Practical lesson. Equation of a plane
So, let the point M(1; 2; -1), which belongs to the plane, is given, and we know two guiding vectors with the coordinates (2; 0; 3) and (-2; 3; 0). The task is to create an equation for this plane. In this case, we need to create a general equation. To do this, we first write the canonical form through the determinant, and then proceed to the general equation of the plane.
To solve this problem, all you need is to remember what was discussed at the lecture. We deduced the formula. If a point and two guiding vectors are known, the plane is given by an equation of the following form (see the video). A determinant is formed, two rows of which coincide with the coordinates of the guiding vectors of the plane, and one row has the following form: (x-1, y-2, z+1) (we subtract the corresponding coordinate from each variable). Note how easy it is to remember why this particular row should be written. In this case, the point lies on the plane, which means that it must satisfy this equation. If we substitute its coordinates, we get a zero row, which means that the specified determinant will be equal to zero, regardless of other rows, and we will get a valid numerical equality.
So, the first row looks like this (see the video), the remaining rows are (2, 0, 3) and (-2, 3, 0) (they coincide with the coordinates of the guiding vectors). The required equation is automatically obtained, which is called the canonical equation of the plane: (see the video).
To turn to the general equation, we expand this determinant, for example, along the first row. We expand it along the first row: x-1 (plus sign before the element) is multiplied by the determinant obtained by crossing out the first row of the first column, we calculate it 0-9. This means that we multiply it by -9. Next, the second element will give us the second term, but its number is 3, the sum of the indices is odd, and the minus sign appears (do not forget about this). y-2 is multiplied by the determinant (we cross out the row, column and get the determinant) |(2 3) , (-2 0)|. We calculate. It is 6. Finally, the third term with a plus sign, z+1 is multiplied by the determinant |(2, 0), (-2, 3)| (because we cross out the third column and the third row), it is equal to 6. So, once again we remembered how, in our case, the third order determinants are expanded. Next, it remains to open the parentheses, give the following: (x-1)(-9)-(y-2)*6+(z+1)*6. That contains the variables: -9x-6y+6z, and – a free term. Try to calculate it on your own, after that check it with what I’ll get. Find a free term, and then you may have to make a reduction. We check it: a free term is +27. In fact, the equation is made up, but for ease of writing, it makes sense to divide by a common factor, in our case, by -3. We get the equation which is equivalent to this one, i.e. the plane does not change, the equation takes a simpler form. Thus, the general equation looks like this: 3x+2y-2z-9=0. Note that you can easily check whether you are right by substituting the point M in the equation. The substitution must result in a valid numeric equality. Check it yourselves. The problem is solved.
Let’s take another similar task. Let’s define the plane analytically, we know one of its points M(5; -3; 0), provided that this plane is perpendicular to the specified straight line (see the video). It is easy to understand that the plane is defined by such conditions unambiguously. Namely, there is a point, there is a straight line, through the point you can draw exactly one plane perpendicular to this line. At the same time, if we consider a guiding vector for this straight line, since the straight line is perpendicular to the plane, the guiding vector for this straight line is a normal vector for our plane.
So, it turns out that the point lies on the plane, and the vector that determines the direction of the line is the normal vector for our plane. Let’s find its coordinates. We have a straight line set by the canonical equation, so the numbers in the denominator determine the vector a(1; 2; -2). This is the vector we need, which is normal for the plane. That’s it! We can write the equation of the plane. The point that the line is set to is absolutely not important to us. It doesn’t have to be a point where the plane intersects a straight line, it’s some other point. And it has nothing to do with our task.
So, we create the equation of the plane that passes through the point M and has a normal vector a. In order to get such an equation, we write the coordinates of the vector as coefficients before the expressions. What are these expressions? These are the same expressions that we made up in the previous problem: 1(x-5)+2(y+3)-2(z-0), (i.e. we subtract the corresponding coordinates of the point M from the coordinates of the variables). Next, we equate the expression to zero. Again, it is not difficult to understand why this is the case. If we substitute the coordinates of the point M in this equality, all the terms are reset, and we get a valid numerical equality, i.e. the point M lies on our plane. So, and the coefficients 1, 2, -2 are the coordinates of the normal vector. After the transformations, we get a general equation of the following form: x+2y+2z+1=0. That is all. The problem is solved, and the general equation is obtained. Moreover, note that the coordinates, or rather the coefficients for variables – 1, 2, -2 – are the coordinates of the normal vector of this plane.
Well, now let’s turn to the previously stated problem to find the distance from a point to a plane. Please, note that we considered a similar problem in the topic “Equation of a straight line on a plane.” Within that topic, we searched for the distance from a point to a straight line, and this problem is solved in a similar way.
Let’s remember the main steps to solve it and write down a similar formula. There is a certain plane defined by the general equation ax+by+cz+d=0, there is a point M(x0, y0, z0,). The task is to find the distance, i.e. the length of the perpendicular dropped from the point to this plane. What is the idea? Let’s denote the distance by the letter d, i.e. the length of the segment MH, so this distance is found from the following formula. We remember that the plane has a normal vector, its coordinates are n(a, b, c). Since the segment MH is perpendicular to the plane, the vectors MH and n are collinear, which means that their scalar product is either the product of the lengths of these vectors if they are co-directed, or the product with a minus sign if they have opposite directions because the vector n can be completely directed down for this figure. Therefore, the modulus of the scalar product is the product of the lengths |MH*n|. The length of MH is d, and the length of the vector n is written like this: |n|. Further, if the vector n has the coordinates (a, b, c), its length is found as the square root of the sum of the squares of its coordinates: √(a2+b2+c2).
We will only need to find the module of the scalar product. To do this, I suggest that you find the coordinates of the MH vector yourselves. To do this, write down the coordinates of the point H(x,y,z), the coordinates of the point M(x0, y0, z0). Then we multiply these two vectors scalarly, open the parentheses, and you will have an expression. In this expression, you will have the following sum ax+by+cz, it will be equal to -d. After that, the left part is converted to the following form |ax0+by0+cz0+d|. If a minus sign appears suddenly under the module in all terms, we can change everything to a plus. You are to get ax0+by0+cz0+d. I repeat that we performed a similar procedure in detail at one of the previous practical lessons. While I’m writing it down, you can already perform some transformations and make sure that you get exactly this expression. I repeat, you write the coordinates of the vector MH, multiply by the vector n scalarly, give the similar ones, and use the fact that the point H lies on the plane, which means that it satisfies this equation. From this equality, it is easy to express the required distance d. To do this, we divide the left part of the equation by the square root.
So, let’s write down the formula, you can remember, if you have already written it in your notebook, what a similar formula looks like for the distance from a point to a straight line on a plane and compare: d=|ax0+by0+cz0+d|/√(a2+b2+c2).The expression turns out to be exactly the same, the only thing is that one summand at a time is added due to the third coordinate. However, if we want to find the distance from a point to a straight line given in space, there is no such a good formula, but we discussed how to find the specified distance at the corresponding practical lesson. That is, in the end, you should know and be able to calculate the distance from a point to a straight line given on a plane using a similar formula, the distance from a point to a straight line in space, and, what we have now deduced, the distance from a point to a plane.