To determine the parallelism and perpendicularity of planes, as well as to calculate the distances between these geometric objects, it is convenient to use one form or another of numerical functions. For what tasks is it convenient to use the plane equation in segments? In this article, we will consider what it is and how to use it in practical tasks.
What is an equation in segments?
A plane can be defined in three-dimensional space in several ways. In this article, some of them will be given during the solution of problems of various types. Here we give a detailed description of the equation in segments of the plane. It generally has the following form:
x / p + y / q + z / r = 1.
Where the symbols p, q, r are some specific numbers. This equation can be easily translated into a general expression and into other forms of numerical functions for a plane.
The convenience of writing an equation in segments is that it contains the explicit coordinates of the intersection of the plane with the perpendicular coordinate axes. On the x axis, relative to the origin, the plane cuts off a segment of length p, on the y axis, equal to q, and on z, length r.
If any of the three variables is not contained in the equation, then this means that the plane does not pass through the corresponding axis (mathematicians say that it intersects at infinity).
Next, we present several tasks in which we show how to work with this equation.
The connection between the general and in the segments of equations
It is known that the plane is given by the following equality:
2 * x - 3 * y + z - 6 = 0.
It is necessary to write down this general equation of the plane in segments.
When a similar problem arises, one must follow this technique: we transfer the free term to the right side of the equality. Then we divide the entire equation into this term, trying to express it in the form given in the previous paragraph. We have:
2 * x - 3 * y + z = 6 =>
2 * x / 6 - 3 * y / 6 + z / 6 = 1 =>
x / 3 + y / (- 2) + z / 6 = 1.
We obtained in the segments the equation of the plane, which was initially defined in the general form. It is noticeable that the plane cuts off segments with lengths of 3, 2, and 6 for the x, y, and z axes, respectively. The y axis intersects the plane in the negative coordinate range.
When drawing up the equation in segments, it is important that the “+” sign is in front of all variables. Only in this case, the number by which this variable is divided will show the coordinate being cut off on the axis.
Normal vector and point on the plane
It is known that some plane has a direction vector (3; 0; -1). It is also known that it passes through the point (1; 1; 1). You should write an equation in segments for this plane.
To solve this problem, you should first use the general form for this two-dimensional geometric object. The general form is written as:
A * x + B * y + C * z + D = 0.
The first three coefficients are here the coordinates of the vector of the guide, which is specified in the condition of the problem, that is:
A = 3;
B is 0;
C = -1.
It remains to find the free term D. It can be determined by the following formula:
D = -1 * (A * x 1 + B * y 1 + C * z 1 ).
Where the coordinate values with index 1 correspond to the coordinates of a point belonging to the plane. We substitute their values from the conditions of the problem, we obtain:
D = -1 * (3 * 1 + 0 * 1 + (-1) * 1) = -2.
Now you can write down the complete equation:
3 * x - z - 2 = 0.
The methodology for converting this expression into an equation in segments of a plane has already been demonstrated above. Apply it:
3 * x - z = 2 =>
x / (2/3) + z / (- 2) = 1.
The answer to the problem is received. Note that this plane intersects only the x and z axes. For y, it is parallel.
Two lines defining a plane
From the course of spatial geometry, every student knows that two arbitrary straight lines uniquely define a plane in three-dimensional space. We solve a similar problem.
Two equations of lines are known:
(x; y; z) = (1; 0; 0) + α * (2; -1; 0);
(x; y; z) = (1; -1; 0) + β * (- 1; 0; 1).
It is necessary to write in the segments the equation of the plane through these lines passing.
Since both lines must lie in the plane, this means that their vectors (guides) must be perpendicular to the vector (guide) for the plane. At the same time, it is known that the vector product of arbitrary two directed segments gives a result in the form of coordinates of the third, perpendicular to the two initial ones. Given this property, we obtain the coordinates of the vector normal to the desired plane:
[(2; -1; 0) * (- 1; 0; 1)] = (-1; -2; -1).
Since it can be multiplied by an arbitrary number, a new directed segment is formed parallel to the original, then we can replace the sign of the obtained coordinates with the opposite one (multiply by -1), we get:
(1; 2; 1).
We know the direction vector. It remains to take an arbitrary point on one of the lines and draw up a general equation of the plane:
A is 1;
B = 2;
C is 1;
D = -1 * (1 * 1 + 2 * 0 + 3 * 0) = -1;
x + 2 * y + z -1 = 0.
We translate this equality into an expression in segments, we get:
x + 2 * y + z = 1 =>
x / 1 + y / (1/2) + z / 1 = 1.
Thus, the plane intersects all three axes in the positive region of the coordinate system.
Three points and a plane
Like two lines, three points define a plane uniquely in three-dimensional space. We write the corresponding equation in segments if the following coordinates of points lying in the plane are known:
Q (1; -2; 0);
P (2; -3; 0);
M (4; 1; 0).
We proceed as follows: we calculate the coordinates of two arbitrary vectors connecting these points, then we find the vector n¯ normal to the plane, calculating the product of the found directed segments. We get:
QP¯ = P - Q = (1; -1; 0);
QM¯ = M - Q = (2; 4; 0);
n¯ = [QP¯ * QM¯] = [(1; -1; 0) * (2; 4; 0)] = (0; 0; 6).
Take for example the point P, make up the equation of the plane:
A is 0;
B is 0;
C = 6;
D = -1 * (0 * 2 + 0 * (- 3) + 6 * 0) = 0;
6 * z = 0 or z = 0.
We got a simple expression that corresponds to the xy plane in a given rectangular coordinate system. It is impossible to write it in segments, since the x and y axes belong to the plane, and the length of the segment cut off on the z axis is zero (the point (0; 0; 0) belongs to the plane).