that in this way we also constructed the linear angle of the dihedral angle, formed by a plane, formed by lines \(a\) and \(c\), and the plane formed by lines \(b\) and \(c\) . This is the angle \(ACH\) .
Let's find this angle. Since we chose the point \(A\) arbitrarily, let us choose it so that \(OA=2\) . Then in rectangular \(\triangle AOC\) : \[\sin 60^\circ=\dfrac(AC)(OA) \quad\Rightarrow\quad AC=\sqrt3 \quad\Rightarrow\quad OC=\sqrt(OA^2-AC^2)=1.\ ] Since \(OH\) is a bisector, then \(\angle HOC=30^\circ\) , therefore, in a rectangular \(\triangle HOC\) : \[\mathrm(tg)\,30^\circ=\dfrac(HC)(OC)\quad\Rightarrow\quad HC=\dfrac1(\sqrt3).\] Then from the rectangular \(\triangle ACH\) : \[\cos\angle \alpha=\cos\angle ACH=\dfrac(HC)(AC)=\dfrac13 \quad\Rightarrow\quad \cos^(-1)\alpha=3.\]
Answer: 3
Task 4 #2910
Task level: More difficult than the Unified State Exam
The planes \(\pi_1\) and \(\pi_2\) intersect along the straight line \(l\) on which the points \(M\) and \(N\) lie. The segments \(MA\) and \(MB\) are perpendicular to the straight line \(l\) and lie in the planes \(\pi_1\) and \(\pi_2\) respectively, and \(MN = 15\) , \(AN = 39\) , \(BN = 17\) , \(AB = 40\) . Find \(3\cos\alpha\) , where \(\alpha\) is the angle between the planes \(\pi_1\) and \(\pi_2\) .
The triangle \(AMN\) is right-angled, \(AN^2 = AM^2 + MN^2\), whence \ The triangle \(BMN\) is right-angled, \(BN^2 = BM^2 + MN^2\), from which \We write the cosine theorem for the triangle \(AMB\): \ Then \ Since the angle \(\alpha\) between the planes is an acute angle, and \(\angle AMB\) turned out to be obtuse, then \(\cos\alpha=\dfrac5(12)\) . Then \
Answer: 1.25
Task 5 #2911
Task level: More difficult than the Unified State Exam
\(ABCDA_1B_1C_1D_1\) is a parallelepiped, \(ABCD\) is a square with side \(a\), point \(M\) is the base of the perpendicular dropped from the point \(A_1\) to the plane \((ABCD)\) , in addition, \(M\) is the point of intersection of the diagonals of the square \(ABCD\) . It is known that \(A_1M = \dfrac(\sqrt(3))(2)a\). Find the angle between the planes \((ABCD)\) and \((AA_1B_1B)\) . Give your answer in degrees.
Let's construct \(MN\) perpendicular to \(AB\) as shown in the figure.
Since \(ABCD\) is a square with side \(a\) and \(MN\perp AB\) and \(BC\perp AB\) , then \(MN\parallel BC\) . Since \(M\) is the point of intersection of the diagonals of the square, then \(M\) is the midpoint of \(AC\), therefore, \(MN\) is middle line And \(MN =\frac12BC= \frac(1)(2)a\).
\(MN\) is the projection of \(A_1N\) onto the plane \((ABCD)\), and \(MN\) is perpendicular to \(AB\), then, by the theorem of three perpendiculars, \(A_1N\) is perpendicular to \(AB \) and the angle between the planes \((ABCD)\) and \((AA_1B_1B)\) is \(\angle A_1NM\) .
\[\mathrm(tg)\, \angle A_1NM = \dfrac(A_1M)(NM) = \dfrac(\frac(\sqrt(3))(2)a)(\frac(1)(2)a) = \sqrt(3)\qquad\Rightarrow\qquad\angle A_1NM = 60^(\circ)\]
Answer: 60
Task 6 #1854
Task level: More difficult than the Unified State Exam
In a square \(ABCD\) : \(O\) – the point of intersection of the diagonals; \(S\) – does not lie in the plane of the square, \(SO \perp ABC\) . Find the angle between the planes \(ASD\) and \(ABC\) if \(SO = 5\) and \(AB = 10\) .
Right triangles \(\triangle SAO\) and \(\triangle SDO\) are equal in two sides and the angle between them (\(SO \perp ABC\) \(\Rightarrow\) \(\angle SOA = \angle SOD = 90^\circ\); \(AO = DO\) , because \(O\) – point of intersection of the diagonals of the square, \(SO\) – common side) \(\Rightarrow\) \(AS = SD\) \(\Rightarrow\) \(\triangle ASD\) – isosceles. Point \(K\) is the middle of \(AD\), then \(SK\) is the height in the triangle \(\triangle ASD\), and \(OK\) is the height in the triangle \(AOD\) \(\ Rightarrow\) plane \(SOK\) is perpendicular to planes \(ASD\) and \(ABC\) \(\Rightarrow\) \(\angle SKO\) – linear angle equal to the desired dihedral angle.
In \(\triangle SKO\) : \(OK = \frac(1)(2)\cdot AB = \frac(1)(2)\cdot 10 = 5 = SO\)\(\Rightarrow\) \(\triangle SOK\) – isosceles right triangle \(\Rightarrow\) \(\angle SKO = 45^\circ\) .
Answer: 45
Task 7 #1855
Task level: More difficult than the Unified State Exam
In a square \(ABCD\) : \(O\) – the point of intersection of the diagonals; \(S\) – does not lie in the plane of the square, \(SO \perp ABC\) . Find the angle between the planes \(ASD\) and \(BSC\) if \(SO = 5\) and \(AB = 10\) .
Right triangles \(\triangle SAO\) , \(\triangle SDO\) , \(\triangle SOB\) and \(\triangle SOC\) are equal in two sides and the angle between them (\(SO \perp ABC\) \(\Rightarrow\) \(\angle SOA = \angle SOD = \angle SOB = \angle SOC = 90^\circ\); \(AO = OD = OB = OC\), because \(O\) – point of intersection of the diagonals of the square, \(SO\) – common side) \(\Rightarrow\) \(AS = DS = BS = CS\) \(\Rightarrow\) \(\triangle ASD\) and \(\triangle BSC\) are isosceles. Point \(K\) is the middle of \(AD\), then \(SK\) is the height in the triangle \(\triangle ASD\), and \(OK\) is the height in the triangle \(AOD\) \(\ Rightarrow\) plane \(SOK\) is perpendicular to plane \(ASD\) . Point \(L\) is the middle of \(BC\), then \(SL\) is the height in the triangle \(\triangle BSC\), and \(OL\) is the height in the triangle \(BOC\) \(\ Rightarrow\) plane \(SOL\) (aka plane \(SOK\)) is perpendicular to the plane \(BSC\) . Thus, we obtain that \(\angle KSL\) is a linear angle equal to the desired dihedral angle.
\(KL = KO + OL = 2\cdot OL = AB = 10\)\(\Rightarrow\) \(OL = 5\) ; \(SK = SL\) – equal heights isosceles triangles, which can be found using the Pythagorean theorem: \(SL^2 = SO^2 + OL^2 = 5^2 + 5^2 = 50\). It can be noticed that \(SK^2 + SL^2 = 50 + 50 = 100 = KL^2\)\(\Rightarrow\) for a triangle \(\triangle KSL\) the inverse Pythagorean theorem holds \(\Rightarrow\) \(\triangle KSL\) – right triangle \(\Rightarrow\) \(\angle KSL = 90^\ circ\) .
Answer: 90
Preparing students to take the Unified State Exam in mathematics, as a rule, begins with repeating basic formulas, including those that allow you to determine the angle between planes. Despite the fact that this section of geometry is covered in sufficient detail within school curriculum, many graduates need to repeat basic material. Understanding how to find the angle between planes, high school students will be able to quickly calculate the correct answer when solving a problem and count on receiving decent scores on the results of passing the unified state exam.
Main nuances
To ensure that the question of how to find a dihedral angle does not cause difficulties, we recommend following a solution algorithm that will help you cope with Unified State Examination tasks.
First you need to determine the straight line along which the planes intersect.
Then you need to select a point on this line and draw two perpendiculars to it.
Next step- finding trigonometric function dihedral angle formed by perpendiculars. The most convenient way to do this is with the help of the resulting triangle, of which the angle is a part.
The answer will be the value of the angle or its trigonometric function.
Preparing for the exam test with Shkolkovo is the key to your success
During classes on the eve of passing the Unified State Exam, many schoolchildren are faced with the problem of finding definitions and formulas that allow them to calculate the angle between 2 planes. School textbook It’s not always on hand exactly when you need it. And to find necessary formulas and examples of their correct use, including for finding the angle between planes on the Internet online, which sometimes requires spending a lot of time.
The mathematical portal "Shkolkovo" offers new approach to prepare for the state exam. Classes on our website will help students identify the most difficult sections for themselves and fill gaps in knowledge.
We have prepared and clearly presented all the necessary material. Basic Definitions and formulas are presented in the “Theoretical Information” section.
In order to better understand the material, we also suggest practicing the appropriate exercises. Large selection of tasks varying degrees complexity, for example, is presented in the “Catalog” section. All tasks contain a detailed algorithm for finding the correct answer. The list of exercises on the website is constantly supplemented and updated.
While practicing solving problems that require finding the angle between two planes, students have the opportunity to save any task online as “Favorites.” Thanks to this they will be able to return to him required amount time and discuss the progress of its decision with school teacher or a tutor.
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 arbitrary point K draw perpendiculars 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. Turning radius R of point K equal to the value 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 projections of horizontals h 1, h 2 and fronts f 1, f 2, belonging to planesα 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.