Practical lesson 2. The relative position of straight lines.
In this lesson, we are going to consider a number of problems in which you need to make up equations of a straight line based on some data. We are also going to solve problems to find out the relative position of two straight lines.
We consider straight lines set in a three-dimensional space. Let the coordinates of three points A(1; 2; -2), B(2; 3; 4), and C(-4; 1; 2) be known in a three-dimensional coordinate system. Let's make an equation of a straight line that passes through point A and is parallel to the BC vector.
Let's make a schematic drawing (see the video): there are three points A, B, and C. The straight line passes through the point A and has the direction BC or CB, that is, we can take any vector collinear to the BC vector as a direction vector. We are going to find its coordinates by knowing the coordinates of its starting and ending points: the BC vector (-6; -2; -2). We can take a vector that is collinear to it dividing all coordinates by -2. We get the vector a, which is also a direction vector for the required straight line: a(3; 1; 1). We need to make an equation for the straight line that passes through the point A (1; 2; -2) and is parallel to the vector a (3; 1; 1). Since the coordinates of the direction vector are non-zero, we can write the equation in canonical form. The equation is an equation of three fractions. The denominator contains the coordinates of the direction vector, the numerator contains expressions of the form x minus the first coordinate of the point, y minus the second coordinate, and z minus the third coordinate. If we do this, then the point A with the specified coordinates will satisfy this equation. We get 0 in all parts of the equations. So, we substitute the coordinates: (x-1)/3=(y-2)/1=(z+2)/1. The required line is set by this equation, the canonical equation of the line. Thus, point a) is fulfilled, the equation is found.
Let's now make up the equation of a straight line that also passes through the point A and through the point that is the midpoint of the segment BC. As we know, a straight line with two points is defined unambiguously. We need to find the second point, namely the midpoint of the BC segment. We need to find the half-sum of the corresponding coordinates: M(-1; 2; 3). For a straight line to pass through these points M(-1; 2; 3) and A(1; 2; -2), it must have the direction AM or MA. Accordingly, we can easily find the direction vector. We find the coordinates for the vector MA and get an interesting feature: the second coordinate is zero, because the points have the same second coordinates, MA(2; 0; -5). Since the direction vector has a zero coordinate, if we write the equations as we did in the previous paragraph, we will have 0 in the denominator. To prevent this, we set a straight line using parametric equations. We have a point through which a straight line passes, for example, the point M and a direction vector. The coordinates of the points of a straight line have the following form: (see the video). We write down the left parts of the parametric equations. Now we write down the general form of each of the three coordinates. We assume that the line passes through the point M and is parallel to the vector MA. x is expressed as -1+2t, where t has an arbitrary value. Then, similarly, it is 2+0t, that is, just 2. The second coordinate of a straight line is equal to two. Finally, z is 3-5t. The parameter t has an arbitrary real value, that is, it belongs to the set R. Note that for t equal to zero, we get the coordinates of the point (-1, 2, 3) through which this line passes. Thus, the resulting system of equations sets the line. The straight line is said to be parametrically defined.
Let's solve the following problem. We are going to find out the relative position of two lines (x+2)/2=(y-4)/-6=(z-2)/-8, (x-1)/-1=(y+5)/3=(z-k)/-4. Both lines are set by canonical equations; the first line is defined unambiguously, and the second line has the parameter k. We need to find out how these two lines relate to each other for all possible values of this parameter.
We write out the direction vectors of these lines. Numbers in the denominators of the fractions set these direction vectors. The first straight line has a direction vector a(2, -6, -8) and the second straight line has a direction vector b(-1, 3, 4). The coordinates should be proportional. If the coordinates of the vector a are divided by -2, we get a line of the coordinates of the vector b. So the vectors a and b are collinear. Therefore, straight lines have the same direction. What does this mean? Lines are either parallel, so that they do not have common points, then their direction vectors are collinear, or the lines can coincide, that is, any point of one line is also a point of another line.
We need to find out which of these two options is the case. Can both options be true, for different k? Can the lines coincide?
Knowing the equation of straight lines, we can immediately name the points, which lie on these lines. Then, if we have different lines, the vector formed by these points is no longer colleniar to the direction vector of these lines. If the lines coincide, all three vectors are collinear. Thus, we can determine which of the two options is the case.
Which point belongs to the first straight line? We consider the numerators of fractions. The first straight line is set by the point М1 with the coordinates (-2; 4; 2). Note that the first coordinate is -2, because here is the expression x+2. For the second straight line, the point that lies on it has the coordinates (1; -5; k). Let's write down the vector М1М2 and, depending on its collinearity or non-collinearity to vectors a and b, we are going to draw a conclusion.
We find the coordinates according to a well-known rule. We get the coordinates of the vector М1М2 (3; -9; k-2). Can all three vectors be collinear? We consider the first two coordinates, they are known. If we divide by -3, we get the first two coordinates of the vector b, that is, the first two coordinates are proportional. In order for the third coordinates to have the same proportionality coefficient, k-2 must be equal to 4*(-3). In this case, all three vectors are collinear. That is, each coordinate of the vector М1М2 is equal to the coordinate of the vector b multiplied by -3. Thus, if k-2=-12, that is, k=-10, all three vectors are collinear, which means that the second option takes place: the straight lines coincide. If k is not equal to -10, then the vectors a and b are collinear, but the vector М1М2 is not collinear to them, so the first case takes place: lines do not have common points. For all k not equal to -10, straight lines are parallel; they do not have any common points, but they lie on the same plane, because they have the same direction.
Thus, regardless of the parameter k, straight lines must have the same direction, they coincide at -10, and for any other k they are parallel (they do not have common points).
Let's solve a similar problem. We have two straight lines, and they are set by different equations. The first line is set by the canonical equation (x-2=y+1=z), the second line is set by the parametric equation (x=-3, y=5, z=1+t). We need to find out how these straight lines relate and what the angle between them is.
The double equation that sets the first straight line is also the canonical equation of the straight line. However, there are no denominators in this equation because the direction vector has unit coordinates. We can easily write this equation in the following form: (x-2)/1=(y+1)/1=z/1. This means that the direction vector of the first straight line has the coordinates a (1; 1; 1). The point through which the straight line passes has the coordinates M1 (2; -1; 0).
Now let us consider the second straight line. It is set parametrically. The first two equations do not have the parameter t. We can always write the +0t combination. This means that the second straight line passes through the point M2 with the coordinates (-3; 5; 1). This point corresponds to the zero value of the parameter. Now we are going to write down the coordinates of the direction vector of this line, they are found by coefficients for the parameter: b (0; 0; 1). Note that this vector is collinear to the Oz axis, i.e. the second straight line is parallel to the third coordinate axis, the applicate axis.
Now we can make the vector M1M2, and understand whether the lines lie in the same plane. Unlike the previous problem, the vectors a and b are no longer collinear, so there is no parallelism here. There are two options: either the straight lines intersect, or they intercross. To find out, let's make the vector M1M2; it has the coordinates (-5; 6; 1). Now the coordinates of three vectors a, b, and M1M2 determine everything. If they are non-coplanar, the lines intercross; if they are coplanar, the lines intersect. In order to find out whether the vectors are coplanar or not, let's make a determinant out of them and understand whether it is equal to zero or not. We write the vector b in the third line, in the beginning we write the coordinates of the vector a, then the coordinates of the vector M1M2: (see the video). We do not need to find the exact value of this determinant, it is important for us to understand whether it is zero or not. One method is to decompose by the last line, since there are two zeros here, thus we get a second-order determinant. Given the sign, (row 3, column 3) the sum of the indices is even, so 1 is multiplied by the determinant|(1, 1), (-5, 6)|. Since the second-order matrix has non-proportional rows, it means that it is nondegenerate. We don't care what the determinant is, as long as it's not zero. This means that the vectors are non-coplanar, i.e. they are not parallel to the same plane. Thus, the straight lines cross. The lines set by these equations intercross, that is, they lie in different planes. The problem is solved.
Finally, we are going to consider the problem of finding the distance from a point to a straight line on the plane. You need to calculate the distance h from the point M (2; 3; -1) to the line l set by the parametric equations (x=1+t, y=2+t, z=13+4t). We have a formula that allows us to find the specified distance on the plane; it a simple algorithm in the form of a formula. In space, the problem is a bit more complex. The formula has a rather complex form.
There is a straight line, no matter how it is set. We have the straight line set parametrically. We find the point that lies on this line, substitute t equal to zero, and get M(1, 2, 13); the direction vector has the coordinates a(1, 1, 4). We show it in the drawing and draw it from the point K. We mark the point M (2, -3, 1); it is located somewhere outside this line.
We find the KM vector; then we understand how these vectors can help us to solve the problem. Let's denote the KM vector with the letter b. We have the vector a and the second vector b, which we draw from the same point K. We find the vector KM (1; 1; -14).
How can we find the distance from a point to the line? What is the distance? This is the length of the perpendicular drawn from the point M to the straight line. The vectors a and b set a parallelogram. This height (its length is indicated by h) is the height of this parallelogram and is included in the well-known area formula. The area of a parallelogram constructed on vectors a and b is equal to the product of the height and the length of the side to which it is drawn, that is, by the modulus of the vector a: S=h*|a|. We have the vector a, so we can find its length; h is unknown; we need to find it. What about S? Knowing S, we can easily express h: we need to find the area of a parallelogram constructed on two vectors. We can solve it through either a scalar product or a vector product. Let's find the area of the parallelogram using the vector product. The area is numerically equal to the modulus of the vector product a by b, and the vector product itself is calculated using the following formula: see the video. Then, we decompose the determinant for the first line and get decomposition of this vector on the basis. We cover the first line, mentally cross it out, and then gradually find 3 determinants of the second order. The first of them is 18, the second is 18, because the first two lines are the same, but there is a minus sign. The coefficient in front of k is 0, that is 0k. This means that the vector product has the coordinates (18; -18; 0). We put 18 out of the bracket and the vector product can be written as follows: 18 multiplied by the vector with the coordinates (1; -1; 0). It would be more correct to say that we identify this linear combination with the coordinates. The length of this vector, the vector product of a by b is the length of the vector with these coordinates multiplied by 18. The length of the vector is √2 multiplied by 18, so the area of our parallelogram is 18√2.
We can substitute it into the equation S=h*|a| and express the height h. 18√2=h√18. For vector a, we have found the length of the vector a using the well-known formula: the square root of the sum of squares of coordinates. We get √18 or, if you take out 9 from under the root sign, then 3√2. Then, it is easy to express h. The height of the parallelogram is 6. This means that the distance from the point M to the given straight line is 6. The problem is solved.
You need to pay attention to the algorithm. Using this algorithm, you can always solve a similar problem. It doesn't matter how a straight line is set. The main thing is to find its direction and point. Then you make a similar configuration and find the required distance.