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 will consider the issue of finding angles between planes and between them and a straight line.

## Line in space

It is known that absolutely any straight line in the plane can be defined by the following equality:

y=ax + b

Here a and b are some numbers. If we represent a 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 using the concept of "direction vector".

The directing vector of a straight line shows its orientation in space. This parameter belongs to the line. Since there is an infinite set of vectors parallel in space, then in order to uniquely determine the considered geometric object, it is also necessary to know the coordinates of the point belonging to it.

Assume that there ispoint P(x_{0}; y_{0}; z_{0) and direction vector v¯(a; b; c), then the equation of a straight line can be given as follows:}

(x; y; z)=P + αv¯ or

(x; y; z)=(x

_{0}; y_{0}; z_{0) + α(a; b; c)}

This expression is called the parametric vector equation of a straight line. The coefficient α is a parameter that can take absolutely any real values. The coordinates of a line can be represented explicitly by expanding this equality:

x=x

_{0+ αa;}y=y

_{0+ αb;}z=z

_{0+ αc}

## Equation of the plane

There are several forms of writing an equation for a plane in space. Here we will consider one of them, which is most often used when calculating the angles between two planes or between one of them and a straight line.

If some vector n¯(A; B; C) is known, which is perpendicular to the desired plane, and the point P(x_{0}; y_{0}; z_{0), which belongs to it, then the general equation for the latter is:}

Ax + By + Cz + D=0 where D=-1(Ax

_{0}+ By_{0}+ Cz_{0)}

We have omitted the derivation of this expression, which is quite simple. Here we only note that, knowing the coefficients of the variables in the equation of the plane, one 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 inclined and the plane and betweenarbitrary analogues.

## The location of the planes and the formula for the angle between them

Let's say there are two planes. What are the options for their relative position in space. Since the plane has two infinite dimensions and one zero, only two options for their mutual orientation are possible:

- they will be parallel to each other;
- they may overlap.

The angle between planes is the index between their direction vectors, i.e. 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 nonzero, but always sharp. A special case of intersection will be the angle 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 scalar product of these vectors. That is, the formula takes place:}

α=arccos((n

_{1}¯n_{2}¯)/(|n_{1 ¯| |n}2_{¯|))}

Assume that the coordinates of these vectors are: n_{1}¯(a_{1}; b_{1}; c_{1}), n_{2}¯(a_{2}; b_{2}; c_{2). Then, using the formulas for calculating the scalar product and modules of vectors through their coordinates, the expression above can be rewritten as:}

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 modulus in the numerator appeared because to exclude the values of obtuse angles.

## Examples of solving problems to determine 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, the equations of which are:

3x + 4y - z + 3=0;

-x - 2y + 5z +1=0

What is the angle between the planes?

To answer the question of the problem, let's remember that the coefficients of the variables in the general equation of the plane are the coordinates of the guide vector. For the indicated planes we have the following coordinates of their normals:

_{1¯(3; 4; -1); }_{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 formula given in the previous paragraph. We get:

α=arccos(|-16 | / (√26√30) ≈ 55, 05

^{o}

The resulting value corresponds to an acute angle of intersection of the planes specified in the conditiontasks.

Now consider another example. Given two planes:

x + y -3=0;

3x + 3y + 8=0

Do they intersect? Let's write out the values of the coordinates of their direction vectors, calculate their scalar product and modules:

_{1¯(1; 1; 0); }_{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 match each other is easy to check. Let's take for this an arbitrary point belonging to the first of them, for example, P(0; 3; 2). Substitute its coordinates into the second equation, we get:

30 +33 + 8=17 ≠ 0

That is, the point P belongs only to the first plane.

So two planes are parallel when their normals are.

## Plane and straight line

In the case of considering the relative position between a plane and a straight line, there are several more options than with two planes. This fact is connected with the fact that the straight line is a one-dimensional object. Line and 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 canintersect at some angle.

Let's consider the last case first, since it requires the introduction of the concept of the intersection angle.

## Line and plane, the angle between them

If a straight line intersects a plane, then it is called inclined with respect to it. The point of intersection is called the base of the slope. To determine the angle between these geometric objects, it is necessary to lower a straight perpendicular to the plane from any point. Then the point of intersection of the perpendicular with the plane and the place of 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. The acute angle between the line and its projection is the required one.

Somewhat confusing definition of the angle between a plane and an oblique will clarify the figure below.

Here the angle ABO is the angle between the line AB and the 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);}Ax + Bx + Cx + D=0

It is easy to calculate the desired angle for these objects if you find the scalar product between the direction vectors of the line and the plane. The resulting acute angle should be subtracted from 90^{o, then it is obtained between a straight line and a plane.}

The figure above shows the described algorithm for findingconsidered angle. 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, a formula was presented 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(|aA + bB + cC| / (√(a

^{2}+ b^{2}+ c^{2})√(A^{2}+ B^{2}+ C^{2)))}

The modulus in the formula allows only acute 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)).}

## Problem: A plane intersects a straight line

Now let's 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 picture.

You can see 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 a straight line and a plane, we get:

α=arcsin(|1| / (√1√2))=arcsin(1 / √2)=45

^{o}

## Problem: straight line parallel to the plane

Now let's decidesimilar to the previous problem, the question of which is posed differently. The equations of the plane and the straight line are known:

x + y - z - 3=0;

(x; y; z)=(1; 0; 0) + λ(0; 2; 2)

It is necessary to find out if these geometric objects are parallel to each other.

We have two vectors: the direction of the straight line is (0; 2; 2) and the direction of the plane is (1; 1; -1). Find their dot product:

01 + 12 - 12=0

The resulting zero indicates that the angle between these vectors is 90^{o, which proves that the line and the plane are parallel.}

Now let's 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, let's take λ=0, then the point P(1; 0; 0) belongs to the line. Substitute into the equation of the plane P:

1 - 3=-2 ≠ 0

The point P does not belong to the plane, which means that the entire line does not lie in it either.

## 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 of real three-dimensional 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 straight prism it is possible not to use these formulas to determinespecified values, then for any type of pyramid their use is inevitable.

Below, consider an example of using the above theory to determine the angles 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. Need to find two corners:

- between side surface and base;
- between side rib and base.

To solve the problem, you first need to enter the coordinate system and determine the parameters of the corresponding vertices. The figure shows that the origin of coordinates 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 the point B(0; 0; h), and the y-axis at the point with coordinates (0; a/2; 0). It does not cross the x-axis. This means that the equation of the ABC plane can be written as:

y / (a / 2) + z / h=1 or

2hy + az - ah=0

Vector AB¯ is a side edge. Its start and end coordinates are: 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 have found all the necessary equations and vectors. Now it remains to use the considered formulas.

First we calculate in the pyramid the angle between the planes of the baseand side. The corresponding normal vectors are: n_{1}¯(0; 0; 1) and n_{2¯(0; 2h; a). Then the angle will be:}

α=arccos(a / √(4h

^{2}+ a^{2))}

The angle between plane and edge AB will be:

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 get the required angles.