When solving geometric problems in space, there are often those where it is necessary to calculate the angles between different spatial objects. In this article, we consider the question of finding the angles between the planes and between them and the straight line.
Straight in space
It is known that absolutely any line on the plane can be defined by the following equality:
y = a * x + b
Here a and b are some numbers. If we represent the straight line in space with the same expression, then we get a plane parallel to the z axis. For the mathematical definition of the spatial line, a different solution method is used than in the two-dimensional case. It consists in the use of the concept of "directional vector."
The directing vector of the line shows its orientation in space. This parameter belongs to the line. Since there is an infinite set of vectors parallel in space, for the unambiguous determination of the geometric object in question it is also necessary to know the coordinates of the point belonging to it.
Suppose that there is a point P (x 0 ; y 0 ; z 0 ) and a direction vector v¯ (a; b; c), then the equation of the line can be defined as follows:
(x; y; z) = P + α * v¯ or
(x; y; z) = (x 0 ; y 0 ; z 0 ) + α * (a; b; c)
This expression is called the parametric vector equation of the line. The coefficient α is a parameter that can take absolutely any real value. The coordinates of the line can be represented explicitly, revealing this equality:
x = x 0 + α * a;
y = y 0 + α * b;
z = z 0 + α * c
Plane equation
Several forms of writing equations for a plane in space are known . Here we consider one of them, which is most often used in calculating the angles between two planes or between one of them and a straight line.
If a certain vector n¯ (A; B; C) is known, which is perpendicular to the desired plane, and a point P (x 0 ; y 0 ; z 0 ) is specified that belongs to it, then the general equation for the latter has the form:
A * x + B * y + C * z + D = 0, where D = -1 * (A * x 0 + B * y 0 + C * z 0 )
We omitted the conclusion of this expression, which is quite simple. Here we only note that, knowing the coefficients of the variables in the equation of the plane, we can easily find all the vectors that are perpendicular to it. The latter are called normals and are used in calculating the angles between the slant and the plane and between arbitrary analogues.
The location of the planes and the angle formula between them
Suppose there are two planes. What are the options for their mutual arrangement in space. Since the plane has two infinite sizes and one zero, only two options for their mutual orientation are possible:
- they will be parallel to each other;
- they may intersect.
The angle between the planes is the exponent between their direction vectors, that is, between their normals n 1 ¯ and n 2 ¯.
Obviously, if they are parallel to the plane, then the intersection angle is zero between them. If they intersect, then it is non-zero, but always sharp. A special case of intersection will be an angle of 90 o when the planes are mutually perpendicular to each other.
The angle α between n 1 ¯ and n 2 ¯ is easily determined from the product of the scalar of these vectors. That is, the formula holds:
α = arccos ((n 1 ¯ * n 2 ¯) / (| n 1 ¯ | * | n 2 ¯ |))
Suppose that the coordinates of these vectors are as follows: n 1 ¯ (a 1 ; b 1 ; c 1 ), n 2 ¯ (a 2 ; b 2 ; c 2 ). Then, using formulas for calculating the scalar product and the modules of vectors through their coordinates, the expression above can be rewritten in the form:
α = arccos (| a 1 * a 2 + b 1 * b 2 + c 1 * c 2 | / (√ (a 1 2 + b 1 2 + c 1 2 ) * √ (a 2 2 + b 2 2 + c 2 2 )))
The module in the numerator appeared because to exclude the values of obtuse angles.
Examples of solving problems on determining the angle of intersection of planes
Knowing how to find the angle between the planes, we will solve the following problem. Two planes are given whose equations 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, we recall that the coefficients that stand for variables in the equation of the general plane are the coordinates of the vector of the guide. For the indicated planes, we have the following coordinates of their normals:
n 1 ¯ (3; 4; -1);
n 2 ¯ (-1; -2; 5)
Now we find the product scalar 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 in the formula given in the previous paragraph. We get:
α = arccos (| -16 | / (√26 * √30) ≈ 55.05 o
The obtained value corresponds to the acute angle of intersection of the planes specified in the condition of the problem.
Now consider another example. Two planes are given:
x + y -3 = 0;
3 * x + 3 * y + 8 = 0
Do they intersect? We write down the coordinate values of their guide vectors, calculate the scalar product of them and the 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 intersection angle 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 verify. For this, we take an arbitrary point belonging to the first of them, for example, P (0; 3; 2). Substitute its coordinates in the second equation, we get:
3 * 0 +3 * 3 + 8 = 17 ≠ 0
That is, the point P belongs only to the first plane.
Thus, two planes are parallel when their normals are.
Plane and straight
When considering the relative position between a plane and a line, there are several more options than with two planes. This fact is connected with the fact that the line is a one-dimensional object. The line and the 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.
We first consider the latter case, since it requires the introduction of the concept of the angle of intersection.
Line and plane, the value of the angle between them
If a plane intersects a line, then it is called oblique with respect to it. The intersection point is called the base of the sloping. To determine the angle between these geometric objects, it is necessary to lower the straight perpendicular from any point to the plane. Then the point of intersection of the perpendicular with the plane and the intersection of the slope with it form a straight line. The latter is called the projection of the original line onto the plane under consideration. The acute angle between the line and its projection is the desired one.
A somewhat confusing definition of the angle between the plane and the inclined will clarify the figure below.
Here the angle ABO is the angle between the AB line and a plane.
To write a formula for it, consider an example. Let there be a 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
It is easy to calculate the desired angle for these objects if you find the scalar product between the directing vectors of the line and the plane. The resulting acute angle should be subtracted from 90 o , then it is obtained between the line and the plane.
The figure above shows the described algorithm for finding the angle under consideration. 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 .
The above formula was presented that provides an answer to the question of how to find the angle between the planes. Now we give the corresponding expression for the case of a 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 only sharp angles to be calculated. The arcsine function appeared instead of the arccosine due to the use of the corresponding reduction formula between trigonometric functions (cos (β) = sin (90 o-β ) = sin (α)).
Task: the plane crosses the line
Now we show how to work with the above formula. Let's solve the problem: it is necessary to calculate the angle between the y axis and the plane given by the equation:
y - z + 12 = 0
This plane is shown in the figure.
It can be seen that it intersects the y and z axes at the points (0; -12; 0) and (0; 0; 12), respectively, and is parallel to the x axis.
The direction vector of the line y has coordinates (0; 1; 0). A vector perpendicular to a given plane is characterized by coordinates (0; 1; -1). We apply the formula for the angle of intersection of the line and the plane, we get:
α = arcsin (| 1 | / (√1 * √2)) = arcsin (1 / √2) = 45 o
Task: straight line parallel to the plane
Now we solve a similar previous problem, the question of which is posed differently. The known equations of the plane and line:
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 to each other.
We have two vectors: the directing line is (0; 2; 2) and the directing plane is (1; 1; -1). 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 straight line and plane parallelism.
Now we check whether this line is only parallel or also lies in the plane. To do this, select an arbitrary point on the line and check whether it belongs to the plane. For example, we take λ = 0, then the point P (1; 0; 0) belongs to the line. We substitute the plane P into the equation:
1 - 3 = -2 ≠ 0
Point P of the plane does not belong, which means that 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 solving problems are not only of theoretical interest. They are often used to determine important physical quantities of real volumetric figures, such as prisms or pyramids. It is important to be able to determine the angle between the planes when calculating the volumes of figures and the areas of their surfaces. Moreover, if in the case of a direct prism it is possible not to use these formulas to determine the indicated values, then for any kind of pyramid their use is inevitable.
Below we consider an example of using the theory presented to determine the angles of a pyramid with a square base.
Pyramid and its corners
The figure below shows the 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 enter the coordinate system and determine the parameters of the corresponding vertices. The figure shows that the origin coincides with the point in the center of the square base. In this case, the base plane is described by the equation:
z = 0
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). The x axis it does not cross. 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:
AB¯ (-a / 2; -a / 2; h)
We found all the necessary equations and vectors. Now it remains to use the above formulas.
First, we calculate in the pyramid the angle 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 the edge AB will be equal to:
β = arcsin (h / √ (a 2/2 + h 2 ))
It remains to substitute the specific values of the side of the base a and the height h to obtain the necessary angles.