Practical lesson 2. Relative position of planes
Let’s start with the following task. Let’s assume that we have two planes defined by general equations (see the video). The task is to check that these planes intersect in a straight line and find the equation of this line, and, in the end, to write this equation in a canonical form.
Let’s look at the equation of planes. There is one peculiarity here: the first equation does not contain the variable z, but we still consider these equations as an equation with three unknowns – the missing variable z is understood in the sense that the coefficient in front of it is zero.
In order to understand that the planes intersect, we find the normal vector of these planes. For the first plane, the normal vector n1 has the coordinates (1, -1, 0). Let me remind you that the coordinates of the normal vector are nothing more than the coefficients for the variables in the general equation. The coefficient before z is zero, it turns out that the third coordinate is 0. For the second plane, the normal vector will be as follows. There are no zero coefficients here, so all the coordinates of the vector are non-zero. We get (2, 1, 3).
Now, we will determine the relative position of the planes since we know the coordinates of the specified vectors. In order for the planes to intersect, these vectors must be non-collinear. If the vectors are collinear, the planes will either be parallel or coincide. Let’s make sure that the vectors are non-collinear.
Indeed, their coordinates are not proportional: 1 to 2, -1 to 1 – the proportionality coefficients are different. Thus, the vectors are non-collinear, so the planes intersect, namely, intersect in the straight line.
Let’s answer the following question: if the planes intersect, can there be an angle of 90 degrees between them? To check whether the planes are perpendicular, you can find the scalar product of the vectors. Let’s find the scalar product n1n2. Let’s remember the formula: we multiply the corresponding coordinates and add them: 1*2=2, -1*1=-1, it will be 0. We get 1, it is not equal to 0. This means that the vectors are not orthogonal. Therefore, the planes are not perpendicular. If necessary, you can find the angle between the planes, as we have done repeatedly, knowing the scalar product.
Now, let’s answer the second question. Let’s write the equation of the line in which these planes intersect. The set of points of this line is the solution to a system of equations, i.e. we write these equations to the system (see the video).
I repeat that the set of points of their intersection line is the set of solutions to this system. We already know that the planes intersect in the straight line. This means that this system must have infinitely many solutions, i.e. when bringing this system to a step-by-step form, we will have one free variable. Let’s subtract the first two from the second equation to get a step-by-step form: the first equation will be rewritten without changing and the second one in the converted form: 2x-2y=0x, y-(-2y)=3y, 3z will remain, 1-(-2)=3 (see the video). One variable can be expressed from the last equation. We express y in terms of z, we will have a free change of z, we express y in terms of z, it turns out: y=-z-1, then we substitute the value of y in the first equation, from which we express x (see the video). First, you can express x in terms of y (see the video), and then substitute the value of the variable y: -z-1+1. After substituting and giving such values, you get –z. So, z takes an arbitrary value, y is expressed like this in terms of z (see the video), x = -z.
Denoting the variable z in terms of t, we can easily obtain a system of parametric equations for the desired line.
Let me remind you that we are looking for a straight line in which these planes intersect, and I suggest that we first create a parametric equation: x is –t, y= -t-1 and z=t (see the video), we consider t a parameter. We get that our line is given parametrically, and t takes an arbitrary value.
We have just given the straight line parametrically, now we must find the canonical equation, as required in the task. To do this, we need a guiding vector and at least one point, which we can now easily determine. We can find the point by taking any value as the parameter t. For example, if we choose t=0, we get the following solution: x=0, y=-1, z=0. We have one of the points through which our straight line passes – (0, -1, 0). The guiding vector is also easy to find, its coordinates are the coefficients before the parameter t, namely (-1, -1, 1). The point through which the straight line passes is known, the vector is known. We create a canonical equation, I remind you that the canonical equation is the equality of three fractions. In the numerator, we write the expression of the form: we subtract the coordinates of the point from the variable, i.e. x-0, the second fraction y(-1), i.e. y+1, and, finally, z-0, and in the denominators, the coordinates of the guiding vector, namely, -1, -1, 1, are obtained (see the video). In a simpler form, the equation is written in this way (see the video). On the one hand, we have a system of parametric equations of a straight line in which our planes intersect, on the other hand, we have a canonical equation of the same straight line. The problem is solved.
Let’s take the next problem. Again, two planes are given (see the video), and here we need to find the parameters a and b so that the planes are parallel.
In the previous task, the planes intersected in the straight line, now we consider the case of parallelism. As I’ve said, in order to distinguish between the cases where the planes intersect, we must find the normal vectors of these planes and require that they be collinear.
Again, we write the normal vector of the first plane, its coordinates are (2, b, 3), these are the coefficients for variables, and the second vector, which is normal for the second plane, has the coordinates (a, -6, -6). Let’s write down the condition of their collinearity. Since we have non-zero coordinates here, it means that neither a nor b can turn to 0. We write equality relations: 2 belongs to a, b to the number -6, as 3 to the number -6 (see the video), thus, these relations must be equal to -½, so the proportionality coefficient is -½. We find a: we divide 2 by -½, it turns out that a is -4, i.e. 2*2 and, given the “-“, it is -4. We look at the second fraction: b/(-6) is -½, so b=3. This condition allowed us to find the parameters a and b: a is -4, b is 3. As I’ve said, this is a necessary condition, but it may happen that the planes will coincide. In this case, their normal vectors are also collinear, i.e. there you still need to check that the planes do not coincide.
Let’s write out the equations of the planes obtained, substituting the found parameters a and b. The first plane has the equation 2x + 3y + 3z – 5 = 0, the second plane is 4x - 6y - 6z + 2 = 0. Once again, we can check whether we are right where: 2/(-4)= -½, 3/(-6)= -½, i.e. the normal vectors are collinear, but if the planes coincide, this means that the rows made up of all the coefficients of these equations are proportional. “Of all” means, given a free term, i.e. (2, 3, 3, -5) and for the second equation we get the row (-4, -6, -6, 2). It is easy to see that these rows are no longer proportional because the ratio of free terms is -5/2 ≠ -½, i.e. the specified rows made up of all coefficients, including the free term, are not proportional, i.e. -5/2 ≠ -½, which means that the planes do not coincide, i.e. these equations are not equivalent and define different planes.
We conclude that for a = -4, b = -3, our planes are parallel, i.e. they do not have any common points.
Next, we have to answer the second question: we need to find the distance between these planes. To do this, we will use the formula that we derived at the last lesson, namely, the formula for the distance from a point to a plane. Since our planes are parallel, this means that no matter what point in one plane we take, the distance from it to the second plane will be the same. It is the distance between these planes geometrically. Geometrically means that we drop the perpendicular from the point M of one plane to the second plane. You need to find the length of this perpendicular. Let’s remember the formula that gives us the required distance. The formula looks like this (see the video), where a, b, c, d are the coefficients in the equation of the plane to which we find the distance, x0, y0 are the coordinates of the point M.
Both planes are known to us. Let’s take a point from the first plane and find the distance from this point to the second plane. Let the notations of the planes be α, β. We take a point from the plane α and find the distance to β. It is easy to see that if we take the following numbers: x and y are 1 each, and z is 0 instead of x, y, z. We will get a valid numerical equality, which means that the point M with the coordinates (1, 1, 0) belongs to the plane α: 2+3+0-5=0. As β, we choose the second plane, and the plane does not change if we divide both parts of the equation by -2 to get simpler coefficients. In this case, the same plane β can be given by the following equation: we divide by -2, we get 2x + 3y + 3y + 3z-1 = 0. We have the same plane β. Please, note: in this case, we have all four coefficients, let me clarify that both four coefficients are proportional, so these equations define the same plane.
We substitute the coordinates of the point in the formula instead of the variables, it turns out |2 + 3 + 0 – 1| and divide by the sum of the squares of the coefficients, i.e. by the sum of 4 + 9 + 9 (see the video), we calculate: 5 -1, we get 4, 9 + 9 = 18, 18 + 4 = 22, √22. The resulting number – 4/√22 – is the distance from the point M to the plane β, which means it is the distance between the parallel planes α and β. The problem is solved.
Another problem is the relative position of a straight line in a plane. Let’s find out the relative position of the straight line, which is given by the canonical equation, and the plane, which is given by the general equation (see the video). In order to determine this relative position, we need to know the point and the guiding vector for the straight line, and the normal vector for the plane. We can easily find this if we know how a straight line and a plane are defined.
Let’s start with the straight line. Let me remind you that if the straight line is defined by the canonical equation, the numbers in the denominator are the coordinates of the guiding vector (-2, 3, 2). Today, we have already spoken about it. To find the coordinates of the point lying in the line, it is necessary to look at the numerators, and you get the following numbers: (-2, 1, 3). Why? Because if we substitute the specified numbers instead of the variables x, y, z, we get a valid equality. There will be zeros in all parts of the equation because the numerators will turn to 0: if you substitute -2 instead of x, substitute 1 instead of y, and substitute 3 instead of z. So, a is the guiding vector of the straight line, m is the point lying on the straight line, and the straight line can be denoted by the letter l. This straight line l is defined by these objects (see the video).
Now, we consider the plane, denote it α. The plane will have a normal vector n with the coordinates (1, 2, -2) and, if necessary, you can find a point lying on this plane by selecting such values of variables to get zero equality.
Let’s get back to our task. We are required to find out the relative position. To do this, we will follow the following algorithm. Everything is determined by the vectors: if the vector a and vector n are not orthogonal, the straight line intersects the plane in exactly one point; if the vectors are orthogonal, i.e. the angle between them is 90 degrees, so the straight line and the plane either have no common points, i.e. parallel to the plane or the straight line lies on the plane completely. To find out whether the vector is orthogonal or not, we find their scalar product: -2 + 6 – 4, we calculate, we get 0. So, the vectors are orthogonal. Therefore, as I’ve said, either the straight line and the plane are parallel, or, alternatively, the straight line lies on this plane, i.e. either this or that variant (see the video).
Let’s find out which of the two options is the case. To do this, we need the point M. The fact is that if the first option is true, i.e. the line and the plane do not have common points, no point of the line lies on the plane, i.e. does not satisfy this equation, the point M, in particular.
If the straight line lies in the plane, all its points are solutions to the equation that our plane sets, the point M, in particular. We take the coordinate of the point M and substitute it into the equation of the plane: -2 + 2 + 6 -6 = 0. What do we see? It is 0, which means that the point M, being a point of the straight line, lies in the plane. So, of these two options, the first one is discarded.
If the point M lies in the plane, the entire line lies entirely in our plane. What we got: on the one hand, in the beginning, we rejected the case when the straight line and the plane intersect, and secondly, we rejected the option when the straight line and the plane are parallel. We get an unambiguous option: the straight line lies in the plane, which is the answer. So, our line lies in the plane. The problem is solved.