In this lesson we will look at how to use the determinant to create plane equation. If you don’t know what a determinant is, go to the first part of the lesson - “Matrices and determinants”. Otherwise, you risk not understanding anything in today’s material.
Equation of a plane using three points
Why do we need a plane equation at all? It's simple: knowing it, we can easily calculate angles, distances and other crap in problem C2. In general, you cannot do without this equation. Therefore, we formulate the problem:
Task. Three points are given in space that do not lie on the same line. Their coordinates:
M = (x 1, y 1, z 1);
N = (x 2, y 2, z 2);
K = (x 3, y 3, z 3);You need to create an equation for the plane passing through these three points. Moreover, the equation should look like:
Ax + By + Cz + D = 0
where the numbers A, B, C and D are the coefficients that, in fact, need to be found.
Well, how to get the equation of a plane if only the coordinates of the points are known? The easiest way is to substitute the coordinates into the equation Ax + By + Cz + D = 0. You get a system of three equations that can be easily solved.
Many students find this solution extremely tedious and unreliable. Last year's Unified State Examination in mathematics showed that the likelihood of making a computational error is really high.
Therefore, the most advanced teachers began to look for simpler and more elegant solutions. And they found it! True, the technique obtained rather relates to higher mathematics. Personally, I had to rummage through the entire Federal List of Textbooks to make sure that we have the right to use this technique without any justification or evidence.
Equation of a plane through a determinant
Enough of the lyrics, let's get down to business. To begin with, a theorem about how the determinant of a matrix and the equation of the plane are related.
Theorem. Let the coordinates of three points through which the plane must be drawn be given: M = (x 1, y 1, z 1); N = (x 2, y 2, z 2); K = (x 3, y 3, z 3). Then the equation of this plane can be written through the determinant:
As an example, let's try to find a pair of planes that actually occur in problems C2. Look how quickly everything is calculated:
A 1 = (0, 0, 1);
B = (1, 0, 0);
C 1 = (1, 1, 1);
We compose a determinant and equate it to zero:
We expand the determinant:
a = 1 1 (z − 1) + 0 0 x + (−1) 1 y = z − 1 − y;
b = (−1) 1 x + 0 1 (z − 1) + 1 0 y = −x;
d = a − b = z − 1 − y − (−x ) = z − 1 − y + x = x − y + z − 1;
d = 0 ⇒ x − y + z − 1 = 0;
As you can see, when calculating the number d, I “combed” the equation a little so that the variables x, y and z were in the correct sequence. That's all! The plane equation is ready!
Task. Write an equation for a plane passing through the points:
A = (0, 0, 0);
B 1 = (1, 0, 1);
D 1 = (0, 1, 1);
We immediately substitute the coordinates of the points into the determinant:
We expand the determinant again:
a = 1 1 z + 0 1 x + 1 0 y = z;
b = 1 1 x + 0 0 z + 1 1 y = x + y;
d = a − b = z − (x + y ) = z − x − y;
d = 0 ⇒ z − x − y = 0 ⇒ x + y − z = 0;
So, the equation of the plane is obtained again! Again, at the last step we had to change the signs in it to get a more “beautiful” formula. It is not at all necessary to do this in this solution, but it is still recommended - to simplify the further solution of the problem.
As you can see, composing the equation of a plane is now much easier. We substitute the points into the matrix, calculate the determinant - and that’s it, the equation is ready.
This could end the lesson. However, many students constantly forget what is inside the determinant. For example, which line contains x 2 or x 3, and which line contains just x. To really get this out of the way, let's look at where each number comes from.
Where does the formula with the determinant come from?
So, let’s figure out where such a harsh equation with a determinant comes from. This will help you remember it and apply it successfully.
All planes that appear in Problem C2 are defined by three points. These points are always marked on the drawing, or even indicated directly in the text of the problem. In any case, to create an equation we will need to write down their coordinates:
M = (x 1, y 1, z 1);
N = (x 2, y 2, z 2);
K = (x 3, y 3, z 3).
Let's consider another point on our plane with arbitrary coordinates:
T = (x, y, z)
Take any point from the first three (for example, point M) and draw vectors from it to each of the three remaining points. We get three vectors:
MN = (x 2 − x 1 , y 2 − y 1 , z 2 − z 1 );
MK = (x 3 − x 1 , y 3 − y 1 , z 3 − z 1 );
MT = (x − x 1 , y − y 1 , z − z 1 ).
Now let's compose a square matrix from these vectors and equate its determinant to zero. The coordinates of the vectors will become rows of the matrix - and we will get the very determinant that is indicated in the theorem:
This formula means that the volume of a parallelepiped built on the vectors MN, MK and MT is equal to zero. Therefore, all three vectors lie in the same plane. In particular, an arbitrary point T = (x, y, z) is exactly what we were looking for.
Replacing points and lines of a determinant
Determinants have several great properties that make it even easier solution to problem C2. For example, it doesn’t matter to us from which point we draw the vectors. Therefore, the following determinants give the same plane equation as the one above:
You can also swap the lines of the determinant. The equation will remain unchanged. For example, many people like to write a line with the coordinates of the point T = (x; y; z) at the very top. Please, if it is convenient for you:
Some people are confused by the fact that one of the lines contains variables x, y and z, which do not disappear when substituting points. But they shouldn't disappear! Substituting the numbers into the determinant, you should get this construction:
Then the determinant is expanded according to the diagram given at the beginning of the lesson, and the standard equation of the plane is obtained:
Ax + By + Cz + D = 0
Take a look at an example. It's the last one in today's lesson. I will deliberately swap the lines to make sure that the answer will give the same equation of the plane.
Task. Write an equation for a plane passing through the points:
B 1 = (1, 0, 1);
C = (1, 1, 0);
D 1 = (0, 1, 1).
So, we consider 4 points:
B 1 = (1, 0, 1);
C = (1, 1, 0);
D 1 = (0, 1, 1);
T = (x, y, z).
First, let's create a standard determinant and equate it to zero:
We expand the determinant:
a = 0 1 (z − 1) + 1 0 (x − 1) + (−1) (−1) y = 0 + 0 + y;
b = (−1) 1 (x − 1) + 1 (−1) (z − 1) + 0 0 y = 1 − x + 1 − z = 2 − x − z;
d = a − b = y − (2 − x − z ) = y − 2 + x + z = x + y + z − 2;
d = 0 ⇒ x + y + z − 2 = 0;
That's it, we got the answer: x + y + z − 2 = 0.
Now let's rearrange a couple of lines in the determinant and see what happens. For example, let’s write a line with the variables x, y, z not at the bottom, but at the top:
We again expand the resulting determinant:
a = (x − 1) 1 (−1) + (z − 1) (−1) 1 + y 0 0 = 1 − x + 1 − z = 2 − x − z;
b = (z − 1) 1 0 + y (−1) (−1) + (x − 1) 1 0 = y;
d = a − b = 2 − x − z − y;
d = 0 ⇒ 2 − x − y − z = 0 ⇒ x + y + z − 2 = 0;
We got exactly the same plane equation: x + y + z − 2 = 0. This means that it really does not depend on the order of the rows. All that remains is to write down the answer.
So, we are convinced that the equation of the plane does not depend on the sequence of lines. We can carry out similar calculations and prove that the equation of the plane does not depend on the point whose coordinates we subtract from other points.
In the problem considered above, we used the point B 1 = (1, 0, 1), but it was quite possible to take C = (1, 1, 0) or D 1 = (0, 1, 1). In general, any point with known coordinates lying on the desired plane.
To obtain the general equation of a plane, let us analyze the plane passing through a given point.
Let there be three coordinate axes already known to us in space - Ox, Oy And Oz. Hold the sheet of paper so that it remains flat. The plane will be the sheet itself and its continuation in all directions.
Let P arbitrary plane in space. Every vector perpendicular to it is called normal vector to this plane. Naturally, we are talking about a non-zero vector.
If any point on the plane is known P and some normal vector to it, then by these two conditions the plane in space is completely defined(through a given point you can draw a single plane perpendicular to the given vector). The general equation of the plane will be:
So, the conditions that define the equation of the plane are. To get yourself plane equation, having the above form, take on the plane P arbitrary point M with variable coordinates x, y, z. This point belongs to the plane only if vector perpendicular to the vector(Fig. 1). For this, according to the condition of perpendicularity of vectors, it is necessary and sufficient that the scalar product of these vectors be equal to zero, that is
The vector is specified by condition. We find the coordinates of the vector using the formula 
:
.
Now, using the scalar product of vectors formula 
, we express the scalar product in coordinate form:
Since the point M(x; y; z) is chosen arbitrarily on the plane, then the last equation is satisfied by the coordinates of any point lying on the plane P. For a point N, not lying on a given plane, i.e. equality (1) is violated.
Example 1. Write an equation for a plane passing through a point and perpendicular to the vector.
Solution. Let’s use formula (1) and look at it again:
In this formula the numbers A , B And C vector coordinates, and numbers x0 , y0 And z0 - coordinates of the point.
The calculations are very simple: we substitute these numbers into the formula and get
We multiply everything that needs to be multiplied and add just numbers (which do not have letters). Result:
.
The required equation of the plane in this example turned out to be expressed by a general equation of the first degree with respect to variable coordinates x, y, z arbitrary point of the plane.
So, an equation of the form
called general plane equation .
Example 2. Construct in a rectangular Cartesian coordinate system a plane given by the equation 
.
Solution. To construct a plane, it is necessary and sufficient to know any three of its points that do not lie on the same straight line, for example, the points of intersection of the plane with the coordinate axes.
How to find these points? To find the point of intersection with the axis Oz, you need to substitute zeros for X and Y in the equation given in the problem statement: x = y= 0 . Therefore we get z= 6. Thus, the given plane intersects the axis Oz at the point A(0; 0; 6) .
In the same way we find the point of intersection of the plane with the axis Oy. At x = z= 0 we get y= −3, that is, the point B(0; −3; 0) .
And finally, we find the point of intersection of our plane with the axis Ox. At y = z= 0 we get x= 2, that is, a point C(2; 0; 0) . Based on the three points obtained in our solution A(0; 0; 6) , B(0; −3; 0) and C(2; 0; 0) construct the given plane.
Let's now consider special cases of the general plane equation. These are cases when certain coefficients of equation (2) become zero.
1. When D= 0 equation 
defines a plane passing through the origin, since the coordinates of the point 0
(0; 0; 0) satisfy this equation.
2. When A= 0 equation 
defines a plane parallel to the axis Ox, since the normal vector of this plane is perpendicular to the axis Ox(its projection onto the axis Ox equal to zero). Similarly, when B= 0 plane 
parallel to the axis Oy, and when C= 0 plane 
parallel to the axis Oz.
3. When A=D= 0 equation defines a plane passing through the axis Ox, since it is parallel to the axis Ox (A=D= 0). Similarly, the plane passes through the axis Oy, and the plane through the axis Oz.
4. When A=B= 0 equation defines a plane parallel to the coordinate plane xOy, since it is parallel to the axes Ox (A= 0) and Oy (B= 0). Similarly, the plane is parallel to the plane yOz, and the plane is the plane xOz.
5. When A=B=D= 0 equation (or z = 0) defines the coordinate plane xOy, since it is parallel to the plane xOy (A=B= 0) and passes through the origin ( D= 0). Likewise, Eq. y = 0 in space defines the coordinate plane xOz, and the equation x = 0 - coordinate plane yOz.
Example 3. Create an equation of the plane P, passing through the axis Oy and period.
Solution. So the plane passes through the axis Oy. Therefore, in her equation y= 0 and this equation has the form . To determine the coefficients A And C let's take advantage of the fact that the point belongs to the plane P .
Therefore, among its coordinates there are those that can be substituted into the plane equation that we have already derived (). Let's look again at the coordinates of the point:
M0 (2; −4; 3) .
Among them x = 2 , z= 3 . We substitute them into the general equation and get the equation for our particular case:
2A + 3C = 0 .
Leave 2 A on the left side of the equation, move 3 C to the right side and we get
A = −1,5C .
Substituting the found value A into the equation, we get
or .
This is the equation required in the example condition.
Solve the plane equation problem yourself, and then look at the solution
Example 4. Define a plane (or planes, if more than one) with respect to coordinate axes or coordinate planes if the plane(s) is given by the equation.
Solutions to typical problems that occur during tests are in the textbook “Problems on a plane: parallelism, perpendicularity, intersection of three planes at one point.”
Equation of a plane passing through three points
As already mentioned, a necessary and sufficient condition for constructing a plane, in addition to one point and the normal vector, are also three points that do not lie on the same line.
Let three different points , and , not lying on the same line, be given. Since the indicated three points do not lie on the same line, the vectors are not collinear, and therefore any point in the plane lies in the same plane with the points, and if and only if the vectors , and 
coplanar, i.e. then and only when mixed product of these vectors equals zero.
Using the expression for the mixed product in coordinates, we obtain the equation of the plane
(3)
After revealing the determinant, this equation becomes an equation of the form (2), i.e. general equation of the plane.
Example 5. Write an equation for a plane passing through three given points that do not lie on the same straight line:
and determine a special case of the general equation of a line, if one occurs.
Solution. According to formula (3) we have:
Normal plane equation. Distance from point to plane
The normal equation of a plane is its equation, written in the form
In order for a single plane to be drawn through any three points in space, it is necessary that these points do not lie on the same straight line.
Consider the points M 1 (x 1, y 1, z 1), M 2 (x 2, y 2, z 2), M 3 (x 3, y 3, z 3) in the general Cartesian coordinate system.
In order for an arbitrary point M(x, y, z) to lie in the same plane with points M 1, M 2, M 3, it is necessary that the vectors be coplanar.
(
)
= 0
Thus, 
Equation of a plane passing through three points:
Equation of a plane given two points and a vector collinear to the plane.
Let the points M 1 (x 1,y 1,z 1),M 2 (x 2,y 2,z 2) and the vector be given 
.
Let's create an equation for a plane passing through the given points M 1 and M 2 and an arbitrary point M (x, y, z) parallel to the vector 
.
Vectors 
and vector 
must be coplanar, i.e.
(
)
= 0
Plane equation:
Equation of a plane using one point and two vectors,
collinear to the plane.
Let two vectors be given 
And 
, collinear planes. Then for an arbitrary point M(x, y, z) belonging to the plane, the vectors 
must be coplanar.
Plane equation:
Equation of a plane by point and normal vector .
Theorem.
If a point M is given in space 0
(X 0
, y 0
,
z 0
), then the equation of the plane passing through the point M 0
perpendicular to the normal vector
(A,
B,
C) has the form:
A(x – x 0 ) + B(y – y 0 ) + C(z – z 0 ) = 0.
Proof.
For an arbitrary point M(x, y, z) belonging to the plane, we compose a vector. Because vector
is the normal vector, then it is perpendicular to the plane, and, therefore, perpendicular to the vector 
. Then the scalar product
=
0
Thus, we obtain the equation of the plane
The theorem has been proven.
Equation of a plane in segments.
If in the general equation Ax + Bi + Cz + D = 0 we divide both sides by (-D)
,
replacing 
, we obtain the equation of the plane in segments:
The numbers a, b, c are the intersection points of the plane with the x, y, z axes, respectively.
Equation of a plane in vector form.
Where
- radius vector of the current point M(x, y, z),
A unit vector having the direction of a perpendicular dropped onto a plane from the origin.
, and are the angles formed by this vector with the x, y, z axes.
p is the length of this perpendicular.
In coordinates, this equation looks like:
xcos + ycos + zcos - p = 0.
Distance from a point to a plane.
The distance from an arbitrary point M 0 (x 0, y 0, z 0) to the plane Ax+By+Cz+D=0 is:
Example. Find the equation of the plane, knowing that point P(4; -3; 12) is the base of the perpendicular dropped from the origin to this plane.
So A = 4/13; B = -3/13; C = 12/13, we use the formula:
A(x – x 0 ) + B(y – y 0 ) + C(z – z 0 ) = 0.
Example. Find the equation of a plane passing through two points P(2; 0; -1) and
Q(1; -1; 3) perpendicular to the plane 3x + 2y – z + 5 = 0.
Normal vector to the plane 3x + 2y – z + 5 = 0 
parallel to the desired plane.
We get:
Example. Find the equation of the plane passing through points A(2, -1, 4) and
B(3, 2, -1) perpendicular to the plane X + at + 2z – 3 = 0.
The required equation of the plane has the form: A x+B y+C z+ D = 0, normal vector to this plane 
(A, B, C). Vector 
(1, 3, -5) belongs to the plane. The plane given to us, perpendicular to the desired one, has a normal vector 
(1, 1, 2). Because points A and B belong to both planes, and the planes are mutually perpendicular, then
So the normal vector 
(11, -7, -2). Because point A belongs to the desired plane, then its coordinates must satisfy the equation of this plane, i.e. 112 + 71 - 24 +D= 0;D= -21.
In total, we get the equation of the plane: 11 x - 7y – 2z – 21 = 0.
Example. Find the equation of the plane, knowing that point P(4, -3, 12) is the base of the perpendicular dropped from the origin to this plane.
Finding the coordinates of the normal vector 
= (4, -3, 12). The required equation of the plane has the form: 4 x
– 3y
+ 12z+ D = 0. To find the coefficient D, we substitute the coordinates of point P into the equation:
16 + 9 + 144 + D = 0
In total, we get the required equation: 4 x – 3y + 12z – 169 = 0
Example. Given are the coordinates of the vertices of the pyramid A 1 (1; 0; 3), A 2 (2; -1; 3), A 3 (2; 1; 1),
Find the length of edge A 1 A 2.
Find the angle between edges A 1 A 2 and A 1 A 4.
Find the angle between edge A 1 A 4 and face A 1 A 2 A 3.
First we find the normal vector to the face A 1 A 2 A 3 
as a cross product of vectors 
And 
.
=
(2-1;
1-0; 1-3) = (1; 1; -2);
Let's find the angle between the normal vector and the vector 
.
-4
– 4 = -8.
The desired angle between the vector and the plane will be equal to = 90 0 - .
Find the area of face A 1 A 2 A 3.
Find the volume of the pyramid.
Find the equation of the plane A 1 A 2 A 3.
Let's use the formula for the equation of a plane passing through three points.
2x + 2y + 2z – 8 = 0
x + y + z – 4 = 0;
When using the computer version “ Higher mathematics course” you can run a program that will solve the above example for any coordinates of the vertices of the pyramid.
To start the program, double-click on the icon:
In the program window that opens, enter the coordinates of the vertices of the pyramid and press Enter. In this way, all decision points can be obtained one by one.
Note: To run the program, the Maple program ( Waterloo Maple Inc.) of any version, starting with MapleV Release 4, must be installed on your computer.
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 or another type of numerical functions. For what problems is it convenient to use the plane equation in segments? In this article we will look at what it is and how to use it in practical tasks.
What is a line equation?
A plane can be defined in three-dimensional space in several ways. In this article, some of them will be presented while solving problems of various types. Here we will give a detailed description of the equation in segments of the plane. In general, it has the following form:
Where the symbols p, q, r denote some specific numbers. This equation can be easily translated into a general expression and other forms of numerical functions for the plane.
The convenience of writing an equation in segments is that it contains explicit coordinates of the intersection of the plane with perpendicular coordinate axes. On the x axis relative to the origin of coordinates, the plane cuts off a segment of length p, on the y axis - equal to q, on z - with 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).
Relationship between the general and 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, you need to follow this technique: move the free term to the right side of the equality. Then we divide the entire equation by 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 segments the equation of the plane, initially given in general form. It is noticeable that the plane cuts off segments with lengths 3, 2 and 6 for the x, y and z axes, respectively. The y-axis intersects the plane in the negative coordinate region.
When composing an equation in segments, it is important that all variables are preceded by a “+” sign. Only in this case, the number by which this variable is divided will show the coordinate cut off on the axis.
Normal vector and point on the plane
It is known that some plane has (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 must first use a 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 guide vector, which is specified in the problem statement, that is:
It remains to find the free term D. It can be determined using the following formula:
D = -1*(A*x 1 + B*y 1 + C*z 1).
Where coordinate values with index 1 correspond to the coordinates of a point belonging to the plane. We substitute their values from the problem conditions, we get:
D = -1*(3*1 + 0*1 + (-1)*1) = -2.
Now we can write the equation in full:
The technique of converting this expression into an equation in plane segments has already been demonstrated above. Let's apply it:
3*x - z = 2 =>
x/(2/3) + z/(-2) = 1.
The answer to the problem has been received. Note that this plane intersects only the x and z axes. For y it is parallel.
Two straight lines defining a plane
From a course in spatial geometry, every schoolchild knows that two arbitrary straight lines uniquely define a plane in three-dimensional space. Let's solve a similar problem.
There are two known line equations:
(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 down in segments the equation of the plane passing through these lines.
Since both lines must lie in the plane, this means that their vectors (directors) must be perpendicular to the vector (director) for the plane. At the same time, it is known that the vector product of arbitrary two directed segments gives the result in the form of the coordinates of the third, perpendicular to the two original ones. Taking this property into account, 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, in this case a new directed segment is formed, parallel to the original one, then the sign of the obtained coordinates can be replaced with the opposite one (multiplied by -1), we get:
We know the direction vector. It remains to take an arbitrary point on one of the lines and compose a general equation of the plane:
D = -1*(1*1 + 2*0 + 3*0) = -1;
x + 2*y + z -1 = 0.
Translating 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.
Just like two straight lines, three points define a plane uniquely in three-dimensional space. Let us write the corresponding equation in segments if the following coordinates of points lying in the plane are known:
Let's proceed as follows: calculate the coordinates of two arbitrary vectors connecting these points, then find the vector n¯ normal to the plane by calculating the product of the found directed segments. We get:
QP¯ = P - Q = (1; -1; 0);
QM¯ = M - Q = (2; 4; 0);
n¯ = = [(1; -1; 0)*(2; 4; 0)] = (0; 0; 6).
Let's take point P as an example and create an equation for the plane:
D = -1*(0*2 + 0*(-3) + 6*0) = 0;
6*z = 0 or z = 0.
We have a simple expression that corresponds to the xy plane in a given rectangular coordinate system. It cannot be written 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).
