A plane is a geometric object whose properties are used when constructing projections of points and lines, as well as when calculating distances and dihedral angles between elements of volumetric figures. Let's consider in this article, with the help of which equations it is possible to study the location of planes in space.
Plane definition
Each intuitively knows what kind of object will be discussed. From a geometric point of view, a plane is a collection of points, any vectors between which must be perpendicular to a single vector. For example, if there are m different points in space, then m * (m-1) / 2 different vectors can be composed of them, connecting the points in pairs. If all vectors are perpendicular to some one direction, then this is a sufficient condition for all points m to belong to the same plane.
General equation
In spatial geometry, a plane is described using equations that generally contain three unknown coordinates corresponding to the x, y, and z axes. To get the general equation in the coordinates of the plane in space, suppose that there is a vector n¯ (A; B; C) and a point M (x 0 ; y 0 ; z 0 ). Using these two objects, you can uniquely determine the plane.
Indeed, suppose that there is some second point P (x; y; z) whose coordinates are unknown. According to the definition given above, the vector MP¯ must be perpendicular to n¯, that is, the product scalar for them is equal to zero. Then we have the right to write the following expression:
(n¯ * MP¯) = 0 or
A * (xx 0 ) + B * (yy 0 ) + C * (zz 0 ) = 0
Opening the brackets and introducing a new coefficient D, we obtain the expression:
A * x + B * y + C * z + D = 0, where D = -1 * (A * x 0 + B * y 0 + C * z 0 )
This expression is called a general equation for the plane. It is important to remember that the coefficients in front of x, y and z form the coordinates of the vector n¯ (A; B; C) perpendicular to the plane. It coincides with the normal and is a guide for the plane. To determine the general equation, it does not matter where this vector is directed. That is, the planes constructed on the vectors n¯ and -n¯ will be the same.
The figure above shows the plane, the normal vector to it and the perpendicular straight line to the plane.
The cut off by the plane segments on the axes and the corresponding equation
The general equation allows using simple mathematical operations to determine at what points the plane will intersect the coordinate axes. It is important to know this information in order to have an idea of the position in space of the plane, as well as when displaying it in the drawings.
To determine the named intersection points, the equation in segments is used. It is so called because it explicitly contains the values of the lengths of the segments cut off by the plane on the coordinate axes when taking the reference from the point (0; 0; 0). We get this equation.
We write the general expression for the plane in the following form:
A * x + B * y + C * z = -D
The left and right sides can be divided by -D without violating the equality. We have:
A / (- D) * x + B / (- D) * y + C / (- D) * z = 1 or
x / (- D / A) + y / (- D / B) + z / (- D / C) = 1
Denote the denominators of each term by a new symbol, we get:
p is -D / A; q = -D / B; r = -D / C then
x / p + y / q + z / r = 1
This is the equation mentioned above in the segments. It follows from it that the value of the denominator of each term indicates the coordinate of the intersection with the corresponding axis of the plane. For example, it intersects the y axis at the point (0; q; 0). This is easy to understand if we substitute the zero coordinates x and z in the equation.
Note that if in the equation in the segments there will not be any variable, then this means that the plane does not intersect the corresponding axis. For example, the expression is given:
x / p + y / q = 1
This means that the plane will cut off the segments p and q on the x and y axes, respectively, but on the z axis it will be parallel.
The conclusion about the behavior of a plane in the absence of a variable in its equation is also valid for an expression of a general type, as the figure below demonstrates.
Vector parametric equation
There is a third type of equation that allows you to describe in space a plane. It is called parametric vector, because it is defined by two vectors lying in the plane, and two parameters that can take arbitrary independent values. We show how to get this equation.
Suppose that there exists a pair of known vectors u ¯ (a 1 ; b 1 ; c 1 ) and v¯ (a 2 ; b 2 ; c 2 ). If they are not parallel, then using them you can specify a specific plane, if you fix the beginning of one of these vectors at a known point M (x 0 ; y 0 ; z 0 ). If an arbitrary vector MP¯ can be represented as a combination of the linear vectors u¯ and v¯, then this means that the point P (x; y; z) belongs to the same plane as u¯, v¯. Thus, we can write the equality:
MP¯ = α * u¯ + β * v¯
Or writing this equality through the coordinates, we get:
(x; y; z) = (x 0 ; y 0 ; z 0 ) + α * (a 1 ; b 1 ; c 1 ) + β * (a 2 ; b 2 ; c 2 )
The presented equality is a parametric vector equation for a plane. In the space of a vector on the plane, u¯ and v¯ are called generators.
Further, when solving the problem, it will be shown how this equation can be brought to a general form for the plane.
The angle between the planes in space
Intuitively, planes in three-dimensional space can either intersect or not. In the first case, it is of interest to find the angle between them. The calculation of this angle is more difficult than the angle between the lines, since we are talking about a dihedral geometric object. However, the already mentioned vector guide for the plane comes to the rescue.
It is geometrically established that the dihedral angle between two intersecting planes is exactly equal to the angle between their guide vectors. Denote these vectors as n 1 ¯ (a 1 ; b 1 ; c 1 ) and n 2 ¯ (a 2 ; b 2 ; c 2 ). The cosine of the angle between them is determined from the scalar product. That is, the angle in space between the planes can be calculated by the formula:
φ = arccos (| (n 1 ¯ * n 2 ¯) | / (| n 1 ¯ | * | n 2 ¯ |))
Here the module in the denominator is used to discard the value of the obtuse angle (between intersecting planes it is always less than or equal to 90 o ).
In coordinate form, this expression can be rewritten as follows:
φ = 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 )))
Planes perpendicular and parallel
If the planes intersect, and the dihedral angle formed by them is 90 o , then they will be perpendicular. An example of such planes is a rectangular prism or cube. These figures are formed by six planes. At each vertex of the named figures there are three planes perpendicular to each other.
To find out whether the planes under consideration are perpendicular, it is enough to calculate the scalar product of their normal vectors. A sufficient condition for perpendicularity in the space of planes is the zero value of this product.
Parallel are disjoint planes. It is also sometimes said that parallel planes intersect at infinity. The parallelism condition in the space of planes coincides with that for the direction vectors n 1 ¯ and n 2 ¯. You can check it in two ways:
- Calculate the cosine of the dihedral angle (cos (φ)) using the scalar product. If the planes are parallel, then the result is 1.
- Try to represent one vector through another by multiplying by some number, that is, n 1 ¯ = k * n 2 ¯. If this can be done, then the corresponding planes are parallel.
The figure shows two parallel planes.
Now we give examples of solving two interesting problems using the obtained mathematical knowledge.
How to get a general view from a vector equation?
This is a parametric vector expression for a plane. To make it easier to understand the course of operations and the mathematical techniques used, consider a specific example:
(x; y; z) = (1; 2; 0) + α * (2; -1; 1) + β * (0; 1; 3)
We will reveal these expressions and express unknown parameters:
x = 1 + 2 * α;
y = 2 - α + β;
z = α + 3 * β
Then:
α = (x - 1) / 2;
β = y - 2 + (x - 1) / 2;
z = (x - 1) / 2 + 3 * (y - 2 + (x - 1) / 2)
Opening the brackets in the last expression, we get:
z = 2 * x-2 + 3 * y - 6 or
2 * x + 3 * y - z - 8 = 0
We obtained a general form of the equation for the plane defined in the condition of the problem in vector form
How to build a plane through three points?
It is possible to draw a single plane through three points if these points do not belong to a single straight line. The algorithm for solving this problem is as follows:
- find the coordinates of two vectors by connecting the pairwise known points;
- calculate their vector product and get a normal to the plane vector;
- write a general equation using the found vector and any of the three points.
We give a concrete example. Points are given:
R (1; 2; 0), P (0; -3; 4), Q (1; -2; 2)
The coordinates of the two vectors are equal:
RP¯ (-1; -5; 4), PQ¯ (1; 1; -2)
Their vector product will be equal to:
n¯ = [RP¯ * PQ¯] = (6; 2; 4)
Taking the coordinates of the point R, we obtain the desired equation:
6 * x + 2 * y + 4 * z -10 = 0 or
3 * x + y + 2 * z -5 = 0
It is recommended to check the correctness of the result by substituting the coordinates of the remaining two points in this expression:
for P: 3 * 0 + (-3) + 2 * 4 -5 = 0;
for Q: 3 * 1 + (-2) + 2 * 2 -5 = 0
Note that it was possible not to find the vector product, but to immediately write down the equation for the plane in a parametric vector form.